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.

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$g$ 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.