I just cited myself.
from fossilesque@mander.xyz to science_memes@mander.xyz on 27 Jun 09:45
https://mander.xyz/post/14652327

#science_memes

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magic_lobster_party@kbin.run on 27 Jun 10:50 next collapse

If 0.999… < 1, then that must mean there’s an infinite amount of real numbers between 0.999… and 1. Can you name a single one of these?

ns1@feddit.uk on 27 Jun 11:05 next collapse

Sure 0.999…95

Just kidding, the guy on the left is correct.

magic_lobster_party@kbin.run on 27 Jun 11:22 collapse

You got me

Drewfro66@lemmygrad.ml on 27 Jun 11:37 collapse

(0.999… + 1) / 2

WldFyre@lemm.ee on 27 Jun 22:45 collapse

That number happens to be exactly 1

rustyfish@lemmy.world on 27 Jun 10:58 next collapse

Remember when US politicians argued about declaring Pi to 3?

Would have been funny seeing the world go boink in about a week.

Wxnzxn@lemmy.ml on 27 Jun 11:05 next collapse

To everyone who might not have heard about that before: It was an attempt to introduce it as a bill in Indiana:

en.m.wikipedia.org/w/index.php?title=Indiana_pi_b…

ValiantDust@feddit.de on 27 Jun 11:23 collapse

the bill’s language and topic caused confusion; a member proposed that it be referred to the Finance Committee, but the Speaker accepted another member’s recommendation to refer the bill to the Committee on Swamplands, where the bill could “find a deserved grave”.

An assemblyman handed him the bill, offering to introduce him to the genius who wrote it. He declined, saying that he already met as many crazy people as he cared to.

I hope medicine in 1897 was up to the treatment of these burns.

Zagorath@aussie.zone on 27 Jun 11:46 collapse

Do you think Goodwin could treat the burns himself?

ValiantDust@feddit.de on 27 Jun 11:56 collapse

I’m sure he would have believed he could.

[deleted] on 27 Jun 11:16 next collapse

.

rbesfe@lemmy.ca on 27 Jun 16:28 collapse

You didn’t even read the first paragraph of that article LMAO

roguetrick@lemmy.world on 27 Jun 13:10 next collapse

I prefer my pi to be in duodecimal anyway. 3.184809493B should get you to where you need to go.

myslsl@lemmy.world on 28 Jun 08:16 collapse

Some software can be pretty resilient. I ended up watching this video here recently about running doom using different values for the constant pi that was pretty nifty.

BachenBenno@feddit.de on 27 Jun 11:09 next collapse

Mathematics is built on axioms that have nothing to do with numbers yet. That means that things like decimal numbers need definitions. And in the definition of decimals is literally included that if you have only nines at a certain point behind the dot, it is the same as increasing the decimal in front of the first nine by one.

Sop@lemmy.blahaj.zone on 27 Jun 11:34 next collapse

That’s not an axiom or definition, it’s a consequence of the axioms that define arithmetic and can therefore be proven.

JohnDClay@sh.itjust.works on 27 Jun 11:42 collapse

There are versions of math where that isn’t true, with infinitesimals that are not equal to zero. So I think it is an axium rather than a provable conclusion.

JackRiddle@sh.itjust.works on 27 Jun 11:45 next collapse

Those versions have different axioms from which different things can be proven, but we don’t define 9.9 repeating as 1

Feathercrown@lemmy.world on 27 Jun 13:23 collapse

That’s not what “axiom” means

aoidenpa@lemmy.world on 27 Jun 12:31 collapse

That’s not how it’s defined. 0.99… is the limit of a sequence and it is precisely 1. 0.99… is the summation of infinite number of numbers and we don’t know how to do that if it isn’t defined. (0.9 + 0.09 + 0.009…) It is defined by the limit of the partial sums, 0.9, 0.99, 0.999… The limit of this sequence is 1. Sorry if this came out rude. It is more of a general comment.

BachenBenno@feddit.de on 27 Jun 13:53 collapse

I study mathematics at university and I remember it being in the definition, but since it follows from the sum’s limit anyways it probably was just there for claritie’s sake. So I guess we’re both right…

Valthorn@feddit.nu on 27 Jun 11:40 next collapse

x=.9999…

10x=9.9999…

Subtract x from both sides

9x=9

x=1

There it is, folks.

barsoap@lemm.ee on 27 Jun 11:53 next collapse

Somehow I have the feeling that this is not going to convince people who think that 0.9999… /= 1, but only make them madder.

Personally I like to point to the difference, or rather non-difference, between 0.333… and ⅓, then ask them what multiplying each by 3 is.

ColeSloth@discuss.tchncs.de on 27 Jun 22:08 next collapse

I’d just say that not all fractions can be broken down into a proper decimal for a whole number, just like pie never actually ends. We just stop and say it’s close enough to not be important. Need to know about a circle on your whiteboard? 3.14 is accurate enough. Need the entire observable universe measured to within a single atoms worth of accuracy? It only takes 39 digits after the 3.

rockerface@lemm.ee on 28 Jun 01:37 next collapse

pi isn’t even a fraction. like, it’s actually an important thing that it isn’t

I_am_10_squirrels@beehaw.org on 28 Jun 03:16 collapse

pi=c/d

it’s a fraction, just not with integers, so it’s not rational, so it’s not a fraction.

Wandering_Uncertainty@lemmy.world on 28 Jun 02:33 next collapse

The problem is, that’s exactly what the … is for. It is a little weird to our heads, granted, but it does allow the conversion. 0.33 is not the same thing as 0.333… The first is close to one third. The second one is one third. It’s how we express things as a decimal that don’t cleanly map to base ten. It may look funky, but it works.

force@lemmy.world on 28 Jun 05:17 next collapse

Pi isn’t a fraction (in the sense of a rational fraction, an algebraic fraction where the numerator and denominator are both polynomials, like a ratio of 2 integers) – it’s an irrational number, i.e. a number with no fractional form; as opposed to rational numbers, which are defined as being able to be expressed as a fraction. Furthermore, π is a transcendental number, meaning it’s never a solution to f(x) = 0, where f(x) is a non-zero finite-degree polynomial expression with rational coefficients. That’s like, literally part of the definition. They cannot be compared to rational numbers like fractions.

Every rational number (and therefore every fraction) can be expressed using either repeating decimals or terminating decimals. Contrastly, irrational numbers only have decimal expansions which are both non-repeating and non-terminating.

Since |r|<1 → ∑[n=1, ∞] arⁿ = ar/(1-r), and 0.999… is equivalent to that sum with a = 9 and r = 1/10 (visually, 0.999… = 9(0.1) + 9(0.01) + 9(0.001) + …), it’s easy to see after plugging in, 0.999… = ∑[n=1, ∞] 9(1/10)ⁿ = 9(1/10) / (1 - 1/10) = 0.9/0.9 = 1). This was a proof present in Euler’s Elements of Algebra.

ccunning@lemmy.world on 28 Jun 11:08 next collapse

pie never actually ends

I want to go to there.

sp3tr4l@lemmy.zip on 29 Jun 05:33 collapse

There are a lot of concepts in mathematics which do not have good real world analogues.

i, the _imaginary number_for figuring out roots, as one example.

I am fairly certain you cannot actually do the mathematics to predict or approximate the size of an atom or subatomic particle without using complex algebra involving i.

It’s been a while since I watched the entire series Leonard Susskind has up on youtube explaining the basics of the actual math for quantum mechanics, but yeah I am fairly sure it involves complex numbers.

myslsl@lemmy.world on 01 Jul 18:51 collapse

i has nice real world analogues in the form of rotations by pi/2 about the origin (though this depends a little bit on what you mean by “real world analogue”).

Since i=exp(ipi/2), if you take any complex number z and write it in polar form z=rexp(it), then multiplication by i yields a rotation of z by pi/2 about the origin because zi=rexp(it)exp(ipi/2)=rexp(i(t+pi/2)) by using rules of exponents for complex numbers.

More generally since any pair of complex numbers z, w can be written in polar form z=rexp(it), w=uexp(iv) we have wz=(ru)exp(i(t+v)). This shows multiplication of a complex number z by any other complex number w can be thought of in terms of rotating z by the angle that w makes with the x axis (i.e. the angle v) and then scaling the resulting number by the magnitude of w (i.e. the number u)

Alternatively you can get similar conclusions by Demoivre’s theorem if you do not like complex exponentials.

SkyezOpen@lemmy.world on 28 Jun 04:09 next collapse

Oh shit, don’t think I saw that before. That makes it intuitive as hell.

DeanFogg@lemm.ee on 28 Jun 06:03 collapse

Cut a banana into thirds and you lose material from cutting it hence .9999

wholookshere@lemmy.blahaj.zone on 29 Jun 18:49 collapse

That’s not how fractions and math work though.

Shampiss@sh.itjust.works on 27 Jun 12:02 next collapse

Divide 1 by 3: 1÷3=0.3333…

Multiply the result by 3 reverting the operation: 0.3333… x 3 = 0.9999… or just 1

0.9999… = 1

ArchAengelus@lemmy.dbzer0.com on 27 Jun 12:27 next collapse

In this context, yes, because of the cancellation on the fractions when you recover.

1/3 x 3 = 1

I would say without the context, there is an infinitesimal difference. The approximation solution above essentially ignores the problem which is more of a functional flaw in base 10 than a real number theory issue

Shampiss@sh.itjust.works on 27 Jun 13:21 next collapse

The context doesn’t make a difference

In base 10 --> 1/3 is 0.333…

In base 12 --> 1/3 is 0.4

But they’re both the same number.

Base 10 simply is not capable of displaying it in a concise format. We could say that this is a notation issue. No notation is perfect. Base 10 has some confusing implications

ColeSloth@discuss.tchncs.de on 27 Jun 23:00 collapse

They’re different numbers. Base 10 isn’t perfect and can’t do everything just right, so you end up with irrational numbers that go on forever, sometimes.

chaonaut@lemmy.world on 27 Jun 14:07 collapse

This seems to be conflating 0.333…3 with 0.333… One is infinitesimally close to 1/3, the other is a decimal representation of 1/3. Indeed, if 1-0.999… resulted in anything other than 0, that would necessarily be a number with more significant digits than 0.999… which would mean that the failed to be an infinite repetition.

ColeSloth@discuss.tchncs.de on 27 Jun 22:13 collapse

You’re just rounding up an irrational number. You have a non terminating, non repeating number, that will go on forever, because it can never actually get up to its whole value.

WldFyre@lemm.ee on 27 Jun 22:43 next collapse

1/3 is a rational number, because it can be depicted by a ratio of two integers. You clearly don’t know what you’re talking about, you’re getting basic algebra level facts wrong. Maybe take a hint and read some real math instead of relying on your bad intuition.

ColeSloth@discuss.tchncs.de on 27 Jun 22:49 collapse

1/3 is rational.

.3333… is not. You can’t treat fractions the same as our base 10 number system. They don’t all have direct conversions. Hence, why you can have a perfect fraction of a third, but not a perfect 1/3 written out in base 10.

WldFyre@lemm.ee on 27 Jun 23:01 next collapse

0.333… exactly equals 1/3 in base 10. What you are saying is factually incorrect and literally nonsense. You learn this in high school level math classes. Link literally any source that supports your position.

pyre@lemmy.world on 28 Jun 04:59 collapse

.333… is rational.

at least we finally found your problem: you don’t know what rational and irrational mean. the clue is in the name.

Klear@sh.itjust.works on 28 Jun 05:30 collapse

TBH the name is a bit misleading. Same for “real” numbers. And oh so much more so for “normal numbers”.

pyre@lemmy.world on 28 Jun 06:37 collapse

not really. i get it because we use rational to mean logical, but that’s not what it means here. yeah, real and normal are stupid names but rational numbers are numbers that can be represented as a ratio of two numbers. i think it’s pretty good.

