It suffices to construct such a function satisfying $f(ab)=f(a)f(b),\ \forall a,b\in\mathbb Q^+\ (*)$ (this implies $f(1)=1$) and $f(f(x))=\frac 1x,\ \forall x\in\mathbb Q^+\ (**)$. All we need to do is define $f(p_i)$ s.t. $(*)$ whenever $x=p_i$ for some $i\ge 1$, where $(p_n)_{n\ge 1}$ is the sequence of primes, and then extend it to the rest of $\mathbb Q^+$ so that $(**)$ holds. Then it's clear that $(*)$ will automatically hold.

as Grobber said(he didn't prove it) we have $ f(ab) = f(a)f(b)$ , this implies $ f(\prod_{i = 1}^np_i^{\alpha_i}) = \prod_{i = 1}^nf(p_i)^{\alpha_i}$ , hence we must deifne the function on all primes, let $ p_i$ denote the $ i - th$ prime number we define $ f$ as: $ f(p_{2k - 1}) = p_{2k}\ ,\ f(p_{2k}) = \frac {1}{p_{2k - 1}}$ this function satisfies the problem , clearly.

So how do you extend to Q?

$ f(f(y)) = f(1)/y$. This implies that $ f$ is injective. $ f(f(1)) = f(1) \longrightarrow f(1) = 1$ Therefore $ f(f(y)) = 1/y$. Let $ y=f(y)$, so $ f(x/y) = f(x)/f(y)$. Then $ f(1/f(y)) = f(1)/f(f(y)) = y$ From the original equation, letting $ y=1/f(y)$ implies $ f(xy) = f(x)f(y)$. A function on primes like Amir's works.

I don't understand how you can conclude that $ f$ is injective, can anyone please share some light?

triplebig wrote: I don't understand how you can conclude that $ f$ is injective, can anyone please share some light? $ f(y_1) = f(y_2)$ $ \implies$ $ f(xf(y_1)) = f(xf(y_2))$ $ \implies$ $ \frac {f(x)}{y_1} = \frac {f(x)}{y_2}$ $ \implies$ $ y_1 = y_2$ (since $ f(x)\neq 0$)

Got it, thank you for the help

Rofler wrote: So how do you extend to Q? Can sameone answer to this, please ?

Dijkschneier wrote: Rofler wrote: So how do you extend to Q? Can sameone answer to this, please ? There is no need for extension : the problem is just for Q+ and we know that any positive rational may be written in a unique manner as the product of prime numbers raised to integer powers. What do you want more ?

Thank you.

Sorry to revive this topic, but can someone please explain why this doesn't work? : Note that from the above we've already established that $f(xy)=f(x)f(y)$ and $f$ is injective. Also, from $ f(f(y)) = \frac{1}{y} $, we know that $ f $ is surjective, therefore $ f^{-1}(x) $ exists for all positive rationals $ x $. So set $ f^{-1}(y)$ as $ y $ in $ f(f(y)) = \frac{1}{y} \Rightarrow f(y)f^{-1}(y) = 1 $ Thus set $ f^{-1}(x) $ as $ x $ and $ y $ as $ y $ into the original equation and we obtain: $ f(f^{-1}(x)f(y)) = \frac{x}{y} $ $ \Rightarrow f^{-1}(x)f(y) = \frac{x}{y} $ $ \Rightarrow \frac{f(y)}{f(x)} = \frac{x}{y} $ $ \Rightarrow f(x) = \frac{1}{x} $ $\forall$ $ x \in \mathbb{Q}^{+} $ which is obviously not a solution to the equation. Can someone please explain what went wrong? Thanks.

CPT_J_H_Miller wrote: Thus set $ f^{-1}(x) $ as $ x $ and $ y $ as $ y $ into the original equation and we obtain: $ f(f^{-1}(x)f(y)) = \frac{x}{y} $ $ \Rightarrow f^{-1}(x)f(y) = \frac{x}{y} $. The implication is abusive; from $f(A) = B$ you infer $A=B$.

