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### What is the error in this 2=1 "proof"?

Posted: **Mon Feb 08, 2010 7:40 am UTC**

by **Rilian**

I heard about this thing where you take an equilateral triangle with the sides of length one and you fold two of the sides down on to the 3rd side so that there are two little triangles with side length 0.5 and then you do that again and again until infinity and that makes it seem like 2=1, but you're actually dividing by infinity and then multiplying by infinity, so I think the error comes in there because infinity/infinity is an indeterminate form, but I don't know exactly what the problem is.

### Re: What is the error in this 2=1 "proof"?

Posted: **Mon Feb 08, 2010 7:47 am UTC**

by **Alpha Omicron**

Rilian wrote:I heard about this thing where you take an equilateral triangle with the sides of length one and you fold two of the sides down on to the 3rd side so that there are two little triangles with side length 0.5

I'm not sure exactly what is meant here. Do you mean to draw the top point down so it touches the base?

and then you do that again and again until infinity and that makes it seem like 2=1

Why would repeating this process seem to imply that 2=1?

but you're actually dividing by infinity and then multiplying by infinity,

No you're not. And you're abusing the term "infinity" here. In the usual sense, "infinity" is not a number upon which you can perform arithmetic operations.

so I think the error comes in there because infinity/infinity is an indeterminate form, but I don't know exactly what the problem is.

I think the problem is that you don't understand what you mean by infinity, what an indeterminate form is, or what exactly this geometric process entails.

### Re: What is the error in this 2=1 "proof"?

Posted: **Mon Feb 08, 2010 7:51 am UTC**

by **Rilian**

Do you mean to draw the top point down so it touches the base?

Yes. And if you do that "forever" it becomes a flat line.

Saying divide by infinity and multiply by infinity is shorthand for saying divide by a variable that you let approach infinity and multiply by a variable that you let approach infinity.

Infinity/infinity is an indeterminate form.

Have you not had calculus?

### Re: What is the error in this 2=1 "proof"?

Posted: **Mon Feb 08, 2010 7:59 am UTC**

by **Syrin**

Rilian wrote:I heard about this thing where you take an equilateral triangle with the sides of length one and you fold two of the sides down on to the 3rd side so that there are two little triangles with side length 0.5 and then you do that again and again until infinity and that makes it seem like 2=1, but you're actually dividing by infinity and then multiplying by infinity, so I think the error comes in there because infinity/infinity is an indeterminate form, but I don't know exactly what the problem is.

I am rather confused by your one-sentence explanation. I think it would be more enlightening if you write out your thought processes in detail - not to mention that you would likely find the error yourself in this way.

### Re: What is the error in this 2=1 "proof"?

Posted: **Mon Feb 08, 2010 8:12 am UTC**

by **antonfire**

We've

covered this

before.

### Re: What is the error in this 2=1 "proof"?

Posted: **Mon Feb 08, 2010 8:14 am UTC**

by **Syrin**

Ah, I see now what he meant by "folding".

### Re: What is the error in this 2=1 "proof"?

Posted: **Mon Feb 08, 2010 8:15 am UTC**

by **jaap**

The length of the limit of the curves is not the limit of the lengths of the curves.

Length of curves in general is defined through integrals, so you're essentially swapping limits and integrals which you cannot expect to give the same answer without some justification.

Limits curve of a sequence of curves can have properties that none of the individual curves have. For example:

f

_{n}(x) = x

^{n}, defined on [0,1].

Each f

_{n} is continuous. And yet its limit is not: The limit is function f

_{inf} = lim f

_{n}, which is 1 at x=1 and 0 elsewhere.

For a weirder example,

spacefilling curves. Each iteration is a single line. The limit is a section of the plane.

There is absolutely no reason to assume without proof that if the length of each curve in a sequence converges that it converges to the length of the limit curve.

### Re: What is the error in this 2=1 "proof"?

Posted: **Mon Feb 08, 2010 10:00 am UTC**

by **DavCrav**

jaap wrote:There is absolutely no reason to assume without proof that if the length of each curve in a sequence converges that it converges to the length of the limit curve.

