Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Square of an irrational number can be a rational number e.g. $\sqrt{2}$ is irrational but its square is 2 which is rational.

But is there a irrational number square root of which is a rational number?

Is it safe to assume, in general, that $n^{th}$-root of irrational will always give irrational numbers?

share|cite|improve this question
As an interesting side-note, it's possible for an irrational number to an irrational power to be rational. – BlueRaja - Danny Pflughoeft Aug 16 '10 at 23:49
up vote 17 down vote accepted

Obviously, if p is rational, then p2 must also be rational (trivial to prove).

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

Take the contraposition, we see that if x is irrational, then √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 $(\sqrt2)^{-2} = \frac12\in\mathbb Q$). It is true by the statement above — just replace 2 by n.

share|cite|improve this answer
cooool :) Thanks!!! – Pratik Deoghare Aug 16 '10 at 11:21
Yes.You are correct. What word should I have used instead of "negative powers"? – Pratik Deoghare Aug 16 '10 at 11:58
@The: n-th root. – kennytm Aug 16 '10 at 12:06
Oh yes. :) Thank you very much!! ( How could I miss that? ) – Pratik Deoghare Aug 16 '10 at 12:08

This is true precisely because the rationals $\mathbb Q$ comprise a multiplicative subsemigroup of the reals $\mathbb R,$
i.e. the subset of rationals is closed under the multiplication operation of $\mathbb R$. Your statement arises by taking the contrapositive of this statement - which transfers it into an equivalent statement in the complement set $\mathbb R \backslash \mathbb Q$ of irrational reals.

Thus $\rm\quad\quad\quad r_1,\ldots,r_n \in \mathbb Q \;\Rightarrow\; r_1 \cdots r_n \in \mathbb Q$

Contra+ $\rm\quad\; r_1 r_2\cdots r_n \not\in \mathbb Q \;\Rightarrow\; r_1\not\in \mathbb Q \;\:$ or $\rm\;\cdots\;$ or $\rm\;r_n\not\in\mathbb Q$.

Your case $\rm\;\;\; r^n\not\in \mathbb Q \;\Rightarrow\; r\not\in \mathbb Q \;$ is the special constant case $\rm r_i = r$

Obviously the same is true if we replace $\rm\mathbb Q\subset \mathbb R$ by any subsemigroup chain $\rm G\subset H$

The contrapositive form is important in algebra since it characterizes prime ideals in semigroups, rings, etc.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.