How many bijections f: Z rightarrow Z are the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the functional equation $f(x+y) = f(x) + f(y)$ for all $x, y \in Z$. This is Cauchy's functional equation on the set of integers $Z$. The ge

Vedclass · Accepted Answer

Two

Vedclass · Answer

(B) Given the functional equation $f(x+y) = f(x) + f(y)$ for all $x, y \in Z$. This is Cauchy's functional equation on the set of integers $Z$. The general solution is $f(x) = kx$ for some constant $k \in Z$. For $f$ to be a bijection,it must be both injective and surjective. If $f(x) = kx$,then $f$ is injective if $kx_1 = kx_2 \implies x_1 = x_2$,which holds if $k 
eq 0$. For $f$ to be surjective,for any $y \in Z$,there must exist $x \in Z$ such that $f(x) = y$,i.e.,$kx = y$. This implies $x = y/k$. For $x$ to be an integer for every $y \in Z$,$k$ must be a divisor of $1$. Thus,$k = 1$ or $k = -1$. Therefore,the possible functions are $f(x) = x$ and $f(x) = -x$. There are exactly $2$ such bijections.

How many bijections $f: Z \rightarrow Z$ are there such that $f(x+y)=f(x)+f(y)$ for all $x, y \in Z$?

Similar Questions

Let $f : R - \{0, 1\} \rightarrow R$ be a function such that $f(x) + f\left(\frac{1}{1-x}\right) = 1 + x$. Then $f(2)$ is equal to:

Let $f: R \rightarrow R$ be such that $f$ is injective and $f(x)f(y) = f(x+y)$ for all $x, y \in R$. If $f(x), f(y),$ and $f(z)$ are in $GP$, then $x, y,$ and $z$ are in:

If $f(x + y) = f(x)f(y)$ for all $x$ and $y$ and $f(5) = 2$,$f'(0) = 3$,then $f'(5)$ will be

The function $f(x) = x - [x]$,where $[x]$ denotes the greatest integer function,is:

If $a$ and $b$ are two fixed positive integers such that $f(a + x) = b + [b^3 + 1 - 3b^2f(x) + 3b\{f(x)\}^2 - \{f(x)\}^3]^{1/3}$ for all real $x$,then $f(x)$ is a periodic function with period:

Vedclass Test Series

Exam Paper Generator

Online Exam Module