Indeed, it's easy to see with a staircase that goes up 10m and across 10m. No matter how small the steps, the total length is 20m, but the limit is a straight line, whose distance is (about) 14.1m.

### Re: What is the error in this 2=1 "proof"?

Posted: **Mon Feb 08, 2010 5:48 pm UTC**

by **kaimason1**

This is not a limit. The steps don't limit to a line, but instead an infinite number of infinitesimally small steps. Your problem is not that the infinities are incomputible, but rather that you assume they can be ignored. They still exist, so the steps do limit to a length of two. They still aren't a straight line, though.

### Re: What is the error in this 2=1 "proof"?

Posted: **Mon Feb 08, 2010 5:54 pm UTC**

by **NathanielJ**

kaimason1 wrote:This is not a limit. The steps don't limit to a line, but instead an infinite number of infinitesimally small steps. Your problem is not that the infinities are incomputible, but rather that you assume they can be ignored. They still exist, so the steps do limit to a length of two. They still aren't a straight line, though.

So our way to explain why hand-wavey proofs don't work is to combat it with more hand-waveyness?

The limit very much exists in any reasonably-chosen topology. The reason the proof fails is simply that the function that maps a curve to its length is not continuous, as has already been stated.

### Re: What is the error in this 2=1 "proof"?

Posted: **Mon Feb 08, 2010 10:20 pm UTC**

by **TNorthover**

NathanielJ wrote:The limit very much exists in any reasonably-chosen topology. The reason the proof fails is simply that the function that maps a curve to its length is not continuous, as has already been stated.

In some topologies the limit exists, in some topologies length is continuous. The bare problem gives us no particular reason to favour one over the other, but once a choice is made the difficulty disappears.

### Re: What is the error in this 2=1 "proof"?

Posted: **Tue Feb 09, 2010 1:46 am UTC**

by **Kurushimi**

Wait, I'm still confused. For one thing, I have no idea what the opening poster was describing or how it relates to 2 = 1, but from the links to other forum posts posted by antonfire it appears to relate to relating a step-taking function to a straight line. Why exactly is the limit of the step-like function not equivalent to a straight line? I mean, it obviously looks more and more like a straight line the more steps we add. Also, this seems to be a popular problem. Does it have a name I can google?

### Re: What is the error in this 2=1 "proof"?

Posted: **Tue Feb 09, 2010 1:56 am UTC**

by **Syrin**

If you're curious, read the threads. I mean, you had them open, didn't you?

### Re: What is the error in this 2=1 "proof"?

Posted: **Tue Feb 09, 2010 3:57 am UTC**

by **PM 2Ring**

Kurushimi wrote:Why exactly is the limit of the step-like function not equivalent to a straight line? I mean, it obviously looks more and more like a straight line the more steps we add.

If you zoom in though, it still looks step-like, no matter how many times you subdivide the steps. Compare that with what happens when you zoom into a smooth curve, eg a circle, which does become more line-like the further you zoom in.

So we need to look at the local behaviour of the curve to see how the arclength behaves.

### Re: What is the error in this 2=1 "proof"?

Posted: **Tue Feb 09, 2010 9:29 am UTC**

by **DavCrav**

PM 2Ring wrote:Kurushimi wrote:Why exactly is the limit of the step-like function not equivalent to a straight line? I mean, it obviously looks more and more like a straight line the more steps we add.

If you zoom in though, it still looks step-like, no matter how many times you subdivide the steps. Compare that with what happens when you zoom into a smooth curve, eg a circle, which does become more line-like the further you zoom in.

So we need to look at the local behaviour of the curve to see how the arclength behaves.

Sure, this is the problem. It acts like a fractal, and we know that these are screwy. But does converge to a line in any reasonable sense. (This isn't against you, but against someone further up.) It might be better to rotate it so that we don't have lines of infinite gradiant cropping up (i.e., rotate the curve so all lines are diagonals). Then the sequences of lines converges using, say, the 'max-norm', which measures maximal distance from the putative limit, and the 'total variation norm', where we take

[math]\lim_{i\to\infty}\int |f_i(x)-f(x)|\,\mathrm{d}x.[/math]

(These aren't the names of the norms, but I can't remember what they really are.) I think these two can give different answers for stupid curves.

