question

[Universe_Man]

is the square root of a prime number always going to be irrational? just a random question.

[d_leet]

I believe the square root of any natural number that is not a perfect square is irrational.

[quasar987]

I would say yes, because lets denote our prime p. If \sqrt{p}=n, (n\in\mathbb{N}), then it means that p=n² <==> p/n=n (i.e. n divides p ==> p is not prime: a contradiction with the hypothesis). If \sqrt{p}=m/n, (m,n \in \mathbb{N}, n\neq 0,1, then it means that p = m²/n², i.e. p is rationnal ==> p is not prime: a contradiction with the hypothesis. The only remaining possibility is that \sqrt{p} is irrational.

[BSMSMSTMSPHD]

Yes, you're both right.

Here's a proof of the prime case...

Suppose there exists a prime p such that \sqrt{p} \in \mathbb{Q}

Then \sqrt{p} = \frac{a}{b} \ where p,q \in \mathbb{Z}

Now, we can assume that a and b are relatively prime (ie the fraction is in lowest form). This is crucial for the remainder of the proof.

b \cdot \sqrt{p} = a

b^{2} \cdot p = a^{2}

From this last equation, we see that p divides a. Therefore, there exists some integer k such that a = pk. Substituting gives us:

b^{2} \cdot p = {(pk)}^{2} = p^{2} \cdot k^{2}

Thus, b^{2} = p \cdot k^{2}

Now, we have an equation that shows p divides b. So we have reached a contradiction. Since a and b were assumed to be relatively prime, they can't both possibly have p as a divisor. Therefore, no such integers a and b exist.

[BSMSMSTMSPHD]

In general, I think that the nth root of any natural number that is not itself a perfect nth power is an irrational number.

[StatusX]

In general, I think that the nth root of any natural number that is not itself a perfect nth power is an irrational number.

This is true, and in fact the nth root of any rational number which doesn't have a perfect nth power in both the numerator and denominator (when written in lowest terms) is irrational. To see this (as I mentioned in another thread (http://www.physicsforums.com/showthread.php?t=129729)), just note that when you take the nth power of a rational number, you get a rational number with the above property. So conversely, a number which doesn't have this property can't be the nth power of a rational number.