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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 $f(0)=0$.)

Bart Michels
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George R.
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6 Answers6

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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.

Holographer
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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.

mastrok
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  • I would appreciate if the down-voter could explain what goes wrong with my answer :( – mastrok Mar 02 '15 at 09:50
  • 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. – Timbuc Mar 02 '15 at 10:19
  • 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. – mastrok Mar 02 '15 at 12:11
  • 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. – Wei Zhan Mar 02 '15 at 12:18
  • One line before the end, how is that $;f(1)^2+1=f(f(1)+1);$ obtained? I got lost. – Timbuc Mar 02 '15 at 12:19
  • $x_0 f(1) + f(x_0) = f(1\dot f(x_0) + x_0 )$ with $x_0 = f(1)$ and $f(x_0) = f(f(1))=1 $, thus $f(1)^2 + 1 = f(1 + f(1)) $ – mastrok Mar 02 '15 at 12:22
  • Ok. In fact, as my failed answer shows, $;f(1)=x_0\implies f(x_0)=1;$ I'm still lost though in the very last part: putting $;x=1;$ in $;f(yf(x)+x);$ renders $;f(yf(1)+1)=f(y+1);$ , fine. But then this equals $;xf(y)+f(x)=1\cdot f(y)+1=f(y)+1;$ , so how this last equals $;y+1;$ ? – Timbuc Mar 02 '15 at 12:28
  • I am sorry, it is wrong. I typed y without $f$ at the beginning and thought that I obtained the answer. I apologize for that. I should have written it down on a paper... – mastrok Mar 02 '15 at 12:31
  • @mastrok Welcome to the club of mistypings and annexed mistakes. This shouldn't be very tough yet it becomes messy pretty easily with all those parentheses and stuff. – Timbuc Mar 02 '15 at 12:32
  • proof improved. Please have a look and thank you for tolerating my faults. – mastrok Mar 02 '15 at 12:41
  • I cannot see why $f(x)+x$ is onto. Please have a second thought when you make claims. – Wei Zhan Mar 02 '15 at 12:43
  • You are right, I cannot prove it. I was inspired by your reciprocal relation and found another way. – mastrok Mar 02 '15 at 13:23
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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.

Wei Zhan
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    How do you intend to use Induction in a non well-ordered set as $;S;$ ? – Timbuc Mar 02 '15 at 12:22
  • @Timbuc I said Eulidean fashion. See my example. – Wei Zhan Mar 02 '15 at 12:24
  • 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. – Timbuc Mar 02 '15 at 12:30
  • @Timbuc Induct on the absolute value of the denominator. – Wei Zhan Mar 02 '15 at 12:31
  • Still the set isn't well ordered, as we have negative and positive stuff. Perhaps someone else can see it more clearly. – Timbuc Mar 02 '15 at 12:34
  • @Timbuc I'm not sure if you know Euclidean division. Simply $-13/5\rightarrow 2/5$ eliminates negative stuff. – Wei Zhan Mar 02 '15 at 12:36
  • Yes, I've heard of Euclidean division...I still cannot see it. Thank you, though. – Timbuc Mar 02 '15 at 12:36
  • I have heard about that as well. I agree that three actions together generates all rational numbers. I just gave a simpler proof based on your result of $f(1/q) = 1/q$. – mastrok Mar 02 '15 at 13:30
  • @mastrok After you had those relations everything is fine. I just showed the most intuitive method. – Wei Zhan Mar 02 '15 at 13:35
  • thank you for offering me a chance to learn and think. upvoted your answer – mastrok Mar 02 '15 at 13:37
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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}

RDK
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  • From https://math.stackexchange.com/q/4349947/979028. – RDK Jan 06 '22 at 07:58
  • This looks wrong. If we put $x=-1$ into $(1)$, we get $-1-y=f(-1-f(y))$ instead of what is written by you. – Aig Mar 05 '24 at 17:40
  • You've made a mistake in your answer, plugging $x = -1$ in $(1)$, we get $$-1 - y = f(-1-f(y)).$$

    You wrote $y$ instead of $f(y)$ thereby assuming $f(y)=y \forall y$.

     – Akshat Mishra Mar 14 '24 at 11:33
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Putting $x = y = 0$, we get $f(f(0)) = 0.$
Now, putting $x=0$, we get $f(f(0)) = yf(0)$ $\forall y$.
Hence, $f(0) = 0$ as well.

Substituting $y = 0$, we get that $f(f(x)) = x.$

Also, putting $y = f(y)$ in original equation, we get :
$f(xy+f(x)) = f(x)f(y)+x$.
Putting $x=y=1$ in this equation, we get $f(1+f(1)) = f^{2}(1)+1$

Now, putting $x=y=1$ in original equation, we get $f(2f(1)) = f(1)+1$
$\implies 2f(1) = f(f(1) + 1). $
But, $f(f(1)+1) = f^{2}(1)+1 = 2f(1).$
Thus, we get $f(1) = 1.$

Now, assuming $f(n)=n$ and putting $x=1$ and $y=n$ respectively in original equation gives us : $f(n+1) = n+1$. Hence, $f(x) = x$ $\forall x \in \mathbb{N}$ by induction.

Now, putting $x=1,y=-1$ in $f(xy+f(x)) = f(x)f(y) + x$, we get :
$f(f(1)-1) = f(-1) + 1 = 0.$
Hence, $f(-1) = -1.$

Now, putting $x=-1$ and $y=n$ in original equation, we get:
$f(-(n+1)) = -(n+1).$
Hence, $f(x) = x$ $\forall x \in \mathbb{Z}.$

Now, putting $y = \frac{p}{q}$ and $x=q$, we get :
$f(q.f(\frac{p}{q})+q) = p+q$.

Applying f on both sides, we get :
$q.f(\frac{p}{q})+q = p+q.$
$\implies f(\frac{p}{q}) = \frac{p}{q}.$

Hence, $f(x) = x$ $\forall x \in \mathbb{Q}$.

Akshat Mishra
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So far we know $f(f(x)) =x$ and $f(0) = 0$. Differentiating $f(f(x)) = x$ on both sides gives us $f'(f(x))f'(x) = 1$ which implies $f'(0) = \pm1$. Now differentiating the orginal equation on both sides with respect to $y$ (keeping in mind $x$ is a constant with respect to $y$) we get $$f'(xf(y) + f(x))f'(y)x = f(x)$$.

Now plug in $y=0$ to get $f'(f(x))f'(0)x = f(x)$ and substituting the value of $f'(f(x)) = \frac{1}{f'(x)}$ to get $f'(0)x = f(x)f'(x)$. This results to the differential equation $$f'(0)xdx = f(x)d(f(x))$$ and integrating both sides gives us $$f'(0)x^2 = f^2(x)$$

knowing constant of integration is $0$. Thus $f(x) = \pm \sqrt{f(0)}x$, which gives us $f(x) = x$.

AAM
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