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.
Since the function is bijective, we know that it is injective and surjective.
Thus define a function;
"\\displaystyle\nf:\\N\\rightarrow\\N,\\text{ defined by }f(x)=\\begin{cases}x-1 & \\text{if }x\\text{ is even}\\\\\nx+1 & \\text{if }x\\text{ is odd}\\end{cases}"
The above function is bijective.
Thus, we have "\\displaystyle\na=1\\in\\N" in both cases and "\\displaystyle\nb=\\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\na,b\\in \\N\\text{ we have }\\color{blue}a+b=2"
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