The (FALSE) proof that 1 = 2 goes as follows.
Let a = 1 and b = 1.
Then
a = b
If we square both sides of that equation, we get
a^2 = b^2
and if we multiply both sides of that same equation by a, we get
a^2 = ab
So we have these two equations:
a^2 = b^2
a^2 = ab
Let's rearrange both of these, to:
a^2 - b^2 = 0
a^2 - ab = 0
Since both of these equate to 0, they equate to each other.
a^2 - b^2 = a^2 - ab
Factor both sides, to obtain
(a - b) (a + b) = a(a - b)
Now that we have (a - b) on both sides, they effectively cancel out because we divide both sides by (a - b). As a result, we get
a + b = a
We initially assigned a = b = 1; therefore
1 + 1 = 1, and
2 = 1
Here, the value of "a' remained consistent throughout.
How can you divide both sides by zero which is (a-b) and get numbers appearing after that? Any better versions that can trick people?
2006-12-22
13:06:13
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8 answers
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asked by
Anonymous
in
Mathematics