Irrationality of sqrt(23)

Problem Statement

Prove that $\sqrt{23}$ 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 $\sqrt{23}$ is rational. Then it can be written as:

\sqrt{23} = \frac{p}{q}

where $p, q \in \mathbb{Z}$ , $q \neq 0$ , and $\gcd(p,q) = 1$ (the fraction is in lowest terms).

Step 2: Square both sides

\begin{aligned} 23 &= \frac{p^2}{q^2} \\ p^2 &= 23q^2 \end{aligned}

Step 3: Deduce divisibility

From $p^2 = 23q^2$ , we see that $p^2$ is divisible by 23.

Since 23 is prime, if $23 \mid p^2$ , then $23 \mid p$ (by fundamental theorem of arithmetic).

Step 4: Substitute and derive contradiction

Since $23 \mid p$ , write $p = 23k$ for some integer $k$ . Substituting into $p^2 = 23q^2$ :

\begin{aligned} (23k)^2 &= 23q^2 \\ 529k^2 &= 23q^2 \\ 23k^2 &= q^2 \end{aligned}

This shows $q^2$ is divisible by 23. Since 23 is prime, $23 \mid q$ .

Step 5: Establish contradiction

We have shown that both $p$ and $q$ are divisible by 23.

This contradicts our assumption that $\gcd(p,q) = 1$ .

Conclusion

Therefore, $\sqrt{23}$ cannot be expressed as a fraction of integers.

Hence, $\sqrt{23}$ is irrational.

\hfill $\square$

Takeaways

Standard Irrationality Template: Assume rational form $\frac{p}{q}$ in lowest terms, square to get $p^2 = nq^2$ , show both $p$ and $q$ divisible by prime factor, contradict lowest terms assumption
Prime Divisibility: If prime $p$ divides $a^2$ , then $p$ divides $a$ (key property from fundamental theorem of arithmetic)
Lowest Terms: Always start with $\gcd(p,q) = 1$ to enable the contradiction via common factors
Generalization: This method proves $\sqrt{p}$ irrational for any prime $p$

Further Readings

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

HSC Proofs: https://vumaths.com/booklets/hsc-proofs/
HSC Combinatorics: https://vumaths.com/booklets/hsc-combinatorics/
HSC Vectors: https://vumaths.com/booklets/hsc-vectors/

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