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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 f...

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

Explanation

Solution

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=(ab,a+b):a,b∈Qf=\\{\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 ff 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=23,b=25a=\dfrac{2}{3},b=\dfrac{2}{5}
Since 23,25∈Q\dfrac{2}{3},\dfrac{2}{5}\in Q
Putting the values, we get
(ab,a+b)=(23×25,23+25)=(415,25)\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, (415,25)∈f\left( \dfrac{4}{15},\dfrac{2}{5} \right)\in f
Again, we will put a=−23,b=−25a=-\dfrac{2}{3},b=-\dfrac{2}{5}
Since −23,−25∈Q-\dfrac{2}{3},-\dfrac{2}{5}\in Q
Putting the values, we get
(ab,a+b)=(−23×−25,−23−25)=(415,−1615)\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, (415,−1615)∈f\left( \dfrac{4}{15},-\dfrac{16}{15} \right)\in f
We can see that the same element i.e. 415\dfrac{4}{15} has two different images i.e. 1615\dfrac{16}{15} and −1615-\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.