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Irrationality of Non-Perfect-Square Roots

Authors
  • avatar
    Name
    Vu Hung
    Twitter

Problem Statement

Let aa be a positive integer. If aa is not a perfect square, prove that a\sqrt{a} is irrational.


Hints

Attempt the proof independently first. Focus on the key theorem, algebraic transformation, or contradiction setup that links the hypothesis to the target conclusion.


Solutions

Proof by Contradiction

Step 1: Assume the negation

Assume, for contradiction, that a\sqrt{a} is rational. Then we can write

a=pq\sqrt{a} = \frac{p}{q}

where p,qZp, q \in \mathbb{Z}, q0q \neq 0, and gcd(p,q)=1\gcd(p,q) = 1.

Step 2: Square both sides

a=p2q2p2=aq2\begin{aligned} a &= \frac{p^2}{q^2} \\ p^2 &= aq^2 \end{aligned}

Step 3: Compare prime factorizations

By the Fundamental Theorem of Arithmetic, every positive integer has a unique prime factorization.

When an integer is squared, every prime exponent in its factorization becomes even. So:

  • in p2p^2, every prime appears with an even exponent;
  • in q2q^2, every prime appears with an even exponent.

Now the equation

p2=aq2p^2 = aq^2

shows that aq2aq^2 must also have only even prime exponents.

Since q2q^2 already contributes only even exponents, this is possible only if every prime appearing in aa also has an even exponent.

Step 4: Derive the contradiction

But if every prime in the factorization of aa has an even exponent, then aa is a perfect square.

This contradicts the hypothesis that aa is not a perfect square.

Conclusion

Therefore, our assumption was false, and a\sqrt{a} must be irrational.

\hfill \square


Takeaways

  • Parity of Prime Exponents: A perfect square has only even exponents in its prime factorization
  • Why the Method Works: Squaring forces all prime exponents in p2p^2 and q2q^2 to be even, so the same must be true for aa
  • Generalization: This extends proofs like $\sqrt{2}$ is irrational'' or 23\sqrt{23} is irrational'' to any positive integer that is not a square
  • Key Tool: The argument relies on the Fundamental Theorem of Arithmetic and uniqueness of prime factorization

Further Readings

If you found this proof interesting, be sure to check out these relevant HSC booklets to sharpen your reasoning skills:


Connect with me

If you're eager for more HSC Maths insights, be sure to check out my YouTube - HSC Maths Extension 1+2. For deeper dives and regular tips, join my Website - Vu's Maths Hub. Let's tackle these challenging math problems together! You can also catch my daily math content on GitHub.