Question

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

Answer 1

Hint: First we’ll learn the rule for a relation to be a function that if we have a relation $X \to Y$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 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 $Z \times Z$ defined by $f = \{ (ab,a + b):a,b \in Z\} $
We know if we have a relation $X \to Y$ 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 \[a = n\] and \[b = - 1\]
$(ab,a + b) = ( - n,n - 1)$
Now, substituting \[a = -n\] and \[b = 1\]
$(ab,a + b) = ( -n, - n + 1)$
From the above-ordered pair of f from $Z \to Z$, we can say that (-n) is having two images in Z i.e. \[\left( {n - 1} \right)\] and \[\left( { - n + 1} \right)\] and is not following the rule of the relation that a function should have,
Therefore, f is not a function from$Z \to Z$.

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
$g:Z \to Z$,$g = \{ (a + b,ab):a,b \in Z\} $
Then f is called a function if and only if $g({x_1}) = g({x_2}) \Rightarrow {x_1} = {x_2}$