Is the square root of a prime number always going to be irrational?

[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 let's 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 https://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.