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!