Question

Let f be the subset of \[Q\times Q\] defined by \[f=\{\left( ab,a+b \right):a,b\in Q\}\]. Is f a function from Q to Q? Justify your answer.

Answer 1

Hint:
Here we have to check whether f is a function from Q to Q or not. We know that a relation from set P to set Q is said to be a function if and only if every element of set P has unique images in set Q. We will put the value of a and b one by one in the given relation f. After evaluation, if we will get a unique image of the first element then the given relation f will be a function from Q to Q otherwise not.

Complete step by step solution:
Here the given relation is \[f=\{\left( ab,a+b \right):a,b\in Q\}\]. Here Q means the rational numbers i.e. a and b are rational here.
We know that a relation $f$ from set P to set Q is said to be a function if and only if every element of set P has unique images in set Q.
We will put the value of a and b one by one.
We will put $a=\dfrac{2}{3},b=\dfrac{2}{5}$
Since $\dfrac{2}{3},\dfrac{2}{5}\in Q$
Putting the values, we get
\[\left( ab,a+b \right)=\left( \dfrac{2}{3}\times \dfrac{2}{5},\dfrac{2}{3}+\dfrac{2}{5} \right)=\left( \dfrac{4}{15},\dfrac{2}{5} \right)\]
Here, \[\left( \dfrac{4}{15},\dfrac{2}{5} \right)\in f\]
Again, we will put $a=-\dfrac{2}{3},b=-\dfrac{2}{5}$
Since $-\dfrac{2}{3},-\dfrac{2}{5}\in Q$
Putting the values, we get
\[\left( ab,a+b \right)=\left( -\dfrac{2}{3}\times -\dfrac{2}{5},-\dfrac{2}{3}-\dfrac{2}{5} \right)=\left( \dfrac{4}{15},-\dfrac{16}{15} \right)\]
Here, \[\left( \dfrac{4}{15},-\dfrac{16}{15} \right)\in f\]
We can see that the same element i.e. $\dfrac{4}{15}$ has two different images i.e. $\dfrac{16}{15}$ and $-\dfrac{16}{15}$.

Thus, the relation f is not a function as it is not satisfying the conditions.

Note:
Here is the proper explanation of a function:-
Let P and Q are two nonempty sets. Then, a relation f from P to Q is said to be function, if every element in set P has a unique image in Q. Thus, a relation f from P to Q is a function, if no two distinct ordered pairs in f have the same first coordinate.