Question

Let the function g : Z → Z be defined by

g(x) = 7x + 3.

(i) Show that g is one-to-one.

(ii) Show that g is not onto.

Answer 1

$g$ is one-to-one iff $\forall x, y\in\Z, g(x)=g(y)=>x=y.$

Assume $g(x)=g(y).$

Show it must be true that $x=y$

$$g(x)=g(y)=>7x+3=7y+3$$

$$=>7x=7y$$

$$=>x=y$$

Therefore $g$ is one-to-one.

ONTO: Given any $y\in \Z,$ can we find an $x\in \Z$ such that $g(x)=y?$

Counter example:

If $y=0,$ then

$$g(x)=7x+3=0$$

$$7x=-3$$

$$x=-\dfrac{3}{7}$$

But $-\dfrac{3}{7}$ is not an integer. Hence there is no integer $x$ for $g(x)=0$ and so $g$  is not onto.