Answer to Question #294762 in Combinatorics | Number Theory for Denver

Question

Suppose f : N → N is a function defined by ƒ(x) = a(x + b), where a, b E N. If f is a bijective function,

then find the value of a + b.

Answer 1

Solution:

Since the function is bijective, we know that it is injective and surjective.

Thus define a function;

$\displaystyle f:\N\rightarrow\N,\text{ defined by }f(x)=\begin{cases}x-1 & \text{if }x\text{ is even}\\ x+1 & \text{if }x\text{ is odd}\end{cases}$

The above function is bijective.

Thus, we have $\displaystyle a=1\in\N$ in both cases and $\displaystyle b=\begin{cases}-1 & \text{if }x \text{ is even}\\1\in\N&\text{if }x \text{ is odd}\end{cases}$

Hence, for this function with $\displaystyle a,b\in \N\text{ we have }a+b=2$