Klear@sh.itjust.works on 28 Jun 06:43 collapse

I know all of that, but it’s still misleading. It’s not a dumb name by any means, but it still causes confusion often (as evidenced by many comments here)

pyre@lemmy.world on 28 Jun 07:30 collapse

fair enough, but i think the confusion for that commenter comes from a misunderstanding of the definition of the mathematical concept rather than the meaning of the English word. they just think irrational numbers are those that have infinite decimal digits, which is not the definition.

pyre@lemmy.world on 28 Jun 05:00 collapse

non repeating

it’s literally repeating

Blum0108@lemmy.world on 27 Jun 12:58 next collapse

I was taught that if 0.9999… didn’t equal 1 there would have to be a number that exists between the two. Since there isn’t, then 0.9999…=1

wieson@feddit.org on 28 Jun 05:46 collapse

Not even a number between, but there is no distance between the two. There is no value X for 1-x = 0.9~

We can’t notate 0.0~ …01 in any way.

ColeSloth@discuss.tchncs.de on 27 Jun 21:57 next collapse

X=.5555…

10x=5.5555…

Subtract x from both sides.

9x=5

X=1 .5555 must equal 1.

There it isn’t. Because that math is bullshit.

lazyViking@lemmy.world on 27 Jun 22:18 next collapse

Quick maffs

blue@ttrpg.network on 27 Jun 22:34 next collapse

x = 5/9 is not 9/9. 5/9 = .55555…

You’re proving that 0.555… equals 5/9 (which it does), not that it equals 1 (which it doesn’t).

It’s absolutely not the same result as x = 0.999… as you claim.

Redex68@lemmy.world on 27 Jun 22:53 next collapse

?
Where did you get 9x=5 -> x=1
and 5/9 is 0.555… so it checks out.

force@lemmy.world on 28 Jun 05:48 collapse

Lol what? How did you conclude that if 9x = 5 then x = 1? Surely you didn’t pass algebra in high school, otherwise you could see that getting x from 9x = 5 requires dividing both sides by 9, which yields x = 5/9, i.e. 0.555… = 5/9 since x = 0.555….

Also, you shouldn’t just use uppercase X in place of lowercase x or vice versa. Case is usually significant for variable names.

[deleted] on 28 Jun 06:01 next collapse

.

yetAnotherUser@discuss.tchncs.de on 28 Jun 11:49 next collapse

Unfortunately not an ideal proof.

It makes certain assumptions:

  1. That a number 0.999… exists and is well-defined
  2. That multiplication and subtraction for this number work as expected

Similarly, I could prove that the number which consists of infinite 9’s to the left of the decimal separator is equal to -1:

...999.0 = x
...990.0 = 10x

Calculate x - 10x:

x - 10x = ...999.0 - ...990.0
-9x = 9
x = -1

And while this is true for 10-adic numbers, it is certainly not true for the real numbers.

Valthorn@feddit.nu on 28 Jun 14:32 collapse

While I agree that my proof is blunt, yours doesn’t prove that .999… is equal to -1. With your assumption, the infinite 9’s behave like they’re finite, adding the 0 to the end, and you forgot to move the decimal point in the beginning of the number when you multiplied by 10.

x=0.999…999

10x=9.999…990 assuming infinite decimals behave like finite ones.

Now x - 10x = 0.999…999 - 9.999…990

-9x = -9.000…009

x = 1.000…001

Thus, adding or subtracting the infinitesimal makes no difference, meaning it behaves like 0.

Edit: Having written all this I realised that you probably meant the infinitely large number consisting of only 9’s, but with infinity you can’t really prove anything like this. You can’t have one infinite number being 10 times larger than another. It’s like assuming division by 0 is well defined.

0a=0b, thus

a=b, meaning of course your …999 can equal -1.

Edit again: what my proof shows is that even if you assume that .000…001≠0, doing regular algebra makes it behave like 0 anyway. Your proof shows that you can’t to regular maths with infinite numbers, which wasn’t in question. Infinity exists, the infinitesimal does not.

yetAnotherUser@discuss.tchncs.de on 28 Jun 15:35 collapse

Yes, but similar flaws exist for your proof.

The algebraic proof that 0.999… = 1 must first prove why you can assign 0.999… to x.

My “proof” abuses algebraic notation like this - you cannot assign infinity to a variable. After that, regular algebraic rules become meaningless.

The proper proof would use the definition that the value of a limit approaching another value is exactly that value. For any epsilon > 0, 0.999… will be within the epsilon environment of 1 (= the interval 1 ± epsilon), therefore 0.999… is 1.

sp3tr4l@lemmy.zip on 29 Jun 05:16 collapse

The explanation I’ve seen is that … is notation for something that can be otherwise represented as sums of infinite series.

In the case of 0.999…, it can be shown to converge toward 1 with the convergence rule for geometric series.

If |r| < 1, then:

ar + ar² + ar³ + … = ar / (1 - r)

Thus:

0.999… = 9(1/10) + 9(1/10)² + 9(1/10)³ + …

= 9(1/10) / (1 - 1/10)

= (9/10) / (9/10)

= 1

Just for fun, let’s try 0.424242…

0.424242… = 42(1/100) + 42(1/100)² + 42(1/100)³

= 42(1/100) / (1 - 1/100)

= (42/100) / (99/100)

= 42/99

= 0.424242…

So there you go, nothing gained from that other than seeing that 0.999… is distinct from other known patterns of repeating numbers after the decimal point.

Tomorrow_Farewell@hexbear.net on 30 Jun 15:23 collapse

The explanation I’ve seen is that … is notation for something that can be otherwise represented as sums of infinite series

The ellipsis notation generally refers to repetition of a pattern. Either ad infinitum, or up to some terminus. In this case we have a non-terminating decimal.

In the case of 0.999…, it can be shown to converge toward 1

0.999… is a real number, and not any object that can be said to converge. It is exactly 1.

So there you go, nothing gained from that other than seeing that 0.999… is distinct from other known patterns of repeating numbers after the decimal point

In what way is it distinct?
And what is a ‘repeating number’? Did you mean ‘repeating decimal’?

sp3tr4l@lemmy.zip on 30 Jun 21:12 collapse

The ellipsis notation generally refers to repetition of a pattern.

Ok. In mathematical notation/context, it is more specific, as I outlined.

This technicality is often brushed over or over simplified by math teachers and courses until or unless you take some more advanced courses.

Context matters, here’s an example:

Generally, pdf denotes the file format specific to adobe reader, while in the context of many modern online videos/discussions, it has become a colloquialism to be able to discuss (accused or confirmed) pedophiles and be able to avoid censorship or demonetization.

0.999… is a real number, and not any object that can be said to converge. It is exactly 1.

Ok. Never said 0.999… is not a real number. Yep, it is exactly 1 because solving the equation it truly represents, a geometric series, results in 1. This solution is obtained using what is called the convergence theorem or rule, as I outlined.

In what way is it distinct?

0.424242… solved via the convergence theorem simply results in itself, as represented in mathematical nomenclature.

0.999… does not again result in 0.999…, but results to 1, a notably different representation that causes the entire discussion in this thread.

And what is a ‘repeating number’? Did you mean ‘repeating decimal’?

I meant what I said: “know patterns of repeating numbers after the decimal point.”

Perhaps I should have also clarified known finite patterns to further emphasize the difference between rational and irrational numbers.

EDIT: You used a valid and even more mathematically esoteric method to demonstrate the same thing I demonstrated elsewhere in this thread, I have no idea why you are taking issue with what I’ve said.

Tomorrow_Farewell@hexbear.net on 30 Jun 23:10 collapse

Ok. In mathematical notation/context, it is more specific, as I outlined.

It is not. You will routinely find it used in cases where your explanation does not apply, such as to denote the contents of a matrix.

Furthermore, we can define real numbers without defining series. In such contexts, your explanation also doesn’t work until we do defines series of rational numbers.

Ok. Never said 0.999… is not a real number

In which case it cannot converge to anything on account of it not being a function or any other things that can be said to converge.

because solving the equation it truly represents, a geometric series, results in 1

A series is not an equation.

This solution is obtained using what is called the convergence theorem

What theorem? I have never heard of ‘the convergence theorem’.

0.424242… solved via the convergence theorem simply results in itself

What do you mean by ‘solving’ a real number?

0.999… does not again result in 0.999…, but results to 1

In what way does it not ‘result in 0.999…’ when 0.999… = 1?

You seem to not understand what decimals are, because while decimals (which are representations of real numbers) ‘0.999…’ and ‘1’ are different, they both refer to the same real number. We can use expressions ‘0.999…’ and ‘1’ interchangeably in the context of base 10. In other bases, we can easily also find similar pairs of digital representations that refer to the same numbers.

I meant what I said: “know patterns of repeating numbers after the decimal point.”

What we have after the decimal point are digits. OTOH, sure, we can treat them as numbers, but still, this is not a common terminology. Furthermore, ‘repeating number’ is not a term in any sort of commonly-used terminology in this context.

The actual term that you were looking for is ‘repeating decimal’.

Perhaps I should have also clarified known finite patterns to further emphasize the difference between rational and irrational numbers

No irrational number can be represented by a repeating decimal.

sp3tr4l@lemmy.zip on 01 Jul 05:55 collapse

www2.kenyon.edu/Depts/Math/…/GeomSeriesCalcB.pdf

Here’s a standard introduction to the concept of the Convergence/Divergence Theorem of Geometric Series, starts on page 2.

Its quite common for this to be referred to as the convergence test or rule or theorem by teachers and TA’s.

Tomorrow_Farewell@hexbear.net on 01 Jul 06:12 collapse

Now, ask yourself this question, ‘is 0.999…, or any real number for that matter, a series?’. The answer to that question is ‘no’.

You seem to be extremely confused, and think that the terms ‘series’ and ‘the sum of a series’ mean the same thing. They do not. 0.999… is the sum of the series 9/10+9/100+9/1000+…, and not a series itself.

EDIT: Also, the author does abuse the notations somewhat when she says ‘1+1/2+1/4 = 2’ is a geometric series, as the geometric series 1+1/2+1/4+… does not equal 2, because a series is either just a formal sum, a sequence of its terms, or, in German math traditions, a sequence of its partial sums. It is the sum of the series 1+1/2+1/4+… that is equal to 2. The confusion is made worse by the fact that sums of series and the series themselves are often denoted in the same way. However, again, those are different things.
Would you mind providing a snippet with the definition of the term ‘series’ that she provides?

EDIT 2: Notably, that document has no theorem that is called ‘convergence theorem’ or ‘the convergence theorem’. The only theorem that is present there is the one on convergence and divergence of geometric series.

Th4tGuyII@fedia.io on 27 Jun 12:49 next collapse

0.999... / 3 = 0.333...
1 / 3 = 0.333...
Ergo 1 = 0.999...

(Or see algebraic proof by @Valthorn)

If the difference between two numbers is so infinitesimally small they are in essence mathematically equal, then I see no reason to not address then as such.

If you tried to make a plank of wood 0.999...m long (and had the tools to do so), you'd soon find out the universe won't let you arbitrarily go on to infinity. You'd find that when you got to the planck length, you'd have to either round up the previous digit, resolving to 1, or stop at the last 9.

themeatbridge@lemmy.world on 27 Jun 13:12 next collapse

Except it isn’t infinitesimally smaller at all. 0.999… is exactly 1, not at all less than 1. That’s the power of infinity. If you wanted to make a wooden board exactly 0.999… m long, you would need to make a board exactly 1 m long (which presents its own challenges).

Th4tGuyII@fedia.io on 27 Jun 19:51 collapse

It is mathematically equal to one, but it isn't physically one. If you wrote out 0.999... out to infinity, it'd never just suddenly round up to 1.

But the point I was trying to make is that I agree with the interpretation of the meme in that the above distinction literally doesn't matter - you could use either in a calculation and the answer wouldn't (or at least shouldn't) change.

That's pretty much the point I was trying to make in proving how little the difference makes in reality - that the universe wouldn't let you explore the infinity between the two, so at some point you would have to round to 1m, or go to a number 1x planck length below 1m.

themeatbridge@lemmy.world on 27 Jun 19:57 collapse

It is physically equal to 1. Infinity goes on forever, and so there is no physical difference.

It’s not that it makes almost no difference. There is no difference because the values are identical. There is no infinity between the two values.

Th4tGuyII@fedia.io on 27 Jun 20:08 collapse

Again, if you started writing 0.999... on a piece of paper, it would never suddenly become 1, it would always be 0.999... - you know that to be true without even trying it.

