Is x Irrational if x² Is Irrational?

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SUMMARY

The discussion centers on proving that if x² is irrational, then x must also be irrational. The proof employs contradiction, starting with the assumption that x is rational, expressed as x = a/b, where a and b are integers. This leads to the conclusion that x² = a²/b², which is a rational number, contradicting the initial premise that x² is irrational. Thus, it is established that x must indeed be irrational.

PREREQUISITES
  • Understanding of rational and irrational numbers
  • Familiarity with proof by contradiction
  • Basic knowledge of algebraic expressions
  • Concept of integers and their properties
NEXT STEPS
  • Study the principles of proof by contradiction in mathematics
  • Explore the properties of rational and irrational numbers
  • Learn about the implications of integer ratios in number theory
  • Investigate other proofs involving irrational numbers, such as the proof of the irrationality of √2
USEFUL FOR

Students of mathematics, particularly those studying number theory and proof techniques, as well as educators looking for clear examples of irrationality proofs.

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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