Answer to Question #289724 in Combinatorics | Number Theory for Avengers

Question #289724

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.



1
Expert's answer
2022-01-31T16:44:11-0500

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

Thus define a function;

f:NN, defined by f(x)={x1if x is evenx+1if x is odd\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 a=1N\displaystyle a=1\in\N in both cases and b={1if x is even1Nif x is odd\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 a,bN we have a+b=2\displaystyle a,b\in \N\text{ we have }\color{blue}a+b=2


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