The difference is virtually nonexistent, and that is what makes them mathematically equal, but there is a difference, otherwise there wouldn't be an infinitely long string of 9s between the two.

themeatbridge@lemmy.world on 27 Jun 21:15 collapse

Sure, but you’re equivocating two things that aren’t the same. Until you’ve written infinity 9s, you haven’t written the number yet. Once you do, the number you will have written will be exactly the number 1, because they are exactly the same. The difference between all the nines you could write in one thousand lifetimes and 0.999… is like the difference between a cup of sand and all of spacetime.

Or think of it another way. Forget infinity for a moment. Think of 0.999… as all the nines. All of them contained in the number 1. There’s always one more, right? No, there isn’t, because 1 contains all of them. There are no more nines not included in the number 1. That’s why they are identical.

Feathercrown@lemmy.world on 27 Jun 13:21 collapse

Math doesn’t care about physical limitations like the planck length.

Th4tGuyII@fedia.io on 27 Jun 19:54 collapse

Any real world implementation of maths (such as the length of an object) would definitely be constricted to real world parameters, and the lowest length you can go to is the Planck length.

But that point wasn't just to talk about a plank of wood, it was to show how little difference the infinite 9s in 0.999... make.

Feathercrown@lemmy.world on 27 Jun 20:16 collapse

Afaik, the Planck Length is not a “real-world pixel” in the way that many people think it is. Two lengths can differ by an amount smaller than the Planck Length. The remarkable thing is that it’s impossible to measure anything smaller than that size, so you simply couldn’t tell those two lengths apart. This is also ignoring how you’d create an object with such a precisely defined length in the first place.

Anyways of course the theoretical world of mathematics doesn’t work when you attempt to recreate it in our physical reality, because our reality has fundamental limitations that you’re ignoring when you make that conversion that make the conversion invalid. See for example the Banach-Tarski paradox, which is utter nonsense in physical reality. It’s not a coincidence that that phenomenon also relies heavily on infinities.

In the 0.999… case, the infinite 9s make all the difference. That’s literally the whole point of having an infinite number of them. “Infinity” isn’t (usually) defined as a number; it’s more like a limit or a process. Any very high but finite number of 9s is not 1. There will always be a very small difference. But as soon as there are infinite 9s, that number is 1 (assuming you’re working in the standard mathematical model, of course).

You are right that there’s “something” left behind between 0.999… and 1. Imagine a number line between 0 and 1. Each 9 adds 90% of the remaining number line to the growing number 0.999… as it approaches one. If you pick any point on this number line, after some number of 9s it will be part of the 0.999… region, no matter how close to 1 it is… except for 1 itself. The exact point where 1 is will never be added to the 0.999… fraction. But let’s see how long that 0.999… region now is. It’s exactly 1 unit long, minus a single 0-dimensional point… so still 1-0=1 units long. If you took the 0.999… region and manually added the “1” point back to it, it would stay the exact same length. This is the difference that the infinite 9s make-- only with a truly infinite number of 9s can we find this property.

HexesofVexes@lemmy.world on 27 Jun 13:50 next collapse

Reals are just point cores of dressed Cauchy sequences of naturals (think of it as a continually constructed set of narrowing intervals “homing in” on the real being constructed). The intervals shrink at the same rate generally.

1!=0.999 iff we can find an n, such that the intervals no longer overlap at that n. This would imply a layer of absolute infinite thinness has to exist, and so we have reached a contradiction as it would have to have a width smaller than every positive real (there is no smallest real >0).

Therefore 0.999…=1.

However, we can argue that 1 is not identity to 0.999… quite easily as they are not the same thing.

This does argue that this only works in an extensional setting (which is the norm for most mathematics).

lemmy_99c4zb3e3@reddthat.com on 27 Jun 15:29 next collapse

extensional setting

Very good answer. The only missing thing is what extensionality is.

https://plato.stanford.edu/entries/logic-intensional/

HexesofVexes@lemmy.world on 27 Jun 21:19 collapse

Thanks for the bedtime reading!

I mostly deal with foundations of analysis, so this could be handy.

pumpkinseedoil@sh.itjust.works on 27 Jun 15:49 collapse

Easiest way to prove it:

1 = 3/3 = 1/3 * 3 = 0.333… * 3 = 0.999…

HexesofVexes@lemmy.world on 27 Jun 21:18 collapse

Ehh, completed infinities give me wind…

jerkface@lemmy.ca on 27 Jun 14:10 next collapse

Okay, but it equals one.

lauha@lemmy.one on 27 Jun 15:00 collapse

No, it equals 0.999…

Karcinogen@discuss.tchncs.de on 27 Jun 15:52 next collapse

2/9 = 0.222… 7/9 = 0.777…

0.222… + 0.777… = 0.999… 2/9 + 7/9 = 1

0.999… = 1

No, it equals 1.

skulblaka@sh.itjust.works on 27 Jun 17:12 next collapse

Sure, when you start decoupling the numbers from their actual values. The only thing this proves is that the fraction-to-decimal conversion is inaccurate. Your floating points (and for that matter, our mathematical model) don’t have enough precision to appropriately model what the value of 7/9 actually is. The variation is negligible though, and that’s the core of this, is the variation off what it actually is is so small as to be insignificant and, really undefinable to us - but that doesn’t actually matter in practice, so we just ignore it or convert it. But at the end of the day 0.999… does not equal 1. A number which is not 1 is not equal to 1. That would be absurd. We’re just bad at converting fractions in our current mathematical understanding.

Edit: wow, this has proven HIGHLY unpopular, probably because it’s apparently incorrect. See below for about a dozen people educating me on math I’ve never heard of. The “intuitive” explanation on the Wikipedia page for this makes zero sense to me largely because I don’t understand how and why a repeating decimal can be considered a real number. But I’ll leave that to the math nerds and shut my mouth on the subject.

barsoap@lemm.ee on 27 Jun 17:35 next collapse

The only thing this proves is that the fraction-to-decimal conversion is inaccurate.

No number is getting converted, it’s the same number in both cases but written in a different representation. 4 is also the same number as IV, no conversion going on it’s still the natural number elsewhere written S(S(S(S(Z)))). Also decimal representation isn’t inaccurate, it just happens to have multiple valid representations for the same number.

A number which is not 1 is not equal to 1.

Good then that 0.999… and 1 are not numbers, but representations.

bitfucker@programming.dev on 27 Jun 19:43 collapse

Lol I fucking love that successor of zero

IntriguedIceberg@lemmy.world on 27 Jun 17:43 next collapse

It still equals 1, you can prove it without using fractions:

x = 0.999…

10x = 9.999…

10x = 9 + 0.999…

10x = 9 + x

9x = 9

x = 1

There’s even a Wikipedia page on the subject

Clinicallydepressedpoochie@lemmy.world on 27 Jun 20:17 next collapse

I hate this because you have to subtract .99999… from 10. Which is just the same as saying 10 - .99999… = 9

Which is the whole controversy but you made it complicated.

It would be better just to have them do the long subtraction

<img alt="" src="https://lemmy.world/pictrs/image/1cd6763c-79b5-4f19-90b6-7906844aaa4e.jpeg">

If they don’t get it and keep trying to show you how you are wrong they will at least be out of your hair until forever.

[deleted] on 27 Jun 20:22 next collapse

.

Laser@feddit.org on 27 Jun 20:35 collapse

You don’t subtract from 10, but from 10x0.999… I mean your statement is also true but it just proves the point further.

[deleted] on 27 Jun 21:17 next collapse

.

Clinicallydepressedpoochie@lemmy.world on 27 Jun 21:17 collapse

No, you do subtract from 9.999999…

ColeSloth@discuss.tchncs.de on 27 Jun 22:22 collapse

Do that same math, but use .5555… instead of .9999…

BeardedGingerWonder@feddit.uk on 27 Jun 23:41 next collapse

Have you tried it? You get 0.555… which kinda proves the point does it not?

Wandering_Uncertainty@lemmy.world on 28 Jun 02:31 collapse

???

Not sure what you’re aiming for. It proves that the setup works, I suppose.

x = 0.555…

10x = 5.555…

10x = 5 + 0.555…

10x = 5+x

9x = 5

x = 5/9

5/9 = 0.555…

So it shows that this approach will indeed provide a result for x that matches what x is supposed to be.

Hopefully it helped?

frezik@midwest.social on 27 Jun 17:48 next collapse

It’s a correct proof.

One way to think about this is that we represent numbers in different ways. For example, 1 can be 1.0, or a single hash mark, or a dot, or 1/1, or 10/10. All of them point to some platonic ideal world version of the concept of the number 1.

What we have here is two different representations of the same number that are in a similar representation. 1 and 0.999… both point to the same concept.

tuna@discuss.tchncs.de on 27 Jun 18:00 next collapse

If they aren’t equal, there should be a number in between that separates them. Between 0.1 and 0.2 i can come up with 0.15. Between 0.1 and 0.15 is 0.125. You can keep going, but if the numbers are equal, there is nothing in between. There’s no gap between 0.1 and 0.1, so they are equal.

What number comes between 0.999… and 1?

(I used to think it was imprecise representations too, but this is how it made sense to me :)

myslsl@lemmy.world on 27 Jun 18:19 next collapse

You are just wrong.

The rigorous explanation for why 0.999…=1 is that 0.999… represents a geometric series of the form 9/10+9/10^2+… by definition, i.e. this is what that notation literally means. The sum of this series follows by taking the limit of the corresponding partial sums of this series (see here) which happens to evaluate to 1 in the particular case of 0.999… this step is by definition of a convergent infinite series.

IsoSpandy@lemm.ee on 28 Jun 03:48 next collapse

My brother. You are scared of infinities. Look up the infinite hotel problem. I will lay it out for you if you are interested.

Image you are incharge of a hotel and it has infinite rooms. Currently your hotel is at full capacity… Meaning all rooms are occupied. A new guest arrives. What do you do? Surely your hotel is full and you can’t take him in… Right? WRONG!!! You tell the resident of room 1 to move to room 2, you tell the resident of room 2 to move to room 3 and so on… You tell the resident of room n to move to room n+1. Now you have room 1 empty

But sir… How did I create an extra room? You didn’t. The question is the same as asking yourself that is there a number for which n+1 doesn’t exist. The answer is no… I can always add 1.

Infinity doesn’t behave like other numbers since it isn’t technically a number.

So when you write 0.99999… You are playing with things that aren’t normal. Maths has come with fuckall ways to deal with stuff like this.

Well you may say, this is absurd… There is nothing in reality that behaves this way. Well yes and no. You know how the building blocks of our universe obey quantum mechanics? The equations contain lots of infinities but only at intermediate steps. You have to “renormalise” them to make them go away. Nature apparently has infinities but likes to hide the from us.

The infinity problem is so fucked up. You know the reason physics people are unable to quantize gravity? Surely they can do the same thing to gravity as they did to say electromagnetic force? NOPE. Gravitation doesn’t normalise. You get left with infinities in your final answer.

Anyways. Keep on learning, the world has a lot of information and it’s amazing. And the only thing that makes us human is the ability to learn and grow from it. I wish you all the very best.

barsoap@lemm.ee on 28 Jun 04:19 collapse

But sir… How did I create an extra room? You didn’t.

When Hilbert runs the hotel, sure, ok. Once he sells the whole thing to an ultrafinitist however you suddenly notice that there’s a factory there and all the rooms are on rails and infinity means “we have a method to construct arbitrarily more rooms”, but they don’t exist before a guest arrives to occupy them.

Tlaloc_Temporal@lemmy.ca on 28 Jun 04:42 collapse

I strongly agree with you, and while the people replying aren’t wrong, they’re arguing for something that I don’t think you said.

1/3 ≈ 0.333… in the same way that approximating a circle with polygons of increasing side number has a limit of a circle, but will never yeild a circle with just geometry.

0.999… ≈ 1 in the same way that shuffling infinite people around an infinite hotel leaves infinite free rooms, but if you try to do the paperwork, no one will ever get anywhere.

Decimals require you to check the end of the number to see if you can round up, but there never will be an end. Thus we need higher mathematics to avoid the halting problem. People get taught how decimals work, find this bug, and then instead of being told how decimals are broken, get told how they’re wrong for using the tools they’ve been taught.

