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Prove If x^2 is irrational then x is irrational

  1. Jan 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove If x^2 is irrational then x is irrational. I can find for example π^2 which is irrational and then π is irrational but I don't know how to approach the proof. Any hint?
     
  2. jcsd
  3. Jan 6, 2012 #2

    Curious3141

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    Try a proof by contradiction. Let's say you have such a rational [itex]x[/itex] where [itex]x^2[/itex] is irrational. Then let [itex]x = \frac{p}{q}[/itex] where p and q are coprime integers (meaning it's a reduced fraction). Now see what form [itex]x^2[/itex] takes. Can you arrive at a contradiction considering that this was supposed to be irrational by the first assumption?
     
  4. Jan 6, 2012 #3
    ok. [itex]x^{2} [/itex] = [itex]\frac{p^{2}}{q^{2}} [/itex]. Now [itex] q^{2}x^{2} = p^{2} \Rightarrow x^{2} \mid p^{2} \Rightarrow x \mid p [/itex] but I can't arrive at [itex]x \mid q[/itex] for the contradiction (when I replace p =xk in the [itex]q^{2}x^{2} = (xk)^{2} [/itex] the x^{2} on both side cancel)
     
  5. Jan 6, 2012 #4

    Curious3141

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    Wouldn't it suffice to observe that [itex]\frac{p^2}{q^2}[/itex] is a reduced rational number since p and q are coprime? Which would imply that [itex]x^2[/itex] is rational as well, which contradicts the original assumption of the irrationality of [itex]x^2[/itex].

    In other words, the negation of the proposition [itex]x^2 \notin \mathbb{Q} \Rightarrow x \notin \mathbb{Q}[/itex] leads to a contradiction. Hence the proposition is true.
     
  6. Jan 6, 2012 #5
    yeah, the observation make sense but I need a lemma which proves that [itex]\frac{p^2}{q^2}[/itex] is reduced form whenever [itex]\frac{p}{q}[/itex] is reduced. How do you do that?
     
  7. Jan 6, 2012 #6

    Curious3141

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    I would've thought that bit's obvious, and would've stated it without proof. If you want to see it more clearly, perhaps express it as [itex]\frac{(p)(p)}{(q)(q)}[/itex]. Neither of the numerator's two factors has any factors in common with either of the denominator's factors (since p and q are coprime by definition), so the fraction is irreducible (nothing to cancel out).

    The only thing I can think of more fundamental than that would be to fully prime-factorise [itex]p^2[/itex] and [itex]q^2[/itex], but this just leads to a more messy yet no more convincing argument.
     
  8. Jan 6, 2012 #7

    Dick

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    Why do you need to show [itex]\frac{p^2}{q^2}[/itex] is in reduced form? It's rational even if it's not in reduced form, isn't it?
     
  9. Jan 6, 2012 #8

    Ray Vickson

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    Why bother? We know [itex] p^2 \mbox{ and } q^2 [/itex] are integers, so [itex] p^2/q^2[/itex] is a ratio of integers, hence a rational number. Who cares if they are coprime?

    RGV
     
  10. Jan 6, 2012 #9
    Thanks everybody!
     
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