Prove If x^2 is irrational then x is irrational

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In summary, the homework statement is trying to 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?
  • #1
basil32
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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?
 
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  • #2
basil32 said:

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 [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?
 
  • #3
Curious3141 said:
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)
 
  • #4
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.
 
  • #5
Curious3141 said:
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?
 
  • #6
basil32 said:
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.
 
  • #7
basil32 said:
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?
 
  • #8
basil32 said:
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
 
  • #9
Thanks everybody!
 

1. What is the statement "Prove If x^2 is irrational then x is irrational" trying to prove?

The statement is trying to prove that if the square of a number is irrational, then the number itself must also be irrational.

2. How does one prove this statement?

This statement can be proven using a proof by contradiction. This means assuming the opposite of what we want to prove, and then showing that it leads to a contradiction, thereby proving the original statement.

3. Can you provide an example to illustrate this statement?

Sure, let's take the number √2 as an example. We know that √2 is irrational because it cannot be expressed as a ratio of two integers. Now, if we square √2, we get 2, which is also irrational. Therefore, based on the statement, we can say that if x^2 is irrational (in this case, 2), then x (in this case, √2) must also be irrational.

4. Is it possible for the statement to be false?

No, the statement is always true. It is a well-known mathematical fact and has been proven by many mathematicians.

5. What are the implications of this statement in mathematics?

This statement has important implications in the field of number theory and algebra. It helps us understand the relationship between irrational numbers and their squares, and also provides a powerful tool for proving other mathematical theorems.

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