If we just accept that decimals fail with infinite steps, the transition to new tools would be so much easier, and reflect the same transition into new tools in other sciences. Like Bohr’s Atom, Newton’s Gravity, Linnaean Taxonomy, or Comte’s Positivism.

skulblaka@sh.itjust.works on 28 Jun 05:06 next collapse

That does very accurately sum up my understanding of the matter, thanks. I haven’t been adding on to any of the other conversation in order to avoid putting my foot in my mouth further, but you’ve pretty much hit the nail on the head here. And the higher mathematics required to solve this halting problem are beyond me.

barsoap@lemm.ee on 28 Jun 05:19 collapse

Decimals require you to check the end of the number to see if you can round up, but there never will be an end.

The character sequence “0.999…” is finite and you know you can round up because you’ve got those three dots at the end. I agree that decimals are a shit representation to formalise rational numbers in but it’s not like using them causes infinite loops. Unless you insist on writing them, that is. You can compute with infinities just fine as long as you keep them symbolic.

That only breaks down with the reals where equality is fundamentally incomputable. Equality of the rationals and approximate equality of reals is perfectly computable though, the latter meaning that you can get equality to arbitrary, but not actually infinite, precision. You can specify a number of digits you want, you can say “don’t take longer than ten seconds to compute”, any kind of bound. Once the precision goes down to plank lengths I think any reasonable engineer would build a bridge with it.

…sometimes I do think that all those formalists with all those fancy rules about fancy limits are actually way more confused about infinity than freshman CS students.

Tlaloc_Temporal@lemmy.ca on 28 Jun 05:37 collapse

Eh, if you need special rules for 0.999… because the special rules for all other repeating decimals failed, I think we should just accept that the system doesn’t work here. We can keep using the workaround, but stop telling people they’re wrong for using the system correctly.

The deeper understanding of numbers where 0.999… = 1 is obvious needs a foundation of much more advanced math than just decimals, at which point decimals stop being a system and are just a quirky representation.

Saying decimals are a perfect system is the issue I have here, and I don’t think this will go away any time soon. Mathematicians like to speak in absolutely terms where everything is either perfect or discarded, yet decimals seem to be too simple and basal to get that treatment. No one seems to be willing to admit the limitations of the system.

barsoap@lemm.ee on 28 Jun 06:14 next collapse

Noone in the right state of mind uses decimals as a formalisation of numbers, or as a representation when doing arithmetic.

But the way I learned decimal division and multiplication in primary school actually supported periods. Spotting whether the thing will repeat forever can be done in finite time. Constant time, actually.

The deeper understanding of numbers where 0.999… = 1 is obvious needs a foundation of much more advanced math than just decimals

No. If you can accept that 1/3 is 0.333… then you can multiply both sides by three and accept that 1 is 0.99999… Primary school kids understand that. It’s a bit odd but a necessary consequence if you restrict your notation from supporting an arbitrary division to only divisions by ten. And that doesn’t make decimal notation worse than rational notation, or better, it makes it different, rational notation has its own issues like also not having unique forms (2/6 = 1/3) and comparisons (larger/smaller) not being obvious. Various arithmetic on them is also more complicated.

The real take-away is that depending on what you do, one is more convenient than the other. And that’s literally all that notation is judged by in maths: Is it convenient, or not.

Tlaloc_Temporal@lemmy.ca on 28 Jun 13:42 collapse

I never commented on the convenience or usefulness of any method, just tried to explain why so many people get stuck on 0.999… = 1 and are so recalcitrant about it.

If you can accept that 1/3 is 0.333… then you can multiply both sides by three and accept that 1 is 0.99999…

This is a workaround of the decimal flaw using algebraic logic. Trying to hold both systems as fully correct leads to a conflic, and reiterating the algebraic logic (or any other proof) is just restating the problem.

The problem goes away easily once we understand the limits of the decimal system, but we need to state that the system is limited! Otherwise we get conflicting answers and nothing makes sense.

barsoap@lemm.ee on 28 Jun 14:09 collapse

The problem goes away easily once we understand the limits of the decimal system, but we need to state that the system is limited!

But the system is not limited: It has a representation for any rational number. Subjectively you may consider it inelegant, you may consider its use in some area inconvenient, but it is formally correct and complete.

I bet there’s systems where rational numbers have unique representations (never looked into it), and I also bet that they’re awkward AF to use in practice.

This is a workaround of the decimal flaw using algebraic logic.

The representation has to reflect algebraic logic, otherwise it would indeed be flawed. It’s the algebraic relationships that are primary to numbers, not the way in which you happen to put numbers onto paper.

And, honestly, if you can accept that 1/3 == 2/6, what’s so surprising about decimal notation having more than one valid representation for one and the same number? If we want our results to look “clean” with rational notation we have to normalise the fraction from 2/6 to 1/3, and if we want them to look “clean” with decimal notation we, well, have to normalise the notation, from 0.999… to 1. Exact same issue in a different system, and noone complains about.

Tlaloc_Temporal@lemmy.ca on 28 Jun 15:01 collapse

Decimals work fine to represent numbers, it’s the decimal system of computing numbers that is flawed. The “carry the 1” system if you prefer. It’s how we’re taught to add/subtract/multiply/divide numbers first, before we learn algebra and limits.

This is the flawed system, there is no method by which 0.999… can become 1 in here. All the logic for that is algebraic or better.

My issue isn’t with 0.999… = 1, nor is it with the inelegance of having multiple represetations of some numbers. My issue lies entirely with people who use algebraic or better logic to fight an elementary arithmetic issue.

People are using the systems they were taught, and those systems are giving an incorrect answer. Instead of telling those people they’re wrong, focus on the flaws of the tools they’re using.

barsoap@lemm.ee on 28 Jun 15:55 next collapse

This is the flawed system, there is no method by which 0.999… can become 1 in here.

Of course there is a method. You might not have been taught in school but you should blame your teachers for that, and noone else. The rule is simple: If you have a nine as repeating decimal, replace it with a zero and increment the digit before that.

That’s it. That’s literally all there is to it.

My issue lies entirely with people who use algebraic or better logic to fight an elementary arithmetic issue.

It’s not any more of an arithmetic issue than 2/6 == 1/3: As I already said, you need an additional normalisation step. The fundamental issue is that rational numbers do not have unique representations in the systems we’re using.

And, in fact, normalisation in decimal representation is way easier, as the only case to worry about is indeed the repeating nine. All other representations are unique while in the fractional system, all numbers have infinitely many representations.

Instead of telling those people they’re wrong, focus on the flaws of the tools they’re using.

Maths teachers are constantly wrong about everything. Especially in the US which single-handedly gave us the abomination that is PEMDAS.

Instead of blaming mathematicians for talking axiomatically, you should blame teachers for not teaching axiomatic thinking, of teaching procedure instead of laws and why particular sets of laws make sense.

That method I described to get rid of the nines is not mathematical insight. It teaches you nothing. You’re not an ALU, you’re capable of so much more than that, capable of deeper understanding that rote rule application. Don’t sell yourself short.


EDIT: Bijective base-10 might be something you want to look at. Also, I was wrong, there’s way more non-unique representations: 0002 is the same as 2. Damn obvious, that’s why it’s so easy to overlook. Dunno whether it easily extends to fractions can’t be bothered to think right now.

Tlaloc_Temporal@lemmy.ca on 29 Jun 01:50 next collapse

I don’t really care how many representations a number has, so long as those representations make sense. 2 = 02 = 2.0 = 1+1 = -1+3 = 8/4 = 2x/x. That’s all fine, we can use the basic rules of decimal notation to understand the first three, basic arithmetic to understand the next three, and basic algebra for the last one.

0.999… = 1 requires more advanced algebra in a pointed argument, or limits and infinite series to resolve, as well as disagreeing with the result of basic decimal notation. It’s steeped in misdirection and illusion like a magic trick or a phishing email.

I’m not blaming mathematicians for this, I am blaming teachers (and popular culture) for teaching that tools are inflexible, instead of the limits of those systems.

In this whole thread, I have never disagreed with the math, only it’s systematic perception, yet I have several people auguing about the math with me. It’s as if all math must be regarded as infinitely perfect, and any unbelievers must be cast out to the pyre of harsh correction. It’s the dogmatic rejection I take issue with.

barsoap@lemm.ee on 29 Jun 06:29 collapse

0.999… = 1 requires more advanced algebra in a pointed argument,

You’re used to one but not the other. You convinced yourself that because one is new or unacquainted it is hard, while the rest is not. The rule I mentioned Is certainly easier that 2x/x that’s actual algebra right there.

It’s as if all math must be regarded as infinitely perfect, and any unbelievers must be cast out to the pyre of harsh correction

Why, yes. I totally can see your point about decimal notation being awkward in places though I doubt there’s a notation that isn’t, in some area or the other, awkward, and decimal is good enough. We’re also used to it, that plays a big role in whether something is judged convenient.

On the other hand 0.9999… must be equal to 1. Because otherwise the system would be wrong: For the system to be acceptable, for it to be infinitely perfect in its consistency with everything else, it must work like that.

And that’s what everyone’s saying when they’re throwing “1/3 = 0.333… now multiply both by three” at you: That 1 = 0.9999… is necessary. That it must be that way. And because it must be like that, it is like that. Because the integrity of the system trumps your own understanding of what the rules of decimal notation are, it trumps your maths teacher, it trumps all the Fields medallists. That integrity is primal, it’s always semantics first, then figure out some syntax to support it (unless you’re into substructural logics, different topic). It’s why you see mathematicians use the term “abuse of notation” but never “abuse of semantics”.

Tlaloc_Temporal@lemmy.ca on 29 Jun 20:18 collapse

Again, I don’t disagree with the math. This has never been about the math. I get that ever model is wrong, but some are useful. Math isn’t taught like that though, and that’s why people get hung up things like this.

Basic decimal notation doesn’t work well with some things, and insinuates incorrect answers. People use the tools they were taught to use. People get told they’re doing it wrong. People give up on math, stop trying to learn, and just go with what they can understand.

If instead we focus on the limitations of some tools and stop hammering people’s faces in with bigger equations and dogma, the world might have more capable people willing to learn.

barsoap@lemm.ee on 30 Jun 04:11 collapse

I get that ever model is wrong, but some are useful.

There is nothing wrong about decimal notation. It is correct. There’s also nothing wrong about Roman numerals… they’re just awkward AF.

Basic decimal notation doesn’t work well with some things, and insinuates incorrect answers.

You could just as well argue that fractional notation “insinuates” that 1/3 + 1/3 = 2/6. You could argue that 8 + 8 is four because that’s four holes there. Lots of things that people can consider more intuitive than the intended meaning. Don’t get me started on English spelling.

Tlaloc_Temporal@lemmy.ca on 30 Jun 14:51 collapse

Neither of those examples use the rules of those system though.

Basic arithmetic on decimap notation is performed by adding/subtracting each digit in each place, or multiplying each digit by each digit then adding those sub totals together, or the yet more complicated long division.

Adding (and by extension multiplying) requires the carry operation, because digits only go up to 9. A string of 9s requires starting at the smallest digit. 0.999… has no smallest digit, thus the carry operation fails to roll it over to 1. It’s a bug that requires more comprehensive methods to understand.

Someone using only basic arithmetic on decimal notation will conclude that 0.999… is not 1. Another person using only geocentrism will conclude that some planets follow spiral orbits. Both conclusions are wrong, but the fault lies with the tools, not the people using them.

barsoap@lemm.ee on 30 Jun 15:41 collapse

0.999… has no smallest digit, thus the carry operation fails to roll it over to 1.

That’s where limits get involved, snatching the carry from the brink of infinity. You could, OTOH, also ignore that and simply accept that it has to be the case because 0.333… * 3. And let me emphasise this doubly and triply: That is a correct mathematical understanding. You don’t need to get limits involved. It doesn’t make it any more correct, or detailed, or anything. Glancing at Occam’s razor, it’s even the preferable explanation: There’s a gazillion overcomplicated and egg-headed ways to write 1 + 1 = 2 (just have a look at the Principia Mathematica), that doesn’t mean that a kindergarten student doesn’t understand the concept correctly. Begone, superfluous sophistication!

(I just noticed that sophistication actually shares a root with sophistry. What a coincidence)

Someone using only basic arithmetic on decimal notation will conclude that 0.999… is not 1.

Doesn’t pass scrutiny, because then either 0.333… /= 1/3 or 3 /= 3 (or both). It simply cannot be the case when looking at the whole system, as opposed to only the single question 0.999… ?= 1 and trying to glean something from that. Context matters: Any answer to that question has to be consistent with all the rest you know about the natural numbers. And only 0.999… = 1 fulfils that.

