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Let $\phi :\mathbb{R} \rightarrow \mathbb{R}$ be a one to one map such that $\phi \left( x+y\right) =\phi \left( x\right) +\phi \left( y\right)$ and $\phi \left( x^{2}\right) =\phi \left( x\right) ^{2}$ for all $x,y\in\mathbb{R}$. Show that $\phi \left( xy\right) =\phi \left( x\right) \phi \left( y\right)$ for all $x,y\in\mathbb{R}$ and $\phi \left( q\right) =q$ for all $p\in\mathbb{Q}$.

My proof. By the assumptions, we have $\phi \left( x+y\right) ^{2}=\phi \left( x\right) ^{2}+2\phi \left( xy\right)+\phi \left( y\right)^{2}$. We also have, $\phi \left( x+y\right) ^{2}=\phi \left( x\right) ^{2}+2\phi \left( xy\right)+\phi \left( y\right) ^{2}$. Hence, we obtain $\phi \left( xy\right) =\phi \left( x\right) \phi \left( y\right)$. We are done.

Now, we will show $\phi \left( q\right) =q$ for all $p\in\mathbb{Q}$.

Claim. $\phi \left( \dfrac {a} {b}\right) =b \phi \left( \dfrac {a} {b^{2}}\right)$ $a,b\in\mathbb{Z}$ (as $b\neq 0$).

My question is: How can I prove this claim? I couldn't show to use induction on $b$.

Note: This question is from my set theory class. Tag can be set theory, but I don't sure.

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    As the answer that should be a comment mentions, you write $\phi(x+y)^2$ twice. Rather you want one to be $\phi((x+y)^2)$ and the other $\phi(x+y)^2$, which are equal by assumptions about $\phi$. – snulty Nov 23 '16 at 22:09
  • Yes, you are right. Buti It already is equal to $\phi \left( \left( x+y\right) \right) ^{2}$. –  Nov 23 '16 at 22:14
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    @snulty thanks. for some reason I clicked the wrong button in my mobile phone. I meant to write a comment -.- – RGS Nov 23 '16 at 22:18

2 Answers2

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Hint: Find $\phi(1)$ and $\phi(0)$. Then work your way to any rational number from there.

Arthur
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  • I found these. But, I think, I should prove this claim. –  Nov 23 '16 at 22:06
  • I found these $\phi \left( 1\right) =1$ and $\phi \left( 0\right) =0$ Also,I found $\phi \left( \dfrac {1} {2}\right) =2\Phi \left( \dfrac {1} {2^{2}}\right)$. Now, I think, I should prove $\phi \left( \dfrac {a} {b}\right) =a\phi \left( \dfrac {a} {b^{2}}\right)$. But, how? –  Nov 23 '16 at 22:12
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    I think you should prove $\phi(n)=n$ instead (with $n$ being integer). From there that $\phi(1/n)=1/n$, and you're basically done. – Arthur Nov 23 '16 at 22:20
  • Claim. $\phi \left( n\right) =n$ for all $n\in\mathbb{Z}^{>0}$. Proof. Induction on $n$. So, it is trivial. But, how can I for all for all $n\in\mathbb{Z}^{<0}$? –  Nov 24 '16 at 01:55
  • I think, we need to have a new claim, i.e., $\phi \left( -x\right) =-\phi \left( x\right)$ for all $x\in\mathbb{R}$. Proof. Since $\phi \left( 0\right) =0$, we have $\phi \left( 0\right) =\phi \left( x+\left( -x\right) \right) =0$. Hence, we have $\phi \left( -x\right) =-\phi \left( x\right)$. We are done. –  Nov 24 '16 at 02:06
  • @Kahler And there you have it. The only thing left is $\phi(1/n)$, which you can do by way of $\phi(n\cdot 1/n)$ and finally $\phi(m/n)=\phi(m\cdot 1/n)$. – Arthur Nov 24 '16 at 08:46
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$$\phi\left(\frac{a}{b}\right) = \phi\left(\frac{ba}{b^2}\right) =\phi\left(\frac{a}{b^2}+\frac{a}{b^2}+\cdots+\frac{a}{b^2}\right) $$ $$=\phi\left(\frac{a}{b^2}\right) +\phi\left(\frac{a}{b^2}\right) +\cdots +\phi\left(\frac{a}{b^2}\right)=b\phi\left(\frac{a}{b^2}\right) $$

B. Goddard
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