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## Here's the Solution to this Question

Let $g$ be a function from $Z^+$ (the set of positive integers) to $Q$ (the set of rational numbers) defined by

$(x, y)\in g\ \text{iff } y=4x-\dfrac{3}{7}\ (g\sube Z^+\times Q)$

and let $f$ be a function on $Z^+$ defined by

$(x, y)\in g\ \text{iff } y=5x^2+2x-3\ (f\sube Z^+\times Z^+)$

Question 6

$g\circ f(x)=g(5x^2+2x-3)$

$=4(5x^2+2x-3)-\dfrac{3}{7}$

$=20x^2+8x-12\dfrac{3}{7}$

$g\circ f:Z^+\to Q$ defined by $g\circ f=20x^2+8x-12\dfrac{3}{7}$

1.  $20x^2+8x-12\dfrac{3}{7}$

Question 7

We consider the statements provided in the different alternatives:

1. $g$ is not defined on the set of real numbers thus $g$ cannot be depicted as a straight line graph.

Only positive integers can be present in the domain of $g.$

It is the case that rdered pairs such as $(1, 3\dfrac{4}{7}), (2, 7\dfrac{4}{7}), (3,11\dfrac{4}{7}),...$ belong to $g$ and these pairs can be presented as dots in a graph.

2. We prove that $g$ is indeed injective:

$g(u)=g(v)=>4u-\dfrac{3}{7}=4v-\dfrac{3}{7}$

$=>4(u-v)=0=>u=v$

$g$ is injective.

3. The function $g$ is NOT surjective.

Counterexample

$y=\dfrac{1}{2}, \dfrac{1}{2}\in Q$

Then

$y=4x-\dfrac{3}{7}=\dfrac{1}{2}=>4x=\dfrac{13}{14}=>x=\dfrac{13}{56}$

$\dfrac{13}{56}\not\in Z^+$

We conclude that $g$ is not surjective.

4. Since $g$ is not surjective, then $g$ is not bijective.

2. $g$ is injective.