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 Post subject: A proof of 2=1 that I just saw for the first time todayPosted: Sun Jul 17, 2011 10:46 am

Joined: Thu Dec 02, 2004 12:09 pm
Location: Missouri
I assume a math puzzle would be considered a Non-Twisty puzzle so here goes.

I was reading Andreas Nortmann's thread 3x3x3's - NOT all of them and that had me looking through the old threads he had posted in back in 2004. It was there that I came across this post by Sandy.

Sandy wrote:
Here's a few stolen from the net:

x^2 = x + x + . . . + x (With x terms on the right, x^2 = x * x.)
2 x = 1 + 1 + . . . + 1 (Take the derivative of each side.)
2 x = x (We added up the x 1's on the right.)
2 = 1 (Divide both sides by x.)
I think I found where Sandy stole it from.

But I can't seem to find the explanation of the error. I believe it's something very closely related to issue discuessed here.

http://en.wikipedia.org/wiki/Mathematical_fallacy#Variable_ambiguity

In the first step you are taking the derivative with respect to x on boths sides but you are treading it as if there is a "constant" x terms on the right side. The amount of terms is actually a variable. At least I think that is the error.

Any better explanation?

Thanks,
Carl

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 Post subject: Re: A proof of 2=1 that I just saw for the first time todayPosted: Sun Jul 17, 2011 10:51 am

Joined: Mon Oct 18, 2010 10:48 am
X=0?

You can go anywhere from there in fewer steps.
3x=x
3=1.

Once you ascertain that x=0, you can no longer divide by it. 2x=x only where x=0, so to divide by x is to divide by zero.

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 Post subject: Re: A proof of 2=1 that I just saw for the first time todayPosted: Sun Jul 17, 2011 11:05 am

Joined: Wed Oct 31, 2007 10:55 am
Location: Hong Kong, China

[
x^2 = x*x
By product rule : d/dx(x*x) = 1*x + x*1 = 2x

In the steps for proving 1=2, only the leading "x" was differentiated, which caused all the problems.
]

Sounds correct to me, but I'm a lot less experienced and knowledgeable on mathematics than a lot of people in here .

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 Post subject: Re: A proof of 2=1 that I just saw for the first time todayPosted: Sun Jul 17, 2011 11:09 am

Joined: Thu Dec 02, 2004 12:09 pm
Location: Missouri
NType3 wrote:
X=0?
I agree division by 0 is at the root of many proofs like this but this particular one doesn't contain any division... or if it does its much better hidden then most. Maybe I need to go back to definition of a derivative... its been quite a while since I took Calc I. I also don't see anything that reqires x to be zero.

Carl

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 Post subject: Re: A proof of 2=1 that I just saw for the first time todayPosted: Sun Jul 17, 2011 11:11 am

Joined: Mon Oct 18, 2010 10:48 am
The laststep is divide both sides by x, which can't be done.

The only value for x for which twice itself is itself (2x=x) is zero.

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 Post subject: Re: A proof of 2=1 that I just saw for the first time todayPosted: Sun Jul 17, 2011 11:18 am

Joined: Thu Dec 02, 2004 12:09 pm
Location: Missouri
NType3 wrote:
The laststep is divide both sides by x, which can't be done.

The only value for x for which twice itself is itself (2x=x) is zero.
I'd argue the error however is made before the last step. Let's write this as:

x^2 = x + x + . . . + x (With x terms on the right, x^2 = x * x.)
2 x ≠ 1 + 1 + . . . + 1 (Take the derivative of each side.)
2 x ≠ x (We added up the x 1's on the right.)
2 ≠ 1 (Divide both sides by x.)

What is the error between steps 1 and 2 that turns the = into a ≠?

Carl

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 Post subject: Re: A proof of 2=1 that I just saw for the first time todayPosted: Sun Jul 17, 2011 11:24 am

Joined: Thu Dec 02, 2004 12:09 pm
Location: Missouri
Marco wrote:
Sounds correct to me, but I'm a lot less experienced and knowledgeable on mathematics than a lot of people in here .
Sounds good to me and I think that agrees with what I said in the first post. You are taking the derivative of the "x" terms but you aren't taking the derivative of the fact they occure "x" times.

Carl

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 Post subject: Re: A proof of 2=1 that I just saw for the first time todayPosted: Sun Jul 17, 2011 11:35 am

Joined: Mon Oct 18, 2010 10:48 am
Regardless of whether or not the calculus is correct, there is an error from 2x=x to 2=1. I do not deny the potential existence of more than one error in the problem, but to go from 2x=x to 2=1 is a fallacy in logic.

