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I would like some input into this question I need to solve.

From my logic: $a = 1$. we get this from $1=2$ ($a=2a$). It appears $b = 0$. because $a = a + b$ ($1=1+0$).

so in the original statement $a = b$ - meaning both $a$ and $b$ are $0$. So all operations would be zero.

My answer would then be : d) Multiplication by zero is meaningless.

Is this correct?

Let a = b
Then a * b = b^2
And, a * a – a * b = a^2 – b^2
So, a * (a – b) = (a + b) * (a – b)
Or, a = a + b
Or a = 2a
So, 1 = 2


a) 1 and 2 are close so what is the big deal?
b) The Laplace transform of 1 and 2 are identical
c) Division by zero is meaningless.
d) Multiplication by zero is meaningless.
e) The reasoning is absurd because no units are specified.
Martin Sleziak
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JL.
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2 Answers2

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Multiplication by zero is not meaningless. Division by zero is. The error occurred when going from $$a(a-b)=(a+b)(a-b)$$ to $$a=a+b,$$ since we can only conclude this if $a-b\ne 0,$ but by assumption $a=b,$ so $a-b=0$. So, while it's certainly true that $a(a-b)=(a+b)(a-b)$--after all, $0=0$)--we can't conclude that $a=a+b$.

Another problem occurred in going from $a=2a$ to $1=2$. The only number that is twice itself is $0$, so if $a=2a$, then $a=0$, and we cannot divide by it.

Cameron Buie
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  • I'm not sure that second problem is really a problem - if the reasoning beforehand had been correct, it would follow by specializing to the case $a=1$. However it does (talking to the OP now) suggest that working through the proof with $a=b=1$ might point out where the problem is. (Maybe that was your point?) – mdp Sep 17 '13 at 16:30
  • Hmmm...that's an interesting idea. I've never seen such an approach before. – Cameron Buie Sep 17 '13 at 16:33
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Think: is it possible to divide by $a-b$?

H.Durham
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