A peculiar functional equations question I found:

Find all functions \(f:\mathbb{Q} \rightarrow [-1,1]\) such that if \(xy=1\) or \(x+y\in\{0,1\}\) for distinct rational \(x\) and \(y\), then

\[f(x) f(y) = -1\]

A peculiar functional equations question I found:

Find all functions \(f:\mathbb{Q} \rightarrow [-1,1]\) such that if \(xy=1\) or \(x+y\in\{0,1\}\) for distinct rational \(x\) and \(y\), then

\[f(x) f(y) = -1\]

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestCall the required functions as \(f_i \quad i=1,2\).

Then,

\[\begin{align} f_i(x)&=f_i(\{x\}) \quad \forall x \not\in \left\{-\frac{1}{2}\right\} \cup \mathbb{Z}\\ f_i(x)&=f_i(1) \quad \forall x \in \mathbb{Z}^+\\ f_i(x)&=f_i(0) \quad \forall x \in \mathbb{Z} - \mathbb{Z}^+\\ f_1(x)&= 1 \quad \forall x \in \left [0,\frac{1}{2} \right) \cup \left\{-\frac{1}{2}\right\} \\ f_1(x)&=-1 \quad \forall x \in \left[\frac{1}{2}, 1 \right]\\ f_2(x)&=-1 \quad \forall x \in \left[0,\frac{1}{2} \right) \cup \left\{-\frac{1}{2}\right\} \\ f_2(x)&=1 \quad \forall x \in \left[\frac{1}{2}, 1 \right] \end{align}\]

@Sharky Kesa – Deeparaj Bhat · 10 months, 3 weeks ago

Log in to reply

@Sharky Kesa What say? – Deeparaj Bhat · 10 months, 3 weeks ago

Log in to reply

It should say \(x + y \in \{0,1\}\). – Jon Haussmann · 10 months, 4 weeks ago

Log in to reply

Put \(x=y=1\) then, we get \((f(1))^2=-1\).

Hence, no such function \(f\) exists.

@Sharky Kesa – Deeparaj Bhat · 10 months, 4 weeks ago

Log in to reply

– Sharky Kesa · 10 months, 3 weeks ago

Sorry, I had meant to write for distinct \(x\) and \(y\).Log in to reply

– Deeparaj Bhat · 10 months, 3 weeks ago

Then, we get \[|f(x)|=1 \quad \forall x \in \mathbb{Q}\\f(x)f(\frac{1}{x})=-1 \quad \forall x \in \mathbb{Q}-\{1,-1\}\\f(x)f(-x)=-1 \quad \forall x \in \mathbb{Q}-\{0\}\\f(x)f(1-x)=-1 \quad \forall x \in \mathbb{Q}-\left\{\frac{1}{2}\right\}\\\implies f(1+x)=f(x) \quad \forall x \in \mathbb{Q}-\left\{0,\frac{-1}{2}\right\}\]Log in to reply

– Sharky Kesa · 10 months, 3 weeks ago

This doesn't give us a rule for \(f\) per se but it's a good starting.Log in to reply

– Deeparaj Bhat · 10 months, 3 weeks ago

Working on figuring \(f(-1/2)\) and on [0,1]...Log in to reply

– Sharky Kesa · 10 months, 3 weeks ago

My hint is to consider continued fractions for all rational values.Log in to reply

Thanks for the hint anyways. – Deeparaj Bhat · 10 months, 3 weeks ago

Log in to reply