Mohammad Cannon

2022-06-19

nth-root of irrational will always give irrational numbers?

Alisa Gilmore

Beginner2022-06-20Added 22 answers

Obviously, if $p$ is rational, then ${p}^{2}$ must also be rational.

$p\in \mathbb{Q}\Rightarrow {p}^{2}\in \mathbb{Q}.$

Take the contraposition, we see that if $x$ is irrational, then $\sqrt{x}$ must also be irrational.

${p}^{2}\notin \mathbb{Q}\Rightarrow p\notin \mathbb{Q}.$

By negative power I assume you mean ($1/n$)-th power (it is obvious that $(\sqrt{2}{)}^{-2}=\frac{1}{2}\in \mathbb{Q}$. It is true by the statement above — just replace ${}^{2}$ by ${}^{\mathrm{n}}$.

$p\in \mathbb{Q}\Rightarrow {p}^{2}\in \mathbb{Q}.$

Take the contraposition, we see that if $x$ is irrational, then $\sqrt{x}$ must also be irrational.

${p}^{2}\notin \mathbb{Q}\Rightarrow p\notin \mathbb{Q}.$

By negative power I assume you mean ($1/n$)-th power (it is obvious that $(\sqrt{2}{)}^{-2}=\frac{1}{2}\in \mathbb{Q}$. It is true by the statement above — just replace ${}^{2}$ by ${}^{\mathrm{n}}$.