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
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(a-b) = 0, therefore you cannot divide both sides by it.
wouldn’t bb be b\^2?
bb is shorter to write
So where exactly was the catch in this question?
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.
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…
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 ;)
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 :)
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 \^\^.
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