### Re: What is the error in this 2=1 "proof"?

Posted: **Tue Feb 09, 2010 9:45 am UTC**

by **TNorthover**

DavCrav wrote:Sure, this is the problem. It acts like a fractal, and we know that these are screwy. But does converge to a line in any reasonable sense. (This isn't against you, but against someone further up.)

That would probably be me. Norms which take derivatives into account are frequently useful (existence theorems for differential equations are probably the simplest example). Step curves would not converge under such a norm,

### Re: What is the error in this 2=1 "proof"?

Posted: **Tue Feb 09, 2010 10:07 am UTC**

by **DavCrav**

Sure. If you are doing certain subjects, you might not like this. But then, since it isn't smooth things are going to get horrible. The same is probably true about something like [imath]\sin(n\pi x)/n[/imath] between 0 and 1, right? (I'm an algebraist so not really used to all this stuff.) This latter example at least solves the non-differentiability of the sawtooth function. Wolfram alpha thinks its arclength is about 2.3 for all n.

And it wasn't against you, but the person a few posts up who just said "This is not a limit. The steps don't limit to a line, but instead an infinite number of infinitesimally small steps." and then rambled on a bit more.

### Re: What is the error in this 2=1 "proof"?

Posted: **Tue Feb 09, 2010 1:51 pm UTC**

by **tastelikecoke**

- amiflo.png (4.15 KiB) Viewed 3092 times

the 2=1 doesn't add up. I mean the geometry and algebra don't add up.

### Re: What is the error in this 2=1 "proof"?

Posted: **Tue Feb 09, 2010 3:33 pm UTC**

by **Shokk**

tastelikecoke wrote:amiflo.png

the 2=1 doesn't add up. I mean the geometry and algebra don't add up.

Well, i think the indication we're supposed to get is that the final straight line you drew is actually a zigzag of infinitely small triangles right next to each other. Doesn't actually mean 2=1, in any case, I think it's about the same issue as the"Infinitesimally small Staircase" thing, which I'm pretty sure there was already a thread about. Correct me if I misunderstand.

### Re: What is the error in this 2=1 "proof"?

Posted: **Tue Feb 09, 2010 5:09 pm UTC**

by **Cryft**

First off:

It is my understanding that if something is "infinitesimally small" then it does not exist. See any .(9)=1 proof for that.

Secondly, I think this problem is being overthought. The limit as n squares in the divided square approaches infinity is still 2 in the same sense that the limit as x approaches 0 of 3x/x is still 3. When x = 3, we have a different situation. Similarly, when n "reaches" infinity, we would have a different situation, but all along the way the total distance traveled is still 2.

### Re: What is the error in this 2=1 "proof"?

Posted: **Tue Feb 09, 2010 5:57 pm UTC**

by **AllSaintsDay**

Hey, look, it's the warning my differential geometry professor gave us the first day of class! Now if only I could find my notes to see what exactly it was he said about it.

### Re: What is the error in this 2=1 "proof"?

Posted: **Wed Feb 10, 2010 4:16 am UTC**

by **Kurushimi**

Cryft wrote:It is my understanding that if something is "infinitesimally small" then it does not exist. See any .(9)=1 proof for that.

I think that understanding is wrong. How can dy/dx exist if neither dy or dx exist?

### Re: What is the error in this 2=1 "proof"?

Posted: **Wed Feb 10, 2010 5:03 am UTC**

by **AllSaintsDay**

Kurushimi wrote:Cryft wrote:It is my understanding that if something is "infinitesimally small" then it does not exist. See any .(9)=1 proof for that.

I think that understanding is wrong. How can dy/dx exist if neither dy or dx exist?

In standard calculus definitions, they don't. dy/dx doesn't get its meaning from "dy" "/" and "dx" anymore than i gets its meaning from combining the meanings of the dot and the line.