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

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

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

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quasar987

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Yes, you're both right.

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

Suppose there exists a prime p such that [tex] \sqrt{p} \in \mathbb{Q} [/tex]

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

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.

[tex] b \cdot \sqrt{p} = a [/tex]

[tex] b^{2} \cdot p = a^{2} [/tex]

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

[tex] b^{2} \cdot p = {(pk)}^{2} = p^{2} \cdot k^{2} [/tex]

Thus, [tex] b^{2} = p \cdot k^{2} [/tex]

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.

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

Suppose there exists a prime p such that [tex] \sqrt{p} \in \mathbb{Q} [/tex]

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

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.

[tex] b \cdot \sqrt{p} = a [/tex]

[tex] b^{2} \cdot p = a^{2} [/tex]

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

[tex] b^{2} \cdot p = {(pk)}^{2} = p^{2} \cdot k^{2} [/tex]

Thus, [tex] b^{2} = p \cdot k^{2} [/tex]

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.

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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.BSMSMSTMSPHD said:

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