Argh... silly mistake... yes it should be $ f(f^{-1}(x)f(y)) = xf(f(y)) = \frac{x}{y} $. Thanks!

I was able to determine the conditions for the function, but not able to construct it. Out of curiosity, how many points would I get for this (on the actual thing I would probably spend time finding it since the conditions take a very small time to find, but...)?

Plugging in x = 1 we get f(f(y)) = f(1)/y and hence f(y1) = f(y2) implies y1 = y2 i.e. that the function is bijective. Plugging in y = 1 gives us f(xf(1)) = f(x) ⇒ xf(1) = x ⇒ f(1) = 1. Hence f(f(y)) = 1/y. Plugging in y = f(z) implies 1/f(z) = f(1/z). Finally setting y = f(1/t) into the original equation gives us f(xt) = f(x)/f(1/t) = f(x)f(t). Conversely, any functional equation on Q+ satisfying (i) f(xt) = f(x)f(t) and (ii) f(f(x)) = 1 x for all x, t ∈ Q+ also satisfies the original functional equation: f(xf(y)) = f(x)f(f(y)) = f(x) y . Hence it suffices to find a function satisfying (i) and (ii). We note that all elements q ∈ Q+ are of the form q = $n i=1 pai i where pi are prime and ai ∈ Z. The criterion (a) implies f(q) = f($n i=1 pai i ) = $n i=1 f(pi)ai . Thus it is sufficient to define the function on all primes. For the function to satisfy (b) it is necessary and sufficient for it to satisfy f(f(p)) = 1 p for all primes p. Let qi denote the i-th smallest prime. We define our function f as follows: f(q2k−1) = q2k, f(q2k) = 1 q2k−1 , k ∈ N . Such a function clearly satisfies (b) and along with the additional condition f(xt) = f(x)f(t) it is well defined for all elements of Q+ and it satisfies the original functional equation.

Amir.S wrote: as Grobber said(he didn't prove it) we have $ f(ab) = f(a)f(b)$ , this implies $ f(\prod_{i = 1}^np_i^{\alpha_i}) = \prod_{i = 1}^nf(p_i)^{\alpha_i}$ , hence we must deifne the function on all primes, let $ p_i$ denote the $ i - th$ prime number we define $ f$ as: $ f(p_{2k - 1}) = p_{2k}\ ,\ f(p_{2k}) = \frac {1}{p_{2k - 1}}$ this function satisfies the problem , clearly. What is the motivation for constructing such a function? Thank you very much!

MathPanda1 wrote: Amir.S wrote: as Grobber said(he didn't prove it) we have $ f(ab) = f(a)f(b)$ , this implies $ f(\prod_{i = 1}^np_i^{\alpha_i}) = \prod_{i = 1}^nf(p_i)^{\alpha_i}$ , hence we must deifne the function on all primes, let $ p_i$ denote the $ i - th$ prime number we define $ f$ as: $ f(p_{2k - 1}) = p_{2k}\ ,\ f(p_{2k}) = \frac {1}{p_{2k - 1}}$ this function satisfies the problem , clearly. What is the motivation for constructing such a function? Thank you very much! First you try out some algebraic methods: none of them are fruitful. Then you note that the problem said construction, which implies a numbertheoretic approach. This immediately applies looking for multiplicity and a way to define f(1), which just happen to be related to each other. (Shows that you are on the right track). Then you obtain the relation: $f(f(p)) = \frac{1}{p}$ for all primes. So it is clear that you cannot manipulate the power of prime p to get from an exponent of 1 to -1 in two steps over $\mathbb{Q^{+}}$. So you have to manipulate the primes in some other method, with a group of elements acting as a medium. (In other words, f(f(p)) maps A $\rightarrow$ B $\rightarrow$ C, where you know that {p} = A, B is unknown, and {$\frac{1}{p}$} = C. This suggests bipartitioning the set of primes, which suggests considering the sets {$p_{2k}$}, {$p_{2k-1}$}, {$\frac{1}{p_{2k}}$} and {$\frac{1}{p_{2k-1}}$}. Playing around with directed arrows that map elements between the sets gives you the function.

