Prove If x^2 is irrational then x is irrational

[basil32]

Homework Statement

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?

 

[Curious3141]

Homework Statement

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?

Try a proof by contradiction. Let's say you have such a rational x where x^2 is irrational. Then let x = \frac{p}{q} where p and q are coprime integers (meaning it's a reduced fraction). Now see what form x^2 takes. Can you arrive at a contradiction considering that this was supposed to be irrational by the first assumption?

 

[basil32]

Try a proof by contradiction. Let's say you have such a rational x where x^2 is irrational. Then let x = \frac{p}{q} where p and q are coprime integers (meaning it's a reduced fraction). Now see what form x^2 takes. Can you arrive at a contradiction considering that this was supposed to be irrational by the first assumption?

ok. x^{2} = \frac{p^{2}}{q^{2}}. Now q^{2}x^{2} = p^{2} \Rightarrow x^{2} \mid p^{2} \Rightarrow x \mid p but I can't arrive at x \mid q for the contradiction (when I replace p =xk in the q^{2}x^{2} = (xk)^{2} the x^{2} on both side cancel)

 

[Curious3141]

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

In other words, the negation of the proposition x^2 \notin \mathbb{Q} \Rightarrow x \notin \mathbb{Q} leads to a contradiction. Hence the proposition is true.

 

[basil32]

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

In other words, the negation of the proposition x^2 \notin \mathbb{Q} \Rightarrow x \notin \mathbb{Q} leads to a contradiction. Hence the proposition is true.

yeah, the observation make sense but I need a lemma which proves that \frac{p^2}{q^2} is reduced form whenever \frac{p}{q} is reduced. How do you do that?

 

[Curious3141]

yeah, the observation make sense but I need a lemma which proves that \frac{p^2}{q^2} is reduced form whenever \frac{p}{q} is reduced. How do you do that?

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 \frac{(p)(p)}{(q)(q)}. 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 p^2 and q^2, but this just leads to a more messy yet no more convincing argument.

 

[Dick]

yeah, the observation make sense but I need a lemma which proves that \frac{p^2}{q^2} is reduced form whenever \frac{p}{q} is reduced. How do you do that?

Why do you need to show \frac{p^2}{q^2} is in reduced form? It's rational even if it's not in reduced form, isn't it?

 

[Ray Vickson]

yeah, the observation make sense but I need a lemma which proves that \frac{p^2}{q^2} is reduced form whenever \frac{p}{q} is reduced. How do you do that?

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

RGV

 

[basil32]

Thanks everybody!