Why are you making this so complicated?

Tlaloc_Temporal@lemmy.ca on 01 Jul 00:34 collapse

simply accept that it has to be the case because 0.333… * 3. […] That is a correct mathematical understanding

This is my point, using a simple system (basic arithmetic) properly will give bad answers in specifically this situation. A correct mathematical understanding of arithmetic will lead you to say that something funky is going on with 0.999… , and without a more comprehensive understanding of mathematical systems, the only valid conclusions are that 0.999… doesn’t equal 1, or that basic arithmetic is limited.

So then why does everyone loose their heads when this happens? Thousands of people forcing algebra and limits on anyone they so much as suspect could have a reasonable but flawed conclusion, yet this thread is the first time I’ve seen anyone even try to mention the limitations of arithmetic, and they get stomped on.

Why is basic arithmetic so sacred that it must not be besmirched? Why is it so hard for people to admit that some tools have limits? Why is everyone bringing in so many more advanced systems when my entire argument this whole time is that a simple system has limits?

That’s my whole argument. Firstly, that 0.999… catches people because using arithmetic properly leads to an incorrect understanding of repeating decimals. And secondly, that starting with the limits of arithmetic will increase understand with less frustration than throwing more complicated solutions around.

My argument have never been with the math, only with our perceptions of it and how we go about teaching it.

barsoap@lemm.ee on 01 Jul 05:55 collapse

Why is basic arithmetic so sacred that it must not be besmirched?

It isn’t. It’s convenient. Toss it if you don’t want to use it. What’s not an option though is to use it incorrectly, and that would be insisting that 0.999… /= 1, because that doesn’t make any sense.

A notational system doesn’t get to say “well I like to do numbers this way, let’s break all the axioms or arithmetic”. If you say that 0.333… = 1/3, then it necessarily follows that 0.999… = 1. Forget about “but how do I calculate that” think about “does multiplying the same number by the same number yield the same result”.

catches people because using arithmetic properly leads to an incorrect understanding of repeating decimals.

Repeating decimals aren’t apart from decimal arithmetic. They’re a necessary part of it. If you didn’t learn 0.999… = 1, you did not learn decimal arithmetic. And with “necessary” I mean necessary: Any positional system that supports expressing rational numbers will have repeating digits. It’s the trade-off you make, by fixing the divisor (10 in our case), to make numbers easily comparable by size, because no number can divide any number cleanly because there’s an infinite number of primes. Quick, which is the bigger number: 38/127 or 39/131.

Any notational system has its awkward spots. You will not get around awkward spots. Decimal notation has quite few of them, certainly fewer than Roman numerals where being able to do long division earned you a Ph.D. If you can come up with something better be my guest, I already linked you to a starting point.

SmartmanApps@programming.dev on 29 Jun 18:38 collapse

Maths teachers are constantly wrong about everything

Very rarely wrong actually.

the abomination that is PEMDAS

The only people who think there’s something wrong with PEMDAS are people who have forgotten one or more rules of Maths.

barsoap@lemm.ee on 29 Jun 18:45 collapse

www.youtube.com/watch?v=lLCDca6dYpA

…oh wait I remember that Unicody user name. It’s you. Didn’t I already explain to you the difference between syntax and semantics until you gave up. I suggest we don’t do it again but instead, you review the thread.

SmartmanApps@programming.dev on 29 Jun 18:54 collapse

www.youtube.com/watch?v=lLCDca6dYpA

…oh wait I remember that

Well, you seem to have forgotten that the woman in that video isn’t a Maths teacher, which would explain why she’s forgotten the rules of The Distributive Law and Terms.

until you gave up

I didn’t give up, you did.

I suggest we don’t do it again but instead, you review the thread

I suggest you check some Maths textbooks, instead of listening to a Physics major.

SmartmanApps@programming.dev on 29 Jun 18:56 next collapse

P.S. you proved my point

The only people who think there’s something wrong with PEMDAS are people who have forgotten one or more rules of Maths.

barsoap@lemm.ee on 29 Jun 19:05 collapse

Terms.

There! Syntax. We went over this. Seriously, we did, and, no, I got the last word.

I suggest you check some Maths textbooks, instead of listening to a Physics major.

I can check any textbook from any discipline. You know what? I could even ask my school teachers. Because I’m not American and I wasn’t taught shit that doesn’t match up with what professionals are doing.

You’re just another yank drunk on jingoism, “We do it like that, therefore, it is right”.

SmartmanApps@programming.dev on 29 Jun 19:08 collapse

You’re just another yank

BWAHAHAHA! I see you still didn’t learn to check facts first. 😂😂😂

SmartmanApps@programming.dev on 29 Jun 19:09 collapse

P.S.

“We do it like that, therefore, it is right”

Yep, Maths teachers do it right. :-)

SmartmanApps@programming.dev on 29 Jun 18:43 collapse

those systems are giving an incorrect answer

When there’s an incorrect answer it’s because the user has made a mistake.

Instead of telling those people they’re wrong

They were wrong, and I told them where they went wrong (did something to one side of the equation and not the other).

Tlaloc_Temporal@lemmy.ca on 29 Jun 20:30 collapse

The system I’m talking about is elementary decimal notation and basic arithmetic. Carry the 1 and all that. Equations and algebra are more advanced and not taught yet.

There is no method by which basic arithmetic and decimal notation can turn 0.999… into 1. All of the carry methods require starting at the smallest digit, and repeating decimals have no smallest digit.

If someone uses these systems as they were taught, they will get told they’re wrong for doing so. If we focus on that person being wrong, then they’re more likely to give up on math entirely, because they’re wrong for doing as they were taught. If we focus on the limitstions of that system, then they have the explanation for the error, and an understanding of why the more complicated system is preferable.

All models are wrong, but some are useful.

SmartmanApps@programming.dev on 30 Jun 04:21 collapse

not taught yet

What do you mean not taught yet? There’s nothing in the meme to indicate this is a primary school problem. In fact it explicitly has a picture of an adult, so high school Maths is absolutely on the table.

There is no method by which basic arithmetic and decimal notation can turn 0.999… into 1.

In high school we teach that they are the same thing. i.e. limits of accuracy, 1 isn’t the same thing as 1.000…, but rather 1+/- some limit of accuracy (usually 1/2). Of course in programming it matters if you’re talking about an integer 1 or a floating point 1.

If someone uses these systems as they were taught, they will get told they’re wrong for doing so

The only people I’ve seen get things wrong is people not using the systems correctly (such as the alleged “proof” in this thread, which broke several rules of Maths and as such didn’t prove anything), and it’s a teacher’s job to point out how to use them correctly.

Tlaloc_Temporal@lemmy.ca on 30 Jun 15:02 collapse

What do you mean not taught yet?

I mean those more advanced methods are taught after basic arithmetic. There are plenty of adults that operate primarily with 5th grade math, and a scary number of them do finances…

limits of accuracy

This isn’t about limits of accuracy, we’re working with abstract values and ideal systems. Any inaccuracies must be introduced by those systems.

If you think the system isn’t at fault here, please show me how basic arithmetic can make 0.999… into 1. Show me how the carry method deals with Infinity correctly. If every error is just using the system incorrectly, then a correct use of the system must be applicable to everything, right? You shouldn’t need a new system like algebra to be correct, right?

SmartmanApps@programming.dev on 30 Jun 16:11 collapse

This isn’t about limits of accuracy

According to who? Where does it say what it’s about? It doesn’t.

please show me how basic arithmetic can make 0.999

You still haven’t shown why you’re limiting yourself to basic arithmetic. There isn’t anything at all in the meme to indicate it’s about basic arithmetic only. It’s just some Maths statements with no context given.

then a correct use of the system must be applicable to everything, right?

Different systems for different applications. Sometimes multiple systems for one problem (e.g. proofs).

You shouldn’t need a new system like algebra to be correct, right?

Limits of accuracy isn’t algebra.

Tlaloc_Temporal@lemmy.ca on 01 Jul 00:06 collapse

This isn’t about limits of accuracy

According to who?

According to me, talking about the origin of the 0.999… issue of the original comment, the “conversion of fractions to decimals”, or using basic arithmetic to manipulate values into repeating decimals. This has been my position the entire time. If this was about the limits of accuracy, then it would be impossible to solve the 0.999… = 1 issue. Yet it is possible, our accuracy isn’t limited in this fashion.

You still haven’t shown why you’re limiting yourself to basic arithmetic.

Because that’s where the entire 0.999… = 1 originates. You’ll never even see 0.999… without using basic addition on each digit individually, especially if you use fractions the entire time. Thus 0.999… is an artifact of basic arithmetic, a flaw of that system.

Different systems for different applications.

Then you agree that not every system is applicable everywhere! Even if you use that system perfectly, you’ll still end up with the wrong answer! Thus the issue isn’t someone using the system incorrectly, it’s a limitation of the system that they used. The correct response to this isn’t throwing heaps of other systems at the person, it’s communicating the limit of that system.

If someone is trying to hammer a screw, chastising them for their swinging technique then using your personal impact wrench in front of them isn’t going to help. They’re just going to hit you with the hammer, and continue using the tools they have. Explaining that a hammer can’t do the twisting motion needed for screws, then handing them a screwdriver will get you both much farther.

Limits of accuracy isn’t algebra.

It never was, and neither is the problem we’ve been discussing. You can talk about glue, staples, clamps, rivets, and bolts as much as you like, people with hammers are still going to hit screws.

SmartmanApps@programming.dev on 01 Jul 05:39 collapse

According to me, talking about the origin of the 0.999…

Right. So not according to the meme, which doesn’t tell us where the 0.999… comes from. Nor the 1 - could be an integer, floating point, or an estimation. Thanks for playing.

apolo399@lemmy.world on 28 Jun 17:13 collapse

The system works perfectly, it just looks wonky in base 10. In base 3 0.333… looks like 0.1, exactly 0.1

Tlaloc_Temporal@lemmy.ca on 29 Jun 01:31 collapse

Oh the fundamental math works fine, it’s the imperfect representation that is infinite decimals that is flawed. Every base has at least one.

Crozekiel@lemmy.zip on 27 Jun 17:40 next collapse

That’s the best explanation of this I’ve ever seen, thank you!

zarkanian@sh.itjust.works on 27 Jun 21:45 collapse

That’s more convoluted than the 1/3, 2/3, 3/3 thing.

3/3 = 0.99999…

3/3 = 1

If somebody still wants to argue after that, don’t bother.

Dagrothus@reddthat.com on 28 Jun 08:54 collapse

Nah that explanation is basically using an assumption to prove itself. You need to first prove that 1/3 does in fact equal .3333… which can be done using the ‘convoluted’ but not so convoluted proof

ytg@sopuli.xyz on 27 Jun 21:09 next collapse

Similarly, 1/3 = 0.3333…
So 3 times 1/3 = 0.9999… but also 3/3 = 1

Another nice one:

Let x = 0.9999… (multiply both sides by 10)
10x = 9.99999… (substitute 0.9999… = x)
10x = 9 + x (subtract x from both sides)
9x = 9 (divide both sides by 9)
x = 1

zarkanian@sh.itjust.works on 27 Jun 21:43 collapse

My favorite thing about this argument is that not only are you right, but you can prove it with math.

ColeSloth@discuss.tchncs.de on 27 Jun 22:18 next collapse

Except it doesn’t. The math is wrong. Do the exact same formula, but use .5555… instead of .9999…

Guess it turns out .5555… is also 1.

WldFyre@lemm.ee on 27 Jun 22:35 next collapse

Lol you can’t do math apparently, take a logic course sometime

Let x=0.555…

10x=5.555…

10x=5+x

9x=5

x=5/9=0.555…

Reddfugee42@lemmy.world on 28 Jun 00:14 next collapse

Oh honey

pyre@lemmy.world on 28 Jun 04:52 collapse

you have to do this now

SmartmanApps@programming.dev on 29 Jun 18:17 collapse

you can prove it with math

Not a proof, just wrong. In the “(substitute 0.9999… = x)” step, it was only done to one side, not both (the left side would’ve become 9.99999), therefore wrong.

zarkanian@sh.itjust.works on 29 Jun 19:01 next collapse

They multiplied both sides by 10.

0.9999… times 10 is 9.9999…

X times 10 is 10x.

