Question

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.

Answer 1

Answer:

its true

Step-by-step explanation:

Answer 2

$\large{\underline{\rm{\purple{\bf{Given:-}}}}}$

a and b are real numbers such that a² = b² then a = b

$\large{\underline{\rm{\purple{\bf{To \: Find:-}}}}}$

Whether the given statement is true or false.

$\large{\underline{\rm{\purple{\bf{Solution:-}}}}}$

The given statement can be written in the form of ‘if then’ is given below

If a and b are real numbers such that a² = b² then a = b

Let p: a and b are real numbers such that a² = b²

$\longrightarrow \sf q: \: ab$

The given statement has to be proved false. To show this, two real numbers, a and b, with a² = b² are required such that a ≠ b

Let us consider a = 1 and b = – 1

$\implies \sf a^{2}=(1)^{2}$

$\implies \sf 1 \: and \: b^{2}=(-1)^{2}$

$\implies \sf 1$

Hence, a² = b²

However, a ≠ b

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