Show that the statement "For any real numbers $a$ and $b$,$a^{2}=b^{2}$ implies that $a=b$" is not true by giving a counter-example.

(A) The given statement can be written in the form of "if-then" as follows:
If $a$ and $b$ are real numbers such that $a^{2}=b^{2}$,then $a=b$.
Let $p$ be the statement: $a$ and $b$ are real numbers such that $a^{2}=b^{2}$.
Let $q$ be the statement: $a=b$.
To prove the statement is false,we need to find a case where $p$ is true but $q$ is false (i.e.,$a^{2}=b^{2}$ but $a \neq b$).
Let $a=1$ and $b=-1$.
Then $a^{2}=(1)^{2}=1$ and $b^{2}=(-1)^{2}=1$.
Thus,$a^{2}=b^{2}$ holds true.
However,$a=1$ and $b=-1$,so $a \neq b$.
Since we found a case where $a^{2}=b^{2}$ but $a \neq b$,the statement is false.