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Let $f:\mathbb R\to \mathbb R$ with $f(x)f(y) - f(xy) = x + y$ for every $x,y \in R$.

Prove that:

a)$f(0) = 1$

b)$f(x) = x + 1$

My solution:

a) $f(x)f(y) - f(xy) = x + y$

$f(0)f(0) - f(0\cdot0) = 0 + 0$

$f(0)^2 - f(0) = 0$

$f(0)[f(0) - 1] = 0$

Therefore $f(0) = 0$ or $f(0) = 1$

Since I must prove that $f(0) = 1$, I should reject the first solution. I can't figure out how though...

b) $f(x)f(y) - f(xy) = x + y$

By replacing $y$ with $0$

$f(x)f(0) - f(0\cdot0) = x + 0$

$f(x)\cdot1 - 1 = x$

$f(x) = x + 1$

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  • $\begingroup$ Look what happens in your calculations in bart b) if you assume $f(0)=0$. Can you derive a contradiction? $\endgroup$
    – Tintarn
    Aug 27 '15 at 22:00
  • $\begingroup$ I thought about that, $x = 0$ comes up. $\endgroup$
    – AQUATH
    Aug 27 '15 at 22:02
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If you take f(0) to be 0, replacing x by 0 and y by 1 we get:

f(1)·f(0) - f(0·1) = 0 + 1

f(1)·0 - 0 = 1

0 = 1

Which is a contradiction. Therefore, f(0) must be 1.

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  • $\begingroup$ 5 seconds faster :( $\endgroup$
    – sranthrop
    Aug 27 '15 at 22:02
  • $\begingroup$ I'm really sorry :O $\endgroup$
    – gonthalo
    Aug 27 '15 at 22:07
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Your solution looks fine. For the rest of (a), take $x=0$ and $y=1$. Then $f(0)f(1)-f(0)=1$, which does not work for $f(0)=0$.

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