Suppose $f(x) + f(y) = \frac{1}{f(xy)}$ holds for all $x,y$ and $f(x) > 0$ for all $x$. which are real. I understand that you can choose for example, $x=1$ and $y=4$ to get $f(1) + f(4) = \frac{1}{f(4)}$ but that is not very useful. What I did was chose $x=1$ and $y=1$ to then make a quadratic and get $f(1) = \frac{1}{\sqrt{2}}$. Now I would like to ask the question, why are we allowed to substitute only for $1$ variable at a time? Do we not have to chose the value of $x$ and $y$ each time? What I mean by this is we can write $f(1) + f(y) = \frac{1}{f(y)}$, then solve for $f(y)$, which is equal to $\frac{1}{\sqrt{2}}$, then we can conclude that $f(x)$ is constant for all $x$.
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4Your solution is correct. What exactly is unclear to you? – Martin R Aug 02 '23 at 18:16
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How are we allowed to substitute for $x$ but not for $y$, i.e why is it that we can leave one variable free? – Nav Bhatthal Aug 02 '23 at 18:39
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You're not technically leaving one variable free. You're proving $\forall y ~\left (f(1)+f(y)=\dfrac{1}{f(y)} \right )$. – Robert Shore Aug 02 '23 at 18:59
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2@NavBhatthal "$f(x) + f(y) = \frac{1}{f(xy)}$ holds *for all* $x,y$" $;-;$ Then it holds for $,x=1,$ and *all* $,y,$. – dxiv Aug 02 '23 at 22:19