Prove If x^2 is irrational then x is irrational


by basil32
Tags: irrational, prove
basil32
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Jan6-12, 02:29 AM
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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?
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Curious3141
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Jan6-12, 04:44 AM
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Quote Quote by basil32 View Post
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?
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?
basil32
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Jan6-12, 06:49 AM
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Quote Quote by Curious3141 View Post
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?
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)

Curious3141
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Jan6-12, 06:59 AM
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Prove If x^2 is irrational then x is irrational


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.
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Jan6-12, 07:09 AM
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Quote Quote by Curious3141 View Post
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.
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?
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Jan6-12, 08:25 AM
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Quote Quote by basil32 View Post
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?
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.
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Jan6-12, 08:39 AM
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Quote Quote by basil32 View Post
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?
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?
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Jan6-12, 10:39 AM
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Quote Quote by basil32 View Post
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?
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
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Jan6-12, 10:44 AM
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