# Show that the statement “For any real numbers a and b, a2 = b2

Question:

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

Solution:

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 $p: a$ and $b$ are real numbers such that $a^{2}=b^{2}$.

qa = b

The given statement has to be proved false. For this purpose, it has to be proved that if $p$, then $\sim q$. To show this, two real numbers, $a$ and $b$, with $a^{2}=b^{2}$ are required such that $a \neq b$.

Let a = 1 and b = –1

$a^{2}=(1)^{2}=1$ and $b^{2}=(-1)^{2}=1$

$\therefore a^{2}=b^{2}$

However, $a \neq b$

Thus, it can be concluded that the given statement is false.