Is x Irrational if x² Is Irrational?

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Homework Help Overview

The discussion revolves around the proof that if \( x^2 \) is irrational, then \( x \) must also be irrational. The subject area pertains to number theory and properties of rational and irrational numbers.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore proof by contradiction as a potential method. Questions arise regarding the initial assumptions about \( x \) being rational and how to express it in terms of integers. There is uncertainty about how to proceed to reach a contradiction.

Discussion Status

Some participants have provided guidance on the structure of the proof and the implications of assuming \( x \) is rational. Multiple interpretations of the proof steps are being explored, with no explicit consensus reached yet.

Contextual Notes

Participants note the importance of defining rational numbers clearly, emphasizing that \( a \) and \( b \) must be integers and \( b \) cannot be zero. There is an ongoing discussion about the implications of these definitions on the proof.

kmeado07
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Homework Statement



Prove that if x^2 is irrational then x must be irrational.

Homework Equations





The Attempt at a Solution



Maybe do proof by contradiction. I'm not really sure where to start.
 
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Proof by contradiction sounds good. What if x is rational?
 
"Suppose x is rational. Then x= __________"
 
so i let x= a/b

then obviously x^2 = a^2/b^2

im not sure how to continue to reach the contradiction
 
Assume x^2 is irrational and x is NOT irrational. You've shown that in such a case, x^2 is rational. That's a contradiction. x^2 can't be both rational and irrational. Therefore x must be irrational.
 
kmeado07 said:
so i let x= a/b

then obviously x^2 = a^2/b^2

im not sure how to continue to reach the contradiction

It is not enough just to say "let x= a/b" without saying what a and b are. A number is rational if and only if it can be expressed as a ratio of integers: a/b where a and b are integers (and b is not 0). If x is rational, the x= a/b where a and b are integers. You arrive at the fact that x2= a2/b2, also a ratio of integers. That itself contradicts the hypothesis, that x2 is irrational.
 

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