Question

Which of the following steps prove that $\sqrt{7}$ is an irrational number.

• A. Consider $\sqrt{7}=\frac{p}{q}$ in reduced form, where $p\&q$ are integers and $q\ne 0$
• B. $p^2=7q^2\Rightarrow q^2=\frac{p^2}{7}$ Hence, p is a multiple of 7 i.e. $p=7m$
• C. $49m^2=7q^2\Rightarrow 7m^2=q^2$ Hence, q is a multiple of 7.
• D. Both $p\&q$ have a common factor 7. 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 7. This contradicts our assumption that $\frac{p}{q}$ is in reduced form.

All steps given in options are correct to prove that $\sqrt{7}$ is an irrational number.

$\underset{–}{\text{Step:1}}$ Consider $\sqrt{7}=\frac{p}{q}$ in reduced form, where $p\&q$ are integers and $q\ne 0$

$\underset{–}{\text{Step:2}}$ $p^2=7q^2\Rightarrow q^2=\frac{p^2}{7}$
here $q^2$ is an integer that means $p$ is completely divisible by 7 Hence, p is a multiple of 7 i.e. $p=7m$

$\underset{–}{\text{Step:3}}$ $(7m{)}^2=7q^2\Rightarrow 49m^2=7q^2$
$\Rightarrow 7m^2=q^2$ Here, $q^2$ is multiple of 7 hence, $q$(integer) must be multple of 7.

$\underset{–}{\text{Step:4}}$ Both $p\&q$ have a common factor 7. This contradicts our assumption that $\frac{p}{q}$ is in reduced form.