Answer to Question #218882 in Discrete Mathematics for Nicholas

Question

Let g be a function from Z+ (the set of positive integers) to Q (the set of rational numbers) defined by (x, y) element of g iff y = (g is a subset of Z+ mapped with Q) and let f be a function on Z + defined by (x, y) element of f iff y = 5x2 + 2x – 3 (f subset of Z+ mapped with Z+)

Which one of the following statements regarding the function g is TRUE? 

(Remember, g is a subset of Z+ mapped with Q.)

1. g can be presented as a straight line graph.

2. g is injective.

3. g is surjective.

4. g is bijective.

Answer 1

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

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

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 ordered 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.  

Answer

2. $g$ is injective.