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**Hint:**The set $Q$ is a set of all the rational numbers. For checking whether the given subset is a function or not, we need to use the definition of a function. A function is a relation which gives a unique output for a given input. So we have to choose two rational numbers, one for $a$ and the other for $b$, and check whether for the input $ab$ the output $a+b$ is unique or not. If it gives a unique output for all the values of a given input, then it will be a function.

**Complete step by step solution:**

According to the definition of the set $f=\left[ \left( ab,a+b \right):a,b\in Q \right]$, the input is the product of two rational numbers $a$ and $b$, while the output is the sum of the two numbers.

We know that a function is a relation which gives a unique output for a given input.

Let us pick two rational numbers $2$ and $3$ so that $a=2$ and $b=3$. The input is

$\begin{align}

& \Rightarrow ab=2\times 3 \\

& \Rightarrow ab=6 \\

\end{align}$

And the output is

$\begin{align}

& \Rightarrow a+b=2+3 \\

& \Rightarrow a+b=5 \\

\end{align}$

So for the input $6$, the set $f$ is giving the output of $5$.

Now, let $a=1$ and $b=6$. The input in this case is

$\begin{align}

& \Rightarrow ab=1\times 6 \\

& \Rightarrow ab=6 \\

\end{align}$

And the output is

$\begin{align}

& \Rightarrow a+b=1+6 \\

& \Rightarrow a+b=7 \\

\end{align}$

This time, for the input $6$, the set is giving the output of $7$.

So we see that for the same input of $6$, the set $f$ is giving two different outputs of $5$ and $7$.

Since for a function the output must be unique for a given input, the given set $f$ is contradicting this definition.

**Hence, the set $f$ is not a function.**

**Note:**We must choose the second pair of the values for $a$ and $b$ such that the output $a+b$ changes but the input $ab$ remains the same. Our attempt must not be to prove that the given set is a function, but to prove that it is not a function. This is because if we wish to prove the given set $f$ a function, hundreds of examples would come to our mind. But trying to prove the given set $f$ not a function is not that easy.

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