SmartmanApps@programming.dev on 29 Jun 19:05 collapse

X times 10 is 10x

10x is 9.9999999…

As I said, they didn’t substitute on both sides, only one, thus breaking the rules around rearranging algebra. Anything you do to one side you have to do to the other.

ytg@sopuli.xyz on 30 Jun 17:20 collapse

The substitution property of equality is a part of its definition; you can substitute anywhere.

SmartmanApps@programming.dev on 30 Jun 17:52 collapse

you can substitute anywhere

And if you are rearranging algebra you have to do the exact same thing on both sides, always

SmartmanApps@programming.dev on 30 Jun 17:54 collapse

And if you don’t then you can no longer claim they are still equal.

ytg@sopuli.xyz on 30 Jun 20:14 collapse

For any a, b, c, if a = b and b = c, then a = c, right? The transitive property of equality.
For any a, b, x, if a = b, then x + a = x + b. The substitution property.
By combining both of these properties, for any a, b, x, y, if a = b and y = b + x, it follows that b + x = a + x and y = a + x.

In our example, a is x’ (notice the ') and b is 0.999… (by definition). y is 10x’ and x is 9. Let’s fill in the values.

If x’ = 0.9999… (true by definition) and 10x = 0.999… + 9 (true by algebraic manipulation), then 0.999… + 9 = x’ + 9 and 10x’ = x’ + 9.

if you are rearranging algebra you have to do the exact same thing on both sides

If you actually change any of the sides. Since, after substitution, the numeric value doesn’t change (literally the definition of equality), I don’t have to do anything – as I’m not rearranging. I’m merely presenting the same value in an equivalent manner. By contrast, when multiplying both sides by 10, since multiplication by 10 changes the concrete numeric value, I have to do it on both sides to maintain the equality relation (ditto for subtracting x’). But substitution never changes a numeric value – only rearranges what we already know.


(Edit)

Take the following simple system of equations.

5y = 3
x + y = 6

How would you solve it? Here’s how I would:

\begin{gather*} %% Ignore the LaTeX boilerplate, just so I could render it
\begin{cases}
y = \frac{3}{5} \\ % Isolate y by dividing both sides by 5
x = 6 - y % Subtract y from both sides
\end{cases} \\
x = 6 - \frac{3}{5} \\ % SUBSTITUTE 3/5 for y
x = 5.4 \\
(x, y) = (5.4, 0.6)
\end{gather*}

Here’s how Microsoft Math Solver would do it.

SmartmanApps@programming.dev on 01 Jul 10:40 collapse

10x = 0.999… + 9 (true by algebraic manipulation)

No, you haven’t shown that, because you haven’t shown yet that 9x=9. Welcome to why this doesn’t prove anything. You’re presuming your result, then using it to “prove” your result.

What we know is that the right hand side is 10 times 0.9999…, so if you want to substitute x=0.99999… into the right hand side, then the right hand side becomes 10x (or 9x+x)… which only shows what we already know - 10x=10x. Welcome to the circularity of what you’re trying to achieve. You can’t use something you haven’t yet proven, to prove something you haven’t yet proven.

ColeSloth@discuss.tchncs.de on 27 Jun 22:15 collapse

If you can’t do it without fractions or a … then it can’t be done.

WldFyre@lemm.ee on 27 Jun 22:31 next collapse

1/3=0.333…

2/3=0.666…

3/3=0.999…=1

ColeSloth@discuss.tchncs.de on 27 Jun 22:54 collapse

Fractions and base 10 are two different systems. You’re only approximating what 1/3 is when you write out 0.3333…

The … is because you can’t actually make it correct in base 10.

WldFyre@lemm.ee on 27 Jun 23:03 next collapse

The fractions are still in base 10 lmfao literally what the fuck are you talking about and where are getting this from?

You keep getting basic shit wrong, and it makes you look dumb. Stop talking and go read a wiki.

myslsl@lemmy.world on 28 Jun 03:59 next collapse

What exactly do you think notations like 0.999… and 0.333… mean?

ColeSloth@discuss.tchncs.de on 28 Jun 17:05 collapse

That it repeats forever, to no end. Because it can never actually be correct, just that the difference becomes insignificant.

Bertuccio@lemmy.world on 28 Jun 17:53 next collapse

Numbers are not their notations.

myslsl@lemmy.world on 28 Jun 18:48 collapse

My degree is in mathematics. This is not how these notations are usually defined rigorously.

The most common way to do it starts from sequences of real numbers, then limits of sequences, then sequences of partial sums, then finally these notations turn out to just represent a special kind of limit of a sequence of partial sums.

If you want a bunch of details on this read further:

A sequence of real numbers can be thought of as an ordered nonterminating list of real numbers. For example: 1, 2, 3, … or 1/2, 1/3, 1/4, … or pi, 2, sqrt(2), 1000, 543212345, … or -1, 1, -1, 1, … Formally a sequence of real numbers is a function from the natural numbers to the real numbers.

A sequence of partial sums is just a sequence whose terms are defined via finite sums. For example: 1, 1+2, 1+2+3, … or 1/2, 1/2 + 1/4, 1/2 + 1/4 + 1/8, … or 1, 1 + 1/2, 1 + 1/2 + 1/3, … (do you see the pattern for each of these?)

The notion of a limit is sort of technical and can be found rigorously in any calculus book (such as Stewart’s Calculus) or any real analysis book (such as Rudin’s Principles of Mathematical Analysis) or many places online (such as Paul’s Online Math Notes). The main idea though is that sometimes sequences approximate certain values arbitrarily well. For example the sequence 1, 1/2, 1/3, 1/4, … gets as close to 0 as you like. Notice that no term of this sequence is actually 0. As another example notice the terms of the sequence 9/10, 9/10 + 9/100, 9/10 + 9/100 + 9/1000, … approximate the value 1 (try it on a calculator).

I want to stop here to make an important distinction. None of the above sequences are real numbers themselves because lists of numbers (or more formally functions from N to R) are not the same thing as individual real numbers.

Continuing with the discussion of sequences approximating numbers, when a sequence, call it A, approximates some number L, we say “A converges”. If we want to also specify the particular number that A converges to we say “A converges to L”. We give the number L a special name called “the limit of the sequence A”.

Notice in particular L is just some special real number. L may or may not be a term of A. We have several examples of sequences above with limits that are not themselves terms of the sequence. The sequence 0, 0, 0, … has as its limit the number 0 and every term of this sequence is also 0. The sequence 0, 1, 0, 0, … where only the second term is 1, has limit 0 and some but not all of its terms are 0.

Suppose we define a sequence a1, a2, a3, … where each of the an numbers is one of the numbers from 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9. It can be shown that any sequence of the form a1/10, a1/10 + a2/100, a1/10 + a2/100 + a3/1000, … converges (it is too technical for me to show this here but this is explained briefly in Rudin ch 1 or Hrbacek/Jech’s Introduction To Set Theory).

As an example if each of the an values is 1 our sequence of partial sums above simplifies to 0.1,0.11,0.111,… if the an sequence is 0, 2, 0, 2, … our sequence of partial sums is 0.0, 0.02, 0.020, 0.0202, …

We define the notation 0 . a1 a2 a3 … to be the limit of the sequence of partial sums a1/10, a1/10 + a2/100, a1/10 + a2/100 + a3/1000, … where the an values are all chosen as mentioned above. This limit always exists as specified above also.

In particular 0 . a1 a2 a3 … is just some number and it may or may not be distinct from any term in the sequence of sums we used to define it.

When each of the an values is the same number it is possible to compute this sum explicitly. See here (where a=an, r=1/10 and subtract 1 if necessary to account for the given series having 1 as its first term).

So by definition the particular case where each an is 9 gives us our definition for 0.999…

To recap: the value of 0.999… is essentially just whatever value the (simplified) sequence of partial sums 0.9, 0.99, 0.999, … converges to. This is not necessarily the value of any one particular term of the sequence. It is the value (informally) that the sequence is approximating. The value that the sequence 0.9, 0.99, 0.999, … is approximating can be proved to be 1. So 0.999… = 1, essentially by definition.

ColeSloth@discuss.tchncs.de on 29 Jun 03:06 collapse

That’s is a very precise and very good answer, but im still at a loss as to how all the .9,.99,.999,.9999 eventually just becomes 1.

myslsl@lemmy.world on 29 Jun 04:43 collapse

They don’t eventually become 1. Their limit is 1 but none of the terms themselves are 1.

A sequence, its terms and its limit (if it has one) are all different things. The notation 0.999… represents a limit of a particular sequence, not the sequence itself nor the individual terms of the sequence.

For example the sequence 1, 1/2, 1/3, 1/4, … has terms that get closer and closer to 0, but no term of this sequence is 0 itself.

Look at this graph. If you graph the sequence I just mentioned above and connect each dot you will get the graph shown in this picture (ignoring the portion to the left of x=1).

As you go further and further out along this graph in the positive x direction, the curve that is shown gets closer and closer to the x-axis (where y=0). In a sense the curve is approaching the value y=0. For this curve we could certainly use wordings like “the value the curve approaches” and it would be pretty clear to me and you that we don’t mean the values of the curve itself. This is the kind of intuition that we are trying to formalize when we talk about limits (though this example is with a curve rather than a sequence).

Our sequence 0.9, 0.99, 0.999, … is increasing towards 1 in a similar manner. The notation 0.999… represents the (limit) value this sequence is increasing towards rather than the individual terms of the sequence essentially.

I have been trying to dodge the actual formal definition of the limit of a sequence this whole time since it’s sort of technical. If you want you can check it out here though (note that implicitly in this link the sequence terms and limit values should all be real numbers).

apolo399@lemmy.world on 28 Jun 12:06 collapse

Sure, let’s do it in base 3. 3 in base 3 is 10, and 3^(-1) is 10^(-1), so:

1/3 in base 10 = 1/10 in base 3
0.3… in base 10 = 0.1 in base 3

Multiply by 3 on both sides:

3 × 0.3… in base 10 = 10 × 0.1 in base 3
0.9… in base 10 = 1 in base 3.

But 1 in base 3 is also 1 in base 10, so:

0.9… in base 10 = 1 in base 10

ColeSloth@discuss.tchncs.de on 28 Jun 17:02 collapse

You’re having to use … to make your conversion again. If you need to to an irrational number to make your equation correct, it isn’t really correct.

apolo399@lemmy.world on 28 Jun 17:09 collapse

Do you know what an irrational number is?

Bertuccio@lemmy.world on 27 Jun 23:10 collapse

We’ve found a time traveller from ancient Greece…

Edit: sorry. I mean we’ve found a time traveller from ancient Mesopotamia.

[deleted] on 27 Jun 16:35 next collapse

.

jerkface@lemmy.ca on 28 Jun 14:40 collapse

THAT’S EXACTLY WHAT I SAID.

WolfLink@sh.itjust.works on 27 Jun 14:20 next collapse

en.wikipedia.org/wiki/0.999

CodexArcanum@lemmy.world on 27 Jun 14:21 next collapse

Even the hyperreal numbers *R, which include infinitesimals, define 1 == .999…

pruwybn@discuss.tchncs.de on 27 Jun 16:09 next collapse

I thought the muscular guys were supposed to be right in these memes.

myslsl@lemmy.world on 27 Jun 17:57 next collapse

He is right. 1 approximates 1 to any accuracy you like.

pruwybn@discuss.tchncs.de on 27 Jun 18:03 collapse

Is it true to say that two numbers that are equal are also approximately equal?

mpa92643@lemmy.world on 27 Jun 18:24 next collapse

“Approximately equal” is just a superset of “equal” that also includes values “acceptably close” (using whatever definition you set for acceptable).

Unless you say something like:

a ≈ b ∧ a ≠ b

which implies a is close to b but not exactly equal to b, it’s safe to presume that a ≈ b includes the possibility that a = b.

EatATaco@lemm.ee on 27 Jun 20:30 next collapse

Can I get a citation on this? Because it doesn’t pass the sniff test for me. en.wikipedia.org/wiki/Approximation

mpa92643@lemmy.world on 27 Jun 21:37 collapse

Sure! See ISO-80000-2

Here’s a link: cdn.standards.iteh.ai/…/ISO-80000-2-2019.pdf

<img alt="" src="https://lemmy.world/pictrs/image/7ff33ec7-058d-4885-9ad5-0420198cf271.png">

WldFyre@lemm.ee on 27 Jun 22:30 collapse

ISO is not a source for mathematical definitions

mpa92643@lemmy.world on 27 Jun 23:10 collapse

It’s a definition from a well-respected global standards organization. Can you name a source that would provide a more authoritative definition than the ISO?