Cute problem, nice fit for a $P1$ First we get the obvious substitutions over with- $P(1,x): f(f(x))=\frac{f(1)}{x}$. It's pretty trivial to see that this function is necessarily bijective, as $f(y)=f(t)=> f(xf(y))=f(xf(t))=>y=t$ and surjectivity is direct from $f(f(x))=\frac{f(1)}{x}$. Again, $P(1,x): f(xf(1))=f(x)=>f(1)=1=>f(f(x))=\frac{1}x$ by injectivity.

Here you could already try to hunt for a solution, but unless your intuition is really strong, it'd be quite hard to already hit upon a direct solution- also do try and remember it's an IMO problem, you can't expect to be already done with it in 2 minutes (let's just ignore this year's $P1$ Note

). Playing around with the problem, let's put an extra $f$ around the problem to get $f(f(xf(y)))=f(\frac{f(x)}{y}) =>\frac{1}{xf(y)}=f(\frac{f(x)}{y})$. Replace $x$ with $f(x)$ to get $\frac{1}{f(x)f(y)}=f(\frac{f(f(x))}{y})=f(\frac{1}{xy})$. Naturally, we'd now want to put in $y$ as $\frac{1}{x}$ to get $\frac{1}{f(x) f(\frac{1}x)}=1=> f(\frac{1}{x})=\frac{1}{f(x)}$. Re-substitute to get $f(\frac{1}{x})f(\frac{1}{y})=f(\frac{1}{xy}) =>f(x)f(y)=f(xy)$. Ok this seems like a pretty concrete result- perhaps now we'll be equipped enough to start our construction. Straight off the bat, we see that this must have something to so with the prime factorizations- firstly the function is multiplicative, secondly there doesn't seem to be any purely algebraic solution, and, perhaps most importantly, $Q^+$ is our domain aka integer prime factorizations- this problem is calling for a number-theoretic way to attack the problem. Hmm... so let's just put in $x=p_1^{\alpha_1}p_2^{\alpha_2}...p_l^{\alpha_l}, y=q_1^{\beta_1}q_2^{\beta_2}...q_k^{\beta_k}$ in our original problem- simplifying a little and spamming the multiplicative property, we get $f(f(q_1)^{\beta_1} \cdot f(q_2)^{\beta_2} \cdot ... \cdot f(q_k)^{\beta_k})=\frac{1}{q_1^{\beta_1}q_2^{\beta_2}...q_k^{\beta_k}}$. Hmm... so how do we get the primes to randomly appear in the denominator- oh wait this is trivial, just make a$\pmod{2}$ cycle of sorts- let the $j_{th}$ prime, $p_j$, map to $p_{j+1}$ if, say, the index is odd, and let it map to $\frac{1}{p_{j-1}}$ if even. Now all we have to do is put it back into the equation and check, and this construction indeed seems to work. Hence we're done.

Note

Our function will satisfy $f(f(x))=\frac{1}{x}$ and $f(ab)=f(a)f(b)\forall a,b\in \mathbb{Q^{+}}$. Consider the function that satisfies $f(ab)=f(a)f(b)$ and $f(p_{2k-1})=p_{2k}$, $f(p_{2k})=\frac{1}{p_{2k-1}}$, and $f\left(\frac{1}{p_{2k-1}}\right)=\frac{1}{p_{2k}}$, where $p_n$ is the $n$th prime for all positive integers $k$. It's easy to see we can extend this to all positive rationals. Now we will show that this satisfies $f(f(x))=\frac{1}{x}$. Call a number 1-over-prime if it can be written as $\frac{1}{p}$ for some prime $p$. Clearly $f(f(m))=\frac{1}{m}$ for all primes and 1-over-primes $m$. Now we have $f(f(x))$ can be written as $f(f(m_1))\cdot f(f(m_2))\cdots f(f(m_n))=\frac{1}{x}$ where the $m_i$ are some primes or 1-over-primes which multiply up to $x$. $\blacksquare$ Now we have $f(xf(y))=f(x)f(f(y))=\frac{f(x)}{y}$, as desired.