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 Post subject: Re: A proof of 2=1 that I just saw for the first time todayPosted: Sun Jul 17, 2011 11:44 am

Joined: Mon Mar 30, 2009 5:13 pm
NType3 wrote:
Regardless of whether or not the calculus is correct, there is an error from 2x=x to 2=1. I do not deny the potential existence of more than one error in the problem, but to go from 2x=x to 2=1 is a fallacy in logic.

I'm sorry to say that's complete rubbish, there's nothing wrong to divide both sides of an equation by x, provided that x is not 0. But I don't think that's the case here, and the correct explanation is given above. It's easy to see that there's no equality in the second equation, so the error is in the first step.

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Last edited by KelvinS on Sun Jul 17, 2011 11:50 am, edited 1 time in total.

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 Post subject: Re: A proof of 2=1 that I just saw for the first time todayPosted: Sun Jul 17, 2011 11:50 am

Joined: Fri Feb 08, 2008 1:47 am
Location: near Utrecht, Netherlands
The problem is in taking the derivative. Writing x^2=x+x+x...+x+x+x is only valid for one certain (integer) x so we must constrain the domain of the function to that specific x. However, then x is not a limit point of the domain and we're not allowed to take the derivative in that point x.

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 Post subject: Re: A proof of 2=1 that I just saw for the first time todayPosted: Sun Jul 17, 2011 11:51 am

Joined: Thu Dec 02, 2004 12:09 pm
Location: Missouri
NType3 wrote:
but to go from 2x=x to 2=1 is a fallacy in logic.
Agreed...

But its the error with the calculus that interests me more. Going back to this...

x^2 = x + x + . . . + x (With x terms on the right, x^2 = x * x.)

We could take that one step further and get this:

x^2 = (1+1+ . . . +1) + (1+1+ . . . +1) + . . . + (1+1+ . . . +1) where each (1+1+ . . . +1) is x terms long too.

Now if you take the derivative of each side you get:

2x = (0+0+ . . . + 0) + (0+0+ . . . + 0) + . . . + (0+0+ . . . + 0)
2x = 0 (adding up the x*x 0's on the right side)

Carl

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Last edited by wwwmwww on Sun Jul 17, 2011 5:31 pm, edited 1 time in total.

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 Post subject: Re: A proof of 2=1 that I just saw for the first time todayPosted: Sun Jul 17, 2011 11:52 am

Joined: Mon Mar 30, 2009 5:13 pm
Spot on Tom (and Carl)!

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 Post subject: Re: A proof of 2=1 that I just saw for the first time todayPosted: Sun Jul 17, 2011 11:58 am

Joined: Mon Oct 18, 2010 10:48 am
Kelvin Stott wrote:
NType3 wrote:
Regardless of whether or not the calculus is correct, there is an error from 2x=x to 2=1. I do not deny the potential existence of more than one error in the problem, but to go from 2x=x to 2=1 is a fallacy in logic.

I'm sorry to say that's complete rubbish, there's nothing wrong to divide both sides of an equation by x, provided that x is not 0. But I don't think that's the case here, and the correct explanation is given above. It's easy to see that there's no equality in the second equation, so the error is in the first step.

2x=x. What values can X equal other than zero?

If there's a problem with the derivative, the proof is incorrect. If there isn't, then the problem lies with 2x=x because it *has* to equal zero. Dividing by x to get 2=1 is dividing by zero.

So, if the calculus is correct, the proof is incorrect. If the calculus isn't correct, the proof isn't correct. Either way, the proof makes no sense.

Kelvin Stott wrote:
provided that x is not 0.

See above. 2x=x. x=?

Edit: Seems like there is a problem with the derivative?

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 Post subject: Re: A proof of 2=1 that I just saw for the first time todayPosted: Sun Jul 17, 2011 12:03 pm

Joined: Mon Mar 30, 2009 5:13 pm
As I said, the problem is NOT caused by dividing both sides by x. The problem is that the equation 2x = x is simply wrong, given that x is supposed to be a general variable, not necessarily 0. And the bad equation is caused by the logic in step 1, as Tom and Carl explained. That's the only error in the logic, and there's nothing at all wrong with the last step!

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 Post subject: Re: A proof of 2=1 that I just saw for the first time todayPosted: Sun Jul 17, 2011 12:08 pm

Joined: Fri Feb 08, 2008 1:47 am
Location: near Utrecht, Netherlands
The step of taking the derivative is bonkers. When you're saying x^2=x+x+x+x you're implicitly saying x is some fixed integer value. This means we may only consider the equality when x is equal to that value, but you can't take a derivative when you've only got one point to look at. Taking the derivative is "calculating the slope" but if you have a single point, it does not define a slope - mathematical explanation below.