There’s no universally correct definition for what the ≈ symbol means, and if you write a paper or a proof or whatever, you’re welcome to define it to mean whatever you want in that context, but citing a professional standards organization seems like a pretty reliable way to find a commonly-accepted and understood definition.

BeardedGingerWonder@feddit.uk on 27 Jun 23:30 collapse

Tbh I’m just impressed you:

A) knew there was an iso standard

  1. went to the effort of locating it

iii) posted it in respectful manner, and

e) are correct.

match@pawb.social on 27 Jun 19:58 collapse

assert np.isClose(3, 3)

myslsl@lemmy.world on 27 Jun 18:36 next collapse

Yes, informally in the sense that the error between the two numbers is “arbitrarily small”. Sometimes in introductory real analysis courses you see an exercise like: “prove if x, y are real numbers such that x=y, then for any real epsilon > 0 we have |x - y| < epsilon.” Which is a more rigorous way to say roughly the same thing. Going back to informality, if you give any required degree of accuracy (epsilon), then the error between x and y (which are the same number), is less than your required degree of accuracy

SpeakerToLampposts@lemmy.world on 27 Jun 19:06 next collapse

I recall an anecdote about a mathematician being asked to clarify precisely what he meant by “a close approximation to three”. After thinking for a moment, he replied “any real number other than three”.

Leate_Wonceslace@lemmy.dbzer0.com on 28 Jun 05:27 collapse

It depends on the convention that you use, but in my experience yes; for any equivalence relation, and any metric of “approximate” within the context of that relation, A=B implies A≈B.

RustyNova@lemmy.world on 28 Jun 10:35 collapse

Nah. They are supposed to not care about stuff and just roll with it without any regrets.

It’s just like the wojak crying with the mask on, but not crying behind it.

There’s plenty of cases of memes where the giga chad is just plainly wrong, but they just don’t care. But it’s not supposed to be in a troll way. The giga chad applies what it believes in. If you want a troll, there’s troll face, who speak with the confidence of a giga chad, but know he is bullshiting

[deleted] on 27 Jun 20:08 next collapse

.

EatATaco@lemm.ee on 27 Jun 20:15 collapse

I think the … here is meant to represent repeating digits.

[deleted] on 27 Jun 21:29 collapse

.

zarkanian@sh.itjust.works on 27 Jun 21:48 collapse

You’re being downvoted because you’re wrong.

My point was only that .9 repeating is still less than 1 in tangible measurement

If it’s less than 1, then it isn’t repeating indefinitely, which is what the ellipsis indicates.

ettyblatant@lemmy.world on 27 Jun 22:36 collapse

I appreciate you taking the time, and I understand where I was wrong. Thank you

EDIT I’m going to take down my comments because it’s affecting my mental health, so for those of you that see it as deleted by creator, I made an incorrect statement, was given information on why I was wrong, and that’s about it. Please just downvote this to oblivion instead, I think I will process this being down voted better

SkyezOpen@lemmy.world on 28 Jun 04:15 collapse

This is the third time today I’ve seen someone admit they were wrong and thank the person that explained it to them. I know lemmy is different from reddit but holy hell that’s unheard of on the internet in general.

Keep being excellent, you and everyone else on this site.

humdrumgentleman@lemmy.world on 27 Jun 20:55 next collapse

This is why we can’t have nice things like dependable protection from fall damage while riding a boat in Minecraft.

someacnt_@lemmy.world on 27 Jun 21:00 next collapse

Are we still doing this 0.999… thing? Why, is it that attractive?

Tlaloc_Temporal@lemmy.ca on 28 Jun 04:06 next collapse

The rules of decimal notation don’t sipport infinite decimals properly. In order for a 9 to roll over into a 10, the next smallest decimal needs to roll over first, therefore an infinite string of anything will never resolve the needed discrete increment.

Thus, all arguments that 0.999… = 1 must use algebra, limits, or some other logic beyond decimal notation. I consider this a bug with decimals, and 0.999… = 1 to be a workaround.

pyre@lemmy.world on 28 Jun 04:48 next collapse

i don’t think any number system can be safe from infinite digits. there’s bound to be some number for each one that has to be represented with them. it’s not intuitive, but that’s because infinity isn’t intuitive. that doesn’t mean there’s a problem there though. also the arguments are so simple i don’t understand why anyone would insist that there has to be a difference.

for me the simplest is:

1/3 = 0.333…

so

3×0.333… = 3×1/3

0.999… = 3/3

Tlaloc_Temporal@lemmy.ca on 28 Jun 05:55 next collapse

Any my argument is that 3 ≠ 0.333…

EDIT: 1/3 ≠ 0.333…

We’re taught about the decimal system by manipulating whole number representations of fractions, but when that method fails, we get told that we are wrong.

In chemistry, we’re taught about atoms by manipulating little rings of electrons, and when that system fails to explain bond angles and excitation, we’re told the model is wrong, but still useful.

This is my issue with the debate. Someone uses decimals as they were taught and everyone piles on saying they’re wrong instead of explaining the limitations of systems and why we still use them.

For the record, my favorite demonstration is useing different bases.

In base 10: 1/3 0.333… 0.333… × 3 = 0.999…

In base 12: 1/3 = 0.4 0.4 × 3 = 1

The issue only appears if you resort to infinite decimals. If you instead change your base, everything works fine. Of course the only base where every whole fraction fits nicely is unary, and there’s some very good reasons we don’t use tally marks much anymore, and it has nothing to do with math.

pyre@lemmy.world on 28 Jun 06:34 next collapse

you’re thinking about this backwards: the decimal notation isn’t something that’s natural, it’s just a way to represent numbers that we invented. 0.333… = 1/3 because that’s the way we decided to represent 1/3 in decimals. the problem here isn’t that 1 cannot be divided by 3 at all, it’s that 10 cannot be divided by 3 and give a whole number. and because we use the decimal system, we have to notate it using infinite repeating numbers but that doesn’t change the value of 1/3 or 10/3.

different bases don’t change the values either. 12 can be divided by 3 and give a whole number, so we don’t need infinite digits. but both 0.333… in decimal and 0.4 in base12 are still 1/3.

there’s no need to change the base. we know a third of one is a third and three thirds is one. how you notate it doesn’t change this at all.

Tlaloc_Temporal@lemmy.ca on 28 Jun 13:49 collapse

I’m not saying that math works differently is different bases, I’m using different bases exactly because the values don’t change. Using different bases restates the equation without using repeating decimals, thus sidestepping the flaw altogether.

My whole point here is that the decimal system is flawed. It’s still useful, but trying to claim it is perfect leads to a conflict with reality. All models are wrong, but some are useful.

pyre@lemmy.world on 28 Jun 14:41 collapse

you said 1/3 ≠ 0.333… which is false. it is exactly equal. there’s no flaw; it’s a restriction in notation that is not unique to the decimal system. there’s no “conflict with reality”, whatever that means. this just sounds like not being able to wrap your head around the concept. but that doesn’t make it a flaw.

Tlaloc_Temporal@lemmy.ca on 28 Jun 15:13 collapse

Let me restate: I am of the opinion that repeating decimals are imperfect representations of the values we use them to represent. This imperfection only matters in the case of 0.999… , but I still consider it a flaw.

I am also of the opinion that focusing on this flaw rather than the incorrectness of the person using it is a better method of teaching.

I accept that 1/3 is exactly equal to the value typically represented by 0.333… , however I do not agree that 0.333… is a perfect representation of that value. That is what I mean by 1/3 ≠ 0.333… , that repeating decimal is not exactly equal to that value.

Leate_Wonceslace@lemmy.dbzer0.com on 30 Jun 15:34 collapse

Any my argument is that 3 ≠ 0.333…

After reading this, I have decided that I am no longer going to provide a formal proof for my other point, because odds are that you wouldn’t understand it and I’m now reasonably confident that anyone who would already understands the fact the proof would’ve supported.

Tlaloc_Temporal@lemmy.ca on 01 Jul 00:59 collapse

Ah, typo. 1/3 ≠ 0.333…

It is my opinion that repeating decimals cannot properly represent the values we use them for, and I would rather avoid them entirely (kinda like the meme).

Besides, I have never disagreed with the math, just that we go about correcting people poorly. I have used some basic mathematical arguments to try and intimate how basic arithmetic is a limited system, but this has always been about solving the systemic problem of people getting caught by 0.999… = 1. Math proofs won’t add to this conversation, and I think are part of the issue.

Is it possible to have a coversation about math without either fully agreeing or calling the other stupid? Must every argument about even the topic be backed up with proof (a sociological one in this case)? Or did you just want to feel superior?

Leate_Wonceslace@lemmy.dbzer0.com on 01 Jul 02:33 collapse

It is my opinion that repeating decimals cannot

Your opinion is incorrect as a question of definition.

I have never disagreed with the math

You had in the previous paragraph.

Is it possible to have a coversation about math without either fully agreeing or calling the other stupid?

Yes, however the problem is that you are speaking on matters that you are clearly ignorant. This isn’t a question of different axioms where we can show clearly how two models are incompatible but resolve that both are correct in their own contexts; this is a case where you are entirely, irredeemably wrong, and are simply refusing to correct yourself. I am an algebraist understanding how two systems differ and compare is my specialty. We know that infinite decimals are capable of representing real numbers because we do so all the time. There. You’re wrong and I’ve shown it via proof by demonstration. QED.

They are just symbols we use to represent abstract concepts; the same way I can inscribe a “1” to represent 3-2={ {} } I can inscribe “.9~” to do the same. The fact that our convention is occasionally confusing is irrelevant to the question; we could have a system whereby each number gets its own unique glyph when it’s used and it’d still be a valid way to communicate the ideas. The level of weirdness you can do and still have a valid notational convention goes so far beyond the meager oddities you’ve been hung up on here. Don’t believe me? Look up lambda calculus.

lightnsfw@reddthat.com on 28 Jun 13:22 collapse

the problem is it makes my brain hurt

pyre@lemmy.world on 28 Jun 14:48 collapse

honestly that seems to be the only argument from the people who say it’s not equal. at least you’re honest about it.

by the way I’m not a mathematically adept person. I’m interested in math but i only understand the simpler things. which is fine. but i don’t go around arguing with people about advanced mathematics because I personally don’t get it.

the only reason I’m very confident about this issue is that you can see it’s equal with middle- or high-school level math, and that’s somehow still too much for people who are too confident about there being a magical, infinitely small number between 0.999… and 1.

lightnsfw@reddthat.com on 28 Jun 14:54 collapse

to be clear I’m not arguing against you or disagreeing the fraction thing demonstrates what you’re saying. It just really bothers me when I think about it like my brain will not accept it even though it’s right in front of me it’s almost like a physical sensation. I think that’s what cognitive dissonance is. Fortunately in the real world this has literally never come up so I don’t have to engage with it.

pyre@lemmy.world on 28 Jun 18:22 collapse

no, i know and understand what you mean. as i said in my original comment; it’s not intuitive. but if everything in life were intuitive there wouldn’t be mind blowing discoveries and revelations… and what kind of sad life is that?

Leate_Wonceslace@lemmy.dbzer0.com on 28 Jun 05:17 collapse

don’t sipport infinite decimals properly

Please explain this in a way that makes sense to me (I’m an algebraist). I don’t know what it would mean for infinite decimals to be supported “properly” or “improperly”. Furthermore, I’m not aware of any arguments worth taking seriously that don’t use logic, so I’m wondering why that’s a criticism of the notation.

Tlaloc_Temporal@lemmy.ca on 28 Jun 06:26 next collapse

Decimal notation is a number system where fractions are accomodated with more numbers represeting smaller more precise parts. It is an extension of the place value system where very large tallies can be expressed in a much simpler form.

One of the core rules of this system is how to handle values larger than the highest digit, and lower than the smallest. If any place goes above 9, set that place to 0 and increment the next place by 1. If any places goes below 0, increment the place by (10) and decrement the next place by one (this operation uses a non-existent digit, which is also a common sticking point).

