Irrationality of sqrt(23)

This classic proof by contradiction explores the irrationality of $\sqrt{23}$, a fundamental concept in the HSC Extension 1 Proofs topic. By assuming a rational representation in lowest terms and leveraging prime divisibility, you'll master a technique that generalises to the irrationality of $\sqrt{p}$ for any prime $p$.

Problem Statement

Prove that $\sqrt{23}$ is irrational.

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Hints

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

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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

Squaring removes the radical and gives us an equation connecting the squares of $p$ and $q$:

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

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.

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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 $p$

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Further Readings

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