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

  1. Sep 2, 2006 #1
    is the square root of a prime number always going to be irrational? just a random question.
     
  2. jcsd
  3. Sep 2, 2006 #2
    I believe the square root of any natural number that is not a perfect square is irrational.
     
  4. Sep 2, 2006 #3

    quasar987

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    I would say yes, because lets denote our prime p. If [itex]\sqrt{p}=n[/itex], ([itex]n\in\mathbb{N}[/itex]), then it means that p=n² <==> p/n=n (i.e. n divides p ==> p is not prime: a contradiction with the hypothesis). If [itex]\sqrt{p}=m/n[/itex], ([itex]m,n \in \mathbb{N}[/itex], [itex]n\neq 0,1[/itex], 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 [itex]\sqrt{p}[/itex] is irrational.
     
  5. Sep 2, 2006 #4
    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.
     
    Last edited: Sep 2, 2006
  6. Sep 2, 2006 #5
    In general, I think that the nth root of any natural number that is not itself a perfect nth power is an irrational number.
     
  7. Sep 2, 2006 #6

    StatusX

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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 thread), 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.
     
    Last edited: Sep 2, 2006
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