Question

Which of the following steps prove that $\sqrt{3}$ is an irrational number?

• A. Consider $\sqrt{3}=\frac{p}{q}$ in reduced form, where $p\&q$ are integers and $q\ne 0$
• B. $p^2=3q^2\Rightarrow q^2=\frac{p^2}{3}$ Hence, p is a multiple of 3 i.e. $p=3m$
• C. $9m^2=3q^2\Rightarrow 3m^2=q^2$ Hence, q is a multiple of 3.
• D. Both $p\&q$ have a common factor 3. This contradicts our assumption that $\frac{p}{q}$ is in reduced form.

Answer 1

Answer: (A), (B), (C), (D)

Both $p\&q$ have a common factor 3. This contradicts our assumption that $\frac{p}{q}$ is in reduced form.

All steps given in options are correct to prove that $\sqrt{3}$ is an irrational number.

$\underset{–}{\text{Step:1}}$ Consider $\sqrt{3}=\frac{p}{q}$ in reduced form, where $p\&q$ are integers and $q\ne 0$

$\underset{–}{\text{Step:2}}$ $p^2=3q^2\Rightarrow q^2=\frac{p^2}{3}$
here $q^2$ is an integer that means $p$ is completely divided by 3 Hence, p is a multiple of 3 i.e. $p=3m$

$\underset{–}{\text{Step:3}}$ $(3m{)}^2=3q^2\Rightarrow 9m^2=3q^2$
$\Rightarrow 3m^2=q^2$ Here, $q^2$ is multiple of 3 hence, $q$(integer) must be multple of 3.

$\underset{–}{\text{Step:4}}$ Both $p\&q$ have a common factor 3. This contradicts our assumption that $\frac{p}{q}$ is in reduced form.