orl wrote: Let $ {\mathbb Q}^ +$ be the set of positive rational numbers. Construct a function $ f : {\mathbb Q}^ + \rightarrow {\mathbb Q}^ +$ such that \[ f(xf(y)) = \frac {f(x)}{y} \]for all $ x$, $ y$ in $ {\mathbb Q}^ +$. Let $P(x,y)$ be the assertion. If $f(y_1)=f(y_2) \implies y_1=y_2$. $P(x,1) \implies f(1)=1$. $P(1,y) \implies f(f(y))= \frac{1}{y}.$ Inputting this to get, $f(\frac{1}{y}=\frac{1}{f(y)}$. $P(x,f(\frac{1}{t})) \implies f(xt)=f(x)f(t) \forall t \in \mathbb{Q^+}.$ Now the functions: (a) $f(xt)=f(x)f(t)$; (b) $f(f(x))=\frac{1}{x}$ solve the equation. A function $f$ mapping through primes, satisfying (a) as, $f(p^{n_1}_1p^{n_2}_1 \dots p^{n_k}_k) = [f(p_1)]^{n_1} [f(p_2]^{n_2} \dots [f(p_k)]^{n_k},$ where $p_j$ is the jth prime and $n_j \in \mathbb{Z}$. Such a function wil satisfy (b) for each prime. A possible solution is: $$\begin{cases} P_{j+1} & \text{ if j is odd, } \\ \frac{1}{P_{j-1}} & \text{ if j is even.} \end{cases}$$Clearly, $f(f(p))=\frac{1}{p}$. Thus $f$ satisfies.

wow quite hard for an imo/1 especially one that was 33 years ago Note that any muliplicative function that satisfies $f(xy)=f(x)f(y)$ and $f(f(y))=1/y$ works, as $$f(xf(y))=f(x)f(f(y))=f(x)/y.$$ Let $p_i$ be the $i$th prime. We claim that the following function works: (i) If $i$ is odd, $f(p_i)=p_{i+1}.$ (ii) If $i$ is even, $f(p_i)=\frac{1}{p_{i-1}}.$ (iii) Otherwise, if $m$ and $n$ are relatively prime positive integers, $$f(m/n)=\frac{\prod_{p}f(p)^{v_p(m)}}{\prod_{p}f(p)^{v_p(n)}}$$ Clearly, this is multiplicative from (iii). Additionally, $$f(f(m/n))=f(\frac{\prod_{p}f(p)^{v_p(m)}}{\prod_{p}f(p)^{v_p(n)}})$$$$=\frac{f(\prod_{p}f(p)^{v_p(m)})}{f(\prod_{p}f(p)^{v_p(n)})}=\frac{\prod_p f(f(p))^{v_p(m)}}{\prod_p f(f(p))^{v_p(n)}}=\frac{1/m}{1/n}=n/m.$$

Very classical, reminded me of 2004 IMOSL A3. Consider the unique function $f$ defined such that $f(xy) = f(x)f(y)$ for all $x, y$, where we define $$f(p_{2i-1}) = p_{2i},$$$$ f(p_{2i}) = \frac 1{p_{2i-1}}, $$and$$ f\left(\frac 1{p_{2i-1}}\right) = \frac 1{p_{2i}},$$$$ f\left(\frac 1{p_{2i}}\right) = p_{2i-1},$$where $2 = p_1 < p_2 < \dots$ are the primes. Note that $f$ is uniquely determined due to prime factorisation, and $f$ is multiplicative and $f(f(x)) = \frac 1x$, therefore we're done. $\square$