This is the decimal system as it is taught originally. One of the consequences of it’s rules is that each digit-wise operation must be performed in order, with a beginning and an end. Thus even getting a repeating decimal is going beyond the system. This is usually taught as special handling, and sometimes as baby’s first limit (each step down results in the same digit, thus it’s that digit all the way down).

The issue happens when digit-wise calculation is applied to infinite decimals. For most operations, it’s fine, but incrementing up can only begin if a digit goes beyong 9, which never happens in the case of 0.999… . Understanding how to resolve this requires ditching the digit-wise method and relearing decimals and a series of terms, and then learning about infinite series. It’s a much more robust and applicable method, but a very different method to what decimals are taught as.

Thus I say that the original bitwise method of decimals has a bug in the case of incrementing infinite sequences. There’s really only one number where this is an issue, but telling people they’re wrong for using the tools as they’ve been taught isn’t helpful. Much better to say that the tool they’re using is limited in this way, then showing the more advanced method.

That’s how we teach Newtonian Gravity and then expand to Relativity. You aren’t wrong for applying newtonian gravity to mercury, but the tool you’re using is limited. All models are wrong, but some are useful.

CertifiedBlackGuy@lemmy.world on 28 Jun 06:37 next collapse

Said a simpler way:

1/3= 0.333…

1/3 + 1/3 = 0.666… = 0.333… + 0.333…

1/3 + 1/3 + 1/3 = 1 = 0.333… + 0.333… + 0.333…

The quirk you mention about infinite decimals not incrementing properly can be seen by adding whole number fractions together.

Leate_Wonceslace@lemmy.dbzer0.com on 28 Jun 12:45 collapse

I can’t help but notice you didn’t answer the question.

each digit-wise operation must be performed in order

I’m sure I don’t know what you mean by digit-wise operation, because my conceptuazation of it renders this statement obviously false. For example, we could apply digit-wise modular addition base 10 to any pair of real numbers and the order we choose to perform this operation in won’t matter. I’m pretty sure you’re also not including standard multiplication and addition in your definition of “digit-wise” because we can construct algorithms that address many different orders of digits, meaning this statement would also then be false. In fact, as I lay here having just woken up, I’m having a difficult time figuring out an operation where the order that you address the digits in actually matters.

Later, you bring up “incrementing” which has no natural definition in a densely populated set. It seems to me that you came up with a function that relies on the notation we’re using (the decimal-increment function, let’s call it) rather than the emergent properties of the objects we’re working with, noticed that the function doesn’t cover the desired domain, and have decided that means the notation is somehow improper. Or maybe you’re saying that the reason it’s improper is because the advanced techniques for interacting with the system are dissimilar from the understanding imparted by the simple techniques.

Tlaloc_Temporal@lemmy.ca on 28 Jun 14:48 collapse

In base 10, if we add 1 and 1, we get the next digit, 2.

In base 2, if we add 1 and 1 there is no 2, thus we increment the next place by 1 getting 10.

We can expand this to numbers with more digits: 111(7) + 1 = 112 = 120 = 200 = 1000

In base 10, with A representing 10 in a single digit: 199 + 1 = 19A = 1A0 = 200

We could do this with larger carryover too: 999 + 111 = AAA = AB0 = B10 = 1110 Different orders are possible here: AAA = 10AA = 10B0 = 1110

The “carry the 1” process only starts when a digit exceeds the existing digits. Thus 192 is not 2Z2, nor is 100 = A0. The whole point of carryover is to keep each digit within the 0-9 range. Furthermore, by only processing individual digits, we can’t start carryover in the middle of a chain. 999 doesn’t carry over to 100-1, and while 0.999 does equal 1 - 0.001, (1-0.001) isn’t a decimal digit. Thus we can’t know if any string of 9s will carry over until we find a digit that is already trying to be greater than 9.

This logic is how basic binary adders work, and some variation of this bitwise logic runs in evey mechanical computer ever made. It works great with integers. It’s when we try to have infinite digits that this method falls apart, and then only in the case of infinite 9s. This is because a carry must start at the smallest digit, and a number with infinite decimals has no smallest digit.

Without changing this logic radically, you can’t fix this flaw. Computers use workarounds to speed up arithmetic functions, like carry-lookahead and carry-save, but they still require the smallest digit to be computed before the result of the operation can be known.

Leate_Wonceslace@lemmy.dbzer0.com on 28 Jun 15:02 collapse

If I remember, I’ll give a formal proof when I have time so long as no one else has done so before me. Simply put, we’re not dealing with floats and there’s algorithms to add infinite decimals together from the ones place down using back-propagation. Disproving my statement is as simple as providing a pair of real numbers where doing this is impossible.

Tlaloc_Temporal@lemmy.ca on 29 Jun 01:35 collapse

Are those algorithms taught to people in school?

Once again, I have no issue with the math. I just think the commonly taught system of decimal arithmetic is flawed at representing that math. This flaw is why people get hung up on 0.999… = 1.

barsoap@lemm.ee on 28 Jun 07:49 collapse

Furthermore, I’m not aware of any arguments worth taking seriously that don’t use logic, so I’m wondering why that’s a criticism of the notation.

If you hear someone shout at a mob “mathematics is witchcraft, therefore, get the pitchforks” I very much recommend taking that argument seriously no matter the logical veracity.

Leate_Wonceslace@lemmy.dbzer0.com on 28 Jun 11:55 collapse

Fair, but that still uses logic, it’s just using false premises. Also, more than the argument what I’d be taking seriously is the threat of imminent violence.

barsoap@lemm.ee on 28 Jun 12:11 collapse

But is it a false premise? It certainly passes Occam’s razor: “They’re witches, they did it” is an eminently simple explanation.

Leate_Wonceslace@lemmy.dbzer0.com on 28 Jun 12:53 collapse

By definition, mathematics isn’t witchcraft (most witches I know are pretty bad at math). Also, I think you need to look more deeply into Occam’s razor.

barsoap@lemm.ee on 28 Jun 13:37 collapse

By definition, all sufficiently advanced mathematics is isomorphic to witchcraft. (*vaguely gestures at numerology as proof*). Also Occam’s razor has never been robust against reductionism: If you are free to reduce “equal explanatory power” to arbitrary small tunnel vision every explanation becomes permissible, and taking, of those, the simplest one probably doesn’t match with the holistic view. Or, differently put: I think you need to look more broadly onto Occam’s razor :)

Leate_Wonceslace@lemmy.dbzer0.com on 28 Jun 05:21 collapse

People generally find it odd and unintuitive that it’s possible to use decimal notation to represent 1 as .9~ and so this particular thing will never go away. When I was in HS I wowed some of my teachers by doing proofs on the subject, and every so often I see it online. This will continue to be an interesting fact for as long as decimal is used as a canonical notation.

someacnt_@lemmy.world on 28 Jun 06:00 collapse

Welp, I see. Still, this is way too much recurting of a pattern.

Etterra@lemmy.world on 27 Jun 23:49 next collapse

Sort of how 0.0000001 = 0

null@slrpnk.net on 28 Jun 00:08 collapse

No, not like that.

LodeMike@lemmy.today on 28 Jun 02:56 next collapse

0.9<overbar.> is literally equal to 1

jonsnothere@beehaw.org on 28 Jun 10:57 next collapse

0.9 is most definitely not equal to 1

mathemachristian@lemm.ee on 28 Jun 11:02 collapse

Hence the overbar. Lemmy should support LaTeX for real though

jonsnothere@beehaw.org on 28 Jun 11:53 collapse

Oh, that’s not even showing as a missing character, to me it just looks like 0.9

At least we agree 0.99… = 1

LodeMike@lemmy.today on 28 Jun 17:29 collapse

Oh lol its rendering as HTML for you.

UnderpantsWeevil@lemmy.world on 28 Jun 14:47 collapse

There’s a Real Analysis proof for it and everything.

Basically boils down to

  • If 0.(9) != 1 then there must be some value between 0.(9) and 1.
  • We know such a number cannot exist, because for any given discrete value (say 0.999…9) there is a number (0.999…99) that is between that discrete value and 0.(9)
  • Therefore, no value exists between 0.(9) and 1.
  • So 0.(9) = 1
LodeMike@lemmy.today on 28 Jun 17:29 next collapse

Even simpler: 1 = 3 * 1/3

1/3 =0.333333…

1/3 + 1/3 + 1/3 = 0.99999999… = 1

UnderpantsWeevil@lemmy.world on 28 Jun 20:20 collapse

Even simpler

0.99999999… = 1

But you’re just restating the premise here. You haven’t proven the two are equal.

1/3 =0.333333…

This step

1/3 + 1/3 + 1/3 = 0.99999999…

And this step

Aren’t well-defined. You’re relying on division short-hand rather than a real proof.

SuperSaiyanSwag@lemmy.zip on 28 Jun 21:31 collapse

ELI5

UnderpantsWeevil@lemmy.world on 28 Jun 23:03 collapse

Mostly boils down to the pedantry of explaining why 1/3 = 0.(3) and what 0.(3) actually means.

beejboytyson@lemmy.world on 29 Jun 04:59 next collapse

That actually makes sense, thank you.

Swedneck@discuss.tchncs.de on 01 Jul 18:20 collapse

the explanation (not proof tbf) that actually satisfies my brain is that we’re dealing with infinite repeating digits here, which is what allows something that on the surface doesn’t make sense to actually be true.

UnderpantsWeevil@lemmy.world on 01 Jul 18:57 collapse

Infinite repeating digits produce what is understood as a Limit. And Limits are fundamental to proof-based mathematics, when your goal is to demonstrate an infinite sum or series has a finite total.

[deleted] on 28 Jun 03:31 next collapse

.

uis@lemm.ee on 28 Jun 09:08 next collapse

0.(9)=0.(3)*3=1/3*3=1

fin@sh.itjust.works on 29 Jun 01:14 collapse

Beautiful. Is that true or am I tricked?

cordlesslamp@lemmy.today on 29 Jun 03:47 collapse

0.3 is definitely NOT equal 1/3.

fin@sh.itjust.works on 29 Jun 08:22 collapse

I guess 0.(3) means infinite 0.33333…. which is equal to 1/3

uis@lemm.ee on 29 Jun 10:16 collapse

Yes, 0.(3) means repeating 3 in fractional part.

And there is entire wiki page on 0.(9)

dylanTheDeveloper@lemmy.world on 28 Jun 10:59 next collapse

I wish computers could calculate infinity

ILikeBoobies@lemmy.ca on 28 Jun 15:49 next collapse

As long as you have it forget the previous digit, you can bring up a new digit infinitely

jerkface@lemmy.ca on 29 Jun 03:47 collapse

Computers can calculate infinite series as well as anyone else

bluewing@lemm.ee on 28 Jun 12:59 next collapse

Meh, close enough.

LordCrom@lemmy.world on 29 Jun 00:05 next collapse

I can honestly say I learned something from the comment section. I was always taught the .9 repeating was not equal to 1 but separated by imaginary i … Or infinitely close to 1 without becoming 1.

cumskin_genocide@lemm.ee on 29 Jun 03:12 next collapse

The only sources I trust are the ones that come from my dreams

Tomorrow_Farewell@hexbear.net on 29 Jun 11:53 next collapse

The decimals ‘0.999…’ and ‘1’ refer to the real numbers that are equivalence classes of Cauchy sequences of rational numbers (0.9, 0.99, 0.999,…) and (1, 1, 1,…) with respect to the relation R: (aRb) <=> (lim(a_n-b_n) as n->inf, where a_n and b_n are the nth elements of sequences a and b, respectively).

For a = (1, 1, 1,…) and b = (0.9, 0.99, 0.999,…) we have lim(a_n-b_n) as n->inf = lim(1-sum(9/10^k) for k from 1 to n) as n->inf = lim(1/10^n) as n->inf = 0. That means that (1, 1, 1,…)R(0.9, 0.99, 0.999,…), i.e. that these sequences belong to the same equivalence class of Cauchy sequences of rational numbers with respect to R. In other words, the decimals ‘0.999…’ and ‘1’ refer to the same real number. QED.

Unbeelievable@beehaw.org on 02 Jul 17:19 collapse

The way I see it is the difference between equal numbers is zero.

The difference between 0.999… and 1 is 0.000…, and since the nines don’t end, the zeros don’t end, so the difference is just zero.

Meaning 0.999… = 1