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101 Nov 04, 2004 at 18:50

here’s an interesting proof. see if you can see where it breaks down:

       a = b     given
aa = ab     multiply both sides by a
aa - bb = ab - bb  subtract (b * b) from both sides
(a + b)(a - b) = b(a - b)  factor both sides
a + b = b     cancel out common factors
b + b = b     substitute b for a (from line 1)
2b = b     combine the left side
2 = 1     divide both sides by b


the above was taken from Paul Bourke’s site

#### 10 Replies

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101 Nov 04, 2004 at 18:59

(a-b) = 0, therefore you cannot divide both sides by it.

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101 Nov 10, 2004 at 04:32

wouldn’t bb be b\^2?

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101 Nov 10, 2004 at 04:48

bb is shorter to write

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102 Nov 10, 2004 at 05:12

So where exactly was the catch in this question?

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101 Nov 10, 2004 at 16:00

He divided by zero, so answers go out the roof at that point. I think there’s some more odd algebra in it, but that’s the first I found.

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102 Nov 10, 2004 at 16:11

Yes, i know what the error in the transformation is but why is he giving it here at DevMaster? I have it in my old 6th grade mathbooks…

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101 Nov 10, 2004 at 17:00

Bah, there are a lot more proofs of 2=1 that are actually a *lot* harder to disprove =)
I’ll see if I can remember/dig-up a few just for kicks ;)

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101 Nov 11, 2004 at 07:36

It is just for the fun here. Just to keep the gray matter working but if u have better examples that are harder to disprove please go ahead and place them here.
Or other strange things that are not correct but appear to be correct i just love them :)

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101 Nov 19, 2004 at 13:05

Another example using complex numbers:

1 = sqrt(1) = sqrt(1\^2) = sqrt((-1)\^2) = sqrt(-1) * sqrt(-1) = i * i = i\^2 = -1

So, always be careful with complex numbers and roots in general, or sth like that could happen \^\^.
Ah and please forgive me the misuse of the equal-sign \^\^.

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101 Jan 21, 2005 at 20:52

Something similiar to the original post, but also flawed by the divide by zero, is this:

2(1-1) = (2-2)
2 = (2-2)/(1-1)
2 = 0


:)