What does f'(a)=b really mean? It means that the limit of f(a)-f(x) divided by a-x as x goes to a equals b.
But what does it mean mean that the limit of g(x) is b as x goes to a? It means that if we're given a "tolerance" epsilon, we can find a second "tolerance" delta such that if a value x in the domain of g is closer than delta to a, g(x) will be closer than epsilon to b.

The problem with evaluating the derivative f(a) for a function whose domain contains only a, is that f(a)-f(x)/a-x has an empty domain. Any point x other than a is not in it because f(x) is not defined, and a is not in it because you can divide by a-a=0.
So that means that regardless of the delta we chose, there are no points closer than delta to a in the domain of the function. So we can pick the limit to be any arbitrary value we want since "if a value x in the domain of g is closer than delta to a, g(x) will be closer than epsilon to b" is now a trivial statement and will always be true.

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 Post subject: Re: A proof of 2=1 that I just saw for the first time todayPosted: Sun Jul 17, 2011 12:14 pm

Joined: Mon Mar 30, 2009 5:13 pm
Very well explained, Tom!

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 Post subject: Re: A proof of 2=1 that I just saw for the first time todayPosted: Sun Jul 17, 2011 12:17 pm

Joined: Sun Jul 08, 2007 7:15 pm
There are two problems. The second is that the last step is absolutely dividing by zero. We can solve the equation 2x = x for x and the solution is x=0. So dividing by x is something we cannot do.

Of course if that were the only problem then this would still have shaken math to its foundations because if that were the only problem we would have just proven that every number is equal to zero. That's no better for math than saying 1=2. (In fact it is exactly the same.)

The first problem is the more interesting one because it is spread out over two lines.
x^2= x+x+...+x (x times) is absolutely true if the value of x is fixed. Pick any value of x and it will work.
x=0 gives you x^2= 0
x=1 gives you x^2= x
x=2 gives you x^2= x + x
x=3 gives you x^2= x + x + x
x=pi gives you x^2= x + x + x + (pi-3)x

For any fixed value of x it is true that x^2= x+x+...+x (x times), but what should be apparent from the above is that we simply cannot break up the function f(x)=x^2 into a bunch of different functions of f(x)=x because how many functions we'd need is in constant flux.

x^2= x+x+...+x (x times) only makes sense if we're talking about these things as numbers. If we're talking about them as functions then it is wrong, wrong, wrong.

And then we derive. We treat these things, that absolutely have to be fixed values as if they were functions. That's the problem. Dividing by zero later on is just a cherry on top.

And I see that other people have addressed this.

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 Post subject: Re: A proof of 2=1 that I just saw for the first time todayPosted: Sun Jul 17, 2011 1:26 pm

Joined: Fri Feb 08, 2008 1:47 am
Location: near Utrecht, Netherlands
Here's my take on 2=1.

Let V be a non-empty set.
Let + be a binary operator on V, such that a+b=b+a for all a,b in V.
Let * be a binary operator on V, such that a*b=b*a for all a,b in V and a*(b+c)=a*b+a*c.
Let 0 be the element of V such that 0*a=0 for any a in V.
Let 1 be the element of V such that 1*a=a for any a in V.
Let 2 = 1 + 1.

Under these axioms, assuming 2=1 does not lead to any contradictions. Why?

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 Post subject: Re: A proof of 2=1 that I just saw for the first time todayPosted: Sun Jul 17, 2011 1:48 pm

Joined: Sun Jul 08, 2007 7:15 pm
I'm a little tired so I might not be thinking straight, but I think it's because it is not claimed that 0 and 1 are distinct there are any elements of the set a and b such that a !=b.

If I plug in 0=1=2, all is well.

Though actually I'm used to 0 being the element such that 0+a=a, the additive identity, which can be used to demonstrate that 0*a= 0. It looks to me like you can reverse that process (though I'm not sure, being tired) so it's the same axioms I'm used to, just written in a different way.

Anyway, it seems to me like 1=2 breaks down when you get into things like trichotmy. Any time you haven't specified that there is more than one element 1=2 is just fine. If 1+1=2=1 then 1=0. Which works, so long as everything else is zero. It breaks down if you have some non-zero element of V, say q, because then 1*q = 0*q = 0 != q and then we have a contradiction. But as long as every element of V is zero, there is absolutely no problem with 1 being zero, and if 1 is zero then 1 absolutely equals 2 (because in that case 2 = 1+1 = 0+0 = 0 = 1).

Or something like that.

Am I making any sense?

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