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Question: Let f be the subset of \(Z \times Z\) defined by \(f = \\{ (ab,a + b):a,b \in Z\\} \). Is f a functi...

Let f be the subset of Z×ZZ \times Z defined by f=(ab,a+b):a,b∈Zf = \\{ (ab,a + b):a,b \in Z\\} . Is f a function from Z to Z: Justify your answer

Explanation

Solution

First we’ll learn the rule for a relation to be a function that if we have a relation X→YX \to Ythen this relation is said to a function if and only if every value of elements of set A has only one image in the set B using this we’ll check whether the f is a function or not.

Complete step-by-step answer:
Given data: f be the subset of Z×ZZ \times Z defined by f=(ab,a+b):a,b∈Zf = \\{ (ab,a + b):a,b \in Z\\}
We know if we have a relation X→YX \to Y then this relation is said to a function if and only if every value of elements of set A has only one image in the set B
Now let a natural number n
Substituting a=na = n and b=−1b = - 1
(ab,a+b)=(−n,n−1)(ab,a + b) = ( - n,n - 1)
Now, substituting a=−na = -n and b=1b = 1
(ab,a+b)=(−n,−n+1)(ab,a + b) = ( -n, - n + 1)
From the above-ordered pair of f from Z→ZZ \to Z, we can say that (-n) is having two images in Z i.e. (n−1)\left( {n - 1} \right) and (−n+1)\left( { - n + 1} \right) and is not following the rule of the relation that a function should have,
Therefore, f is not a function fromZ→ZZ \to Z.

Note: In the above solution we can also comment that relation from set A to set B can be a function if and only if no two elements of set is having a similar pre-image in set A i.e.
If we reverse the order of all the ordered pairs and let the relation be called g then
g:Z→Zg:Z \to Z,g=(a+b,ab):a,b∈Zg = \\{ (a + b,ab):a,b \in Z\\}
Then f is called a function if and only if g(x1)=g(x2)⇒x1=x2g({x_1}) = g({x_2}) \Rightarrow {x_1} = {x_2}