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Let $ f : \mathbb{Q} \rightarrow \mathbb{Q} $ be a function which has the following property:

$$ f(x \cdot f(y) + f(x)) = y \cdot f(x) + x \;,\; \forall \; x, y \in \mathbb{Q} $$

Prove that $ f(x) = x, \; \forall \; x, y \in \mathbb{Q} $.

So far, I've found that $f(f(x)) = x$, $f(0) = 0$ and $f(-1) = -1$.

(For $f(0)=0$, we substitute $x=0$ to arrive at $f(f(0))-yf(0)$ identically $0$ for all rational $y$; for $f(f(x))=x$, we substitute $y=0$ and use $f(0)=0$. For $f(-1) = -1$, substitute $x=y=-1$ to get $f(0)=-f(-1)-1$, and use $f(0)=0$.)

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    $\begingroup$ So far, I've found that f(f(x)) = x, f(0) = 0 and f(-1) = -1. $\endgroup$
    – George R.
    Mar 2 '15 at 9:12
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    $\begingroup$ Considering the attempted answers below, certain steps in tackling this problem can be deceptively simple but actually difficult. So it would be useful if you could write out explicitly how you had arrived at $f(f(x))=x$, $f(0)=0$ and $f(-1)=-1$. $\endgroup$ Mar 2 '15 at 13:41
  • $\begingroup$ @KimJongUn Just simple substitutions. A clarification is good but I see no necessarity, since they are common routines in solving functional equations. $\endgroup$
    – Wei Zhan
    Mar 2 '15 at 13:48
  • $\begingroup$ @WillardZhan Actually, I'll just add them myself. $\endgroup$ Mar 2 '15 at 15:28
  • $\begingroup$ what if f(0) = 1, and f(1) = y...so that f(f(0)) - y.f(0) = 0 is true? $\endgroup$ Mar 2 '15 at 17:25
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The equation we have is $f(xf(y)+f(x))=yf(x)+x$.

Set $x=0$ to get $f(f(0))=yf(0)\; \forall y\implies f(0)=0$. Then set $y=0$ for arbitrary $x$ to get $f(f(x))=x\;\forall x$.

Now let $x=y=1$, to get $f(2f(1))=f(1)+1$. So pick $x=1,y=f(1)$ to get $f(1+f(1))=f(1)^2+1$, from which $f(1)^2+1=f(f(2f(1)))=2f(1)$, so $f(1)=1$.

Now set $x=1,y=f(z)$to get $f(z+1)=f(z)+1$, which can be used inductively to show that $f(n)=n$ for all integer $n$.

Finally, let $y=p/q,x=q$, for $p$ and $q$ nonzero integers. Then $f(qf(p/q)+q)=p+q$, applying $f$ gives $qf(p/q)+q=p+q$, from which $f(p/q)=p/q$, so we are done.

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  • $\begingroup$ I like your solution. But I think you should also check that $f(z-1)=f(z)-1$ to obtain $f(z)=z$ for negative integers? $\endgroup$
    – user60589
    Mar 2 '15 at 17:57
  • $\begingroup$ That's exactly the same relationship: you can use it to go both forwards to positive integers or backwards to negative ones $\endgroup$ Mar 2 '15 at 17:59
  • $\begingroup$ Very nice and clear solution! Thanks! $\endgroup$
    – George R.
    Mar 2 '15 at 18:17
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Since $f(f(x))=x$, $f$ is bijective and there is a $y_0$ such that $f(y_0)=1$. Now we have $$f(x+f(x))=y_0f(x)+x$$ $$f(xf(1)+f(x))=f(x)+x$$ which leads to $y_0f(x)+x=xf(1)+f(x)$ for all $x\in\mathbb{Q}$, or $$(f(1)-1)x=(y_0-1)f(x).$$

Suppose that $y_0\neq 1$, then it turns out that $f(x)$ is a linear function. It is easily verified that in this case the only possibility is $f(x)=x$ which contradicts to $y_0\neq 1$.

Therefore we conclude $y_0=1=f(1)$, and as mastrok noticed we have $f(y+1)=f(y)+1, \forall y$, and equivalently $f(y-1)=f(y)-1,\forall y$.

The next step is for every $x\neq 0$, suppose $f(y)=1/x$ and we get $$yf(x)+x=f(1+f(x))=f(f(x))+1=x+1,$$ and hence $y=1/f(x)$. That is, $f(1/x)=1/f(x),\forall x\neq 0.$

Now consider the set $S=\{x\mid f(x)=x\}$. We already know $0\in S$, and $S$ is closed under operations 'plus one', 'minus one' and 'reciprocal'. Now I think it's easy for yourself to check $S=\mathbb{Q}$ by induction with an Euclidean fashion. There's an example here: $$13/5\rightarrow 3/5 \rightarrow 5/3\rightarrow 2/3\rightarrow 3/2\rightarrow \ldots$$

Update: Since there are ones not familiar with Euclidean algorithm, I try to make it clearer. Use induction on $q$ for fraction $p/q\in\mathbb{Q}$, where $(p,q)=1$ and $q>0$. The Euclidean division turn out that $$p=aq+b,a\in\mathbb{Z}, 0\leq b<q.$$ And $b/q\in S$ since $q/b\in S$ by inductve hypothesis. Now $p/q=b/q+a\in S$ and the induction is done since the initial case $q=1$ is trivial.

