Shivang's function

Algebra Level 5

Consider the set of all functions $f:\mathbb{Q} \rightarrow \mathbb{Q}$ such that

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

Find the sum of all possible (distinct) values of $f(99)$ .

This problem is posed by Shivang J .

The answer is 198.

4 solutions

Shivang Jindal
Sep 13, 2013

For any rational numbers $x,y,$ and $z$ ,we have

$\begin{aligned} f(x+f(y))+z &=f(x+y)+f(y)+z\\ f(f(x+f(y))+z) &=f(f(x+y)+f(y)+z)\\ f(x+f(y)+z)+f(x+f(y)) &=f(x+y+f(y)+z)+f(x+y)\\ f(x+y+z)+f(y)+f(x+y)+f(y) &=f(x+2y+z)+f(y)+f(x+y)\\ f(x+y+z)+f(y) &=f(x+2y+z). \end{aligned}$

This implies $f(a)+f(b)=f(a+b)$ for all rational numbers $a$ and $b$ . By the Cauchy functional equation over the rationals, we have $f(x) = mx$ for some $m \in \mathbb{Q}$ . Substituting back into $f(x+f(y))=f(x+y)+f(y)$ gives $(m^2-2m) y = 0$ , hence we must have $m = 0$ or $m = 2$ . We easily see that both of these are solutions to the original functional equation.

Thus, $f(99) = 198$ and $f(99) = 0$ are the distinct values, and their sum is 198.

Moderator note:

We present the solution written by Shivang.

In this question, the justification of steps is definitely much harder than guessing the numerical answer.

$f(x)$ is not a continual function, so I don't think we can apply Cauchy functional equation to this case.

Từ Thiện Nguyễn Văn - 7 years, 8 months ago

No, it works because we work over the rationals !

Louis Abraham - 7 years, 2 months ago

Prakhar Agarwal
Sep 8, 2013

f(x+f(y))=f(x+y)+f(y) when x=0, f(f(y))=2f(y) => when x=0, f(y)=2y therefore, f(99)=2*99=198

Here's a (hopefully better) solution.

$(x,y)=(x-f(x),x):$ $f(x)=f(2x-f(x))+f(x)$

Therefore we obtain $f(2x-f(x))=0$ .

$(x,y)=(0,x):$ $f(f(y))=2f(y)$

Thus $0=2f(2x-f(x))=f(f(2x-f(x)))=f(0)$ which would imply $f(0)=0$ .

$(x,y)=(-x,x):$ $f(f(x)-x)=f(0)+f(x)=f(x)$ --- (*)

Now let's assume that there exists two different rational numbers $a$ and $b$ s.t. $f(a)=f(b)$ .

$(x,y)=(-a,b):$ $f(f(b)-a)=f(b-a)+f(b)$ equivalent to $f(f(a)-a)=f(b-a)+f(a)$ ---- (0)

(*)+(0): $f(b-a)=0$ for all pairs of distinct rational numbers $a$ and $b$

Therefore there are two cases:

if $f(x)$ is constant, then $f(x)=0$
if $f(x)$ is non constant, then it would be impossible for there to be different rational numbers $a$ and $b$ s.t. $f(a)=f(b)$ . --- (1)

Following that we consider that the function is non constant.

We note that we can express $a=b+f(c)$ where $b,c$ are rational numbers.

If $a=d$ , then $a=b+f(c)=d$ , thus $f(a)=f(b+f(c))=f(b+c)+f(c)=f(b+f(c))=f(d)$ . Thus if $a=d$ , then $f(a)=f(d)$ . --- (2)

(1) and (2) imply that $x=y$ if and only if $f(x)=f(y)$ . --- (3)

(3) and (*) imply that $x = f(x) - x$ , which gives the last solution that $f(x)=2x$ .

Gabriel Tan - 7 years, 9 months ago

This a much better approach, but it is still incomplete. The solution is good until you say that $f(b-a)=0$ for all distinct rational numbers $a$ and $b.$ This is only true for those $a$ and $b$ that $f(a)=f(b)$ . Also, after that you spend some effort on proving an obvious: if $a=d,$ then $f(a)=f(d).$ The converse of that (if $f(a)=f(d)$ then $a=d$ ) is what you really need, but you have not proved it. I am sure this can be fixed, keep thinking (or maybe someone else can lend a hand)?

Alexander Borisov - 7 years, 9 months ago

This isn't true. Your implication is wrong. You need one more thing. Correct conclusion is: When x=0 and there exists k such that f(y)=k then f(y)=2y. This doesn't solve the problem yet. For example there exists a function f(x)=0 that is a solution too, but it adds 0 to the overall result.

Actually, I just guessed there is no other function but I'm curious how to prove there are no other solutions than these two.

Ania Piekarska - 7 years, 9 months ago

Wouldn't such k exist for all y because y is an element of Q , which is the domain ?

Mirza Baig - 7 years, 9 months ago

Ania P. is correct, though her comment contains a misprint: instead of $f(y)=k$ is should be $f(k)=y.$ This is a serious gap, even though it does not affect the numerical answer.

Alexander Borisov - 7 years, 9 months ago

Is it alright to post a solution here? I was about to do it, but then my connection was broken and after refreshing my browser, I'm locked out.

Gabriel Tan - 7 years, 9 months ago

You can. Those writing a solution can't see yours until they submit their own or join discussion by forgoing the submission.

Mirza Baig - 7 years, 9 months ago

Though you won't get any votes for that, it would help me and others.

Mirza Baig - 7 years, 9 months ago

Brian Chen
Sep 14, 2013

Define the auxiliary function $g(x) = f(x) - x$ . (This is very easy to motivate once you've guessed the two solutions; $g(x) = \{-x, x\}$ is a much more symmetric solution set than $f(x) = \{0, 2x\}$ .)

So $f(x) = g(x) + x$ , and we get the equivalent functional equation

$g(x + y + g(y)) + x + y + g(y) = g(x + y) + x + y + g(y) + y$

This is true for all $x, y \in \mathbb{Q}$ . If we substitute $z = x + y$ , so $x = z - y$ , it simplifies to

$g(z + g(y)) = g(z) + y$

for all $z, y \in \mathbb{Q}$ . Let $P(z, y)$ denote this statement.

If $g(y_1) = g(y_2)$ , compare $P(z, y_1)$ and $P(z, y_2)$ for any $z$ to see that $y_1 = y_2$ ; thus $g$ is injective.

By $P(0, 0)$ we have $g(g(0)) = g(0)$ , and thus by injectivity $g(0) = 0$ .

Now by $P(0, y)$ we have $g(g(y)) = y$ for all $y \in \mathbb{Q}$ , so $g$ is an involution.

Then by $P(z, g(y))$ we have $g(z + y) = g(z) + g(y)$ , Cauchy's functional equation, so $g(x) \equiv cx$ for some fixed $c \in \mathbb{Q}$ . The only values that let $g$ be an involution are $c = 1$ and $c = -1$ ; both are easily seen to work.

So the two solutions for $f$ are $f(x) = 0$ and $f(x) = 2x$ , and $f(99)$ is either 0 or 198, so our answer is $\boxed{198}$ .

Arkan Megraoui
Sep 14, 2013

Putting $x=0$ yields $f(f(y))=2f(y)$ . Putting $x=f(y)$ yields $f(x)=2x$ . Hence, $f(99)=\boxed{198}$ .

Shivang's function

The answer is 198.

4 solutions

Moderator note:

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