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# 1 = 2 Proved

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### #41Mr H

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Posted 09 May 2006 - 07:30 PM

QUOTE(thegift_guitar @ May 9 2006, 04:12 PM)

Ok please, for once and for all, ENLIGHTEN US!

To be quite honest I can't remember the exact explanation! !!

We were shown this at uni but have since forgotten what the lecturer said. I've come across this 'proof' a few times since and its driven me nuts trying to figure it out each time!

All I remember is that it is to do with differentiating an infinite series, 'x lots of x', which is dodgy unless it converges.

Apologies that this is not really an adequate explanation but at least I am now not alone in being driven nuts!

P.S. Some of you here are at uni aren't you? Maybe you could ask one of your lecturers?

H tends 2 infinity

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Never argue with an idiot. They drag you down to their level then beat you with experience.

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Posted 10 May 2006 - 04:05 PM

yay........

### #43andymckenzie

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Posted 17 May 2006 - 05:40 PM

I've got a far simpler one:

a=b

multiply both sides by a:

a2 = ab

then add a2 - 2ab to both sides:

a2 + a2 - 2ab = ab + a2 - 2ab

This can be simplified to:

2(a2 - ab) = a2 - ab

Now divide by a2 - ab to get:

2 = 1

### #44The Wedge Effect

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Posted 17 May 2006 - 05:46 PM

There's a flaw in your working. You are implying that a=b in the first line. If that was the case, wouldn't it be logical that a=a? Thus your working for the rest doesn't work out.

### #45Nathan

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Posted 17 May 2006 - 06:18 PM

why would you jus randomly add a2 - 2ab though

### #46The Wedge Effect

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Posted 17 May 2006 - 06:28 PM

If you look at the second line, the expression equals zero when you rearrange to get a²-ab. That's the only reason I can think of, doesn't seem right though. But I still think the expression is wrong as a cannot be equal to b unless b is the same value as a, which'd make it a=a. That completely contradicts what he's trying to say...

### #47George

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Posted 17 May 2006 - 06:31 PM

QUOTE(andymckenzie @ May 17 2006, 06:40 PM)

Now divide by a2 - ab to get:

The problem lies here

### #48dfx

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Posted 18 May 2006 - 06:24 PM

I've seen that alot and it uses euclids first and third theorems to manipulate the equation. It's all very well for a and b HOWEVER the moment you try it out with real numbers you end up dividing by zero at one point and that, my friend, screws you over.

### #49Daron Malakian

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Posted 24 May 2006 - 05:17 PM

a^2 = ab

this equation is contrived

### #50isthis_the_way_to_amarillo?

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Posted 26 May 2006 - 10:46 AM

The problem is in convergence. you are saying that X^2=x.x ie x lots of x
But if you look at this then it only applies for positive integers of x. It does not work for other real numbers. If you then graph this function of x^2 using this definition then it is not a continuous function. You can't differentiate a continuous function and so the mistake is made in trying to differentiate x^2.

Wallllaaaaa HOMMMIEEESSSS!

### #51AM4R

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Posted 26 May 2006 - 12:22 PM

::::::/\M/\R::::::

### #52waitingforan_alibi

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Posted 24 July 2006 - 05:53 PM

QUOTE(andymckenzie @ May 17 2006, 06:40 PM)

a=b

multiply both sides by a:

a2 = ab

then add a2 - 2ab to both sides:

a2 + a2 - 2ab = ab + a2 - 2ab

This can be simplified to:

2(a2 - ab) = a2 - ab

Now divide by a2 - ab to get:

2 = 1

a2 - ab = 0 since
a=b
so a2=ab

so division by zero is the problem with this one too (obvious if you look / even more obvious if you read fermat's last theorem by simon singh )

### #53Dave

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Posted 24 July 2006 - 07:09 PM

i knew i should of read that book

seriously i actually have tht in the house

If i am not here i am somewhere else

### #54John

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Posted 24 July 2006 - 07:49 PM

If you subsitute x for a number, it doesnt seem to work, for me at least

### #55John

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Posted 29 July 2006 - 02:33 PM

I retract what i said it does work, must have been tired lol

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