Answer to Question #284729 in Combinatorics | Number Theory for kool

Question

Let us define a function f : N → Q as follows,

(n-1)/4 n is odd

f(n) = {

(n+1)/2 n is even

Which of the following are true?

f is not one to one.

f is not onto.

f is onto.

f is one to one.

Answer 1

Let us define a function $f : \N → \mathbb Q$ as follows, $f(n) =\begin{cases} \frac{n-1}4, \text{ if } n \text{ is odd}\\ \frac{n+1}2, \text{ if } n \text{ is even} \end{cases}.$

Since $f(7)=\frac{7-1}4=\frac{6}4=\frac{3}2$ and $f(2)=\frac{2+1}2=\frac{3}2,$ we conclude that $f(7)=f(2),$ and hence a function $f : \N → \mathbb Q$ is not one-to-one.

Taking into account that $\frac{n-1}4=\frac{1}8$ implies $n=\frac{1}2+1=\frac{3}2\notin \N$ and $\frac{n+1}2=\frac{1}8$ implies $n=\frac{1}4-1=-\frac{3}4\notin \N,$ we conclude that for $\frac{1}8\in\mathbb Q$ the preimage $f^{-1}(\frac{1}8)=\emptyset,$ and thus a function $f : \N → \mathbb Q$ is not onto.

Therefore, the statements "$f$ is not one to one" and "$f$ is not onto" are true, and the staterments "$f$ is onto" and "$f$ is one to one" are false.