Solution to Let g be a function from Z+ (the set of positive integers) to Q (the … - Sikademy
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Archangel Macsika

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.

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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.


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Question ID: mtid-5-stid-8-sqid-2771-qpid-1328