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    $\begingroup$ How do you intend to use Induction in a non well-ordered set as $\;S\;$ ? $\endgroup$
    – Timbuc
    Mar 2 '15 at 12:22
  • $\begingroup$ @Timbuc I said Eulidean fashion. See my example. $\endgroup$
    – Wei Zhan
    Mar 2 '15 at 12:24
  • $\begingroup$ I still can't see it "Euclidean fashion" or whatever. Either we have an inductive set or we have not one, otherwise I can't see how to end the proof formally. $\endgroup$
    – Timbuc
    Mar 2 '15 at 12:30
  • $\begingroup$ @Timbuc Induct on the absolute value of the denominator. $\endgroup$
    – Wei Zhan
    Mar 2 '15 at 12:31
  • $\begingroup$ Still the set isn't well ordered, as we have negative and positive stuff. Perhaps someone else can see it more clearly. $\endgroup$
    – Timbuc
    Mar 2 '15 at 12:34
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If you have $f(f(x))=x$ it means that the function is onto

Furthermore, assume that $f(y_1) = f(y_2)$ for some $y_1 \neq y_2$

$$ f(-f(y_1)-1 ) =f(-f(y_1)-1 ) $$ $$\Rightarrow -y_1 -1 = -y_2 -1 $$ $$\Rightarrow y_1 = y_2 !! $$

Thus the function is $1$ $to$ $1$, its inverse $f^{-1}$ exists, in addition to $f(f(x))=x$, we have $f^{-1}=f$

From above, you get $f(f(1)+1) = f(1)+1 $

$f^{-1} = f \Rightarrow f(yf(x)+x) = f^{-1}(yf(x)+x) = xf(y) +f(x) $

For some $x_0=f(1)$, $f(x_0)=f(f(1))=1$ and putting $y=1$

$x_0 f(1) + f(x_0) = f(1 f(x_0) + x_0 )$ $\Rightarrow f(1)^2 +1 = f(f(1)+1) = 2f(1) \Rightarrow f(1)=1$

$\Rightarrow f(y+1)=f(y)+1$ by putting $x=1$ into $f(yf(x)+x) = xf(y) +f(x) $

From the original definition, putting $y=1$ gives $f(x+f(x)) = f(x) +x$

Since $f(x)$ is onto, $f(x)+x$ is also onto(to be proved), therefore we can find $x$ such that $y=x+f(x)$ for all $y$ then $f(y) =y$

I cannot prove that it is onto... However, I found another way by looking at the answer of Willard Zhan. He has proven $f(1/q) = 1/q$ for integers $q$

For all $\frac{p}{q}$, it can be always written as $\frac{m+1}{q}$, where $m$ is also an integer.

putting $y=m$ and $x=\frac{1}{q}$ into $f(yf(x)+x)=xf(y)+f(x)$ $$f\left( \frac{m+1}{q}\right)=f\left( \frac{m}{q} +\frac{1}{q}\right) = \frac{f(m)+1}{q} = \frac{m+1}{q}$$ since we have proved $f(m)=m$ for integers. I think it completes the proof.

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  • $\begingroup$ I would appreciate if the down-voter could explain what goes wrong with my answer :( $\endgroup$
    – mastrok
    Mar 2 '15 at 9:50
  • $\begingroup$ In general, I dislike downvotes and don't use to apply them most of the time. This case is one of them, yet you shoud perhaps explain a little more what leads you to conclude $\;f=\text{Id.}_{\Bbb Q}\;$ , even with $\;f^{-1}=f\;$ . There are infinite involutions on the rationals. $\endgroup$
    – Timbuc
    Mar 2 '15 at 10:19
  • $\begingroup$ I agree with your comments, that is not a fact even if some points are specified. I gave a more detailed proof. I would appreciate if you could try to read it. $\endgroup$
    – mastrok
    Mar 2 '15 at 12:11
  • $\begingroup$ I downvoted on your first answer without explanation since I saw no evidence for it to be a right answer. Now for your edited proof, please show how you got $f(f(1)+1)=f(1)+1$ and I'll cancel the downvote, even though this is still not a robust proof. $\endgroup$
    – Wei Zhan
    Mar 2 '15 at 12:18
  • $\begingroup$ One line before the end, how is that $\;f(1)^2+1=f(f(1)+1)\;$ obtained? I got lost. $\endgroup$
    – Timbuc
    Mar 2 '15 at 12:19
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\begin{align} P(0, y): \; & f(f(0))=yf(0)=0. \\ \Rightarrow \; & f(0)=0. \\ \ \\ P(x, 0): \; & f(f(x))=x. \\ \ \\ f(P(x, y)): \; & f(x)+xf(y)=f(x+yf(x)). \\ f(y) \to y: \; & f(x)+xy=f(x+f(x)f(y)). \tag 1 \label 1 \\ \ \\ P(-1, -1): \; & f(f(-1)-f(-1))=-1-f(-1). \\ \Rightarrow \; & f(-1)=-1. \\ \ \\ x \to -1 \text{ from } \color {#4995CF} {(\ref 1)}: \; & -1-y=f(-1-y). \\ y \to -x-1: \; & f(x)=x. \end{align}

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