Revisiting irrational numbers

A rational number is one that can be written as $\dfrac{p}{q}$, where $p, q$ are integers and $q \ne 0$. An irrational number is a real number that cannot be written this way. The classical examples are $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$, $\pi$, and $e$. In this section we revisit why such numbers exist and learn the standard proof technique: proof by contradiction.

The key lemma

Lemma. Let $p$ be a prime number. If $p$ divides $a^2$, where $a$ is a positive integer, then $p$ also divides $a$.

This is a direct consequence of the Fundamental Theorem of Arithmetic. Write $a$ as a product of primes: $a = q_1 q_2 \cdots q_n.$ Then $a^2 = q_1^2 q_2^2 \cdots q_n^2.$ By the uniqueness part of FTA, the prime factorisation of $a^2$ uses exactly the primes that appear in $a$. So if $p \mid a^2$, then $p$ must equal some $q_i$, i.e., $p \mid a$.

This is a one-line consequence of FTA but it does all the heavy lifting in the irrationality proofs.

Theorem: $\sqrt{2}$ is irrational

Theorem. $\sqrt{2}$ is irrational.

Proof by contradiction. Suppose, on the contrary, that $\sqrt{2}$ is rational. Then we can write $\sqrt{2} = \frac{p}{q},$ where $p, q$ are integers, $q \ne 0$, and $\gcd(p, q) = 1$ (we have cancelled all common factors).

Squaring: $2 = \dfrac{p^2}{q^2}$, so $p^2 = 2 q^2$.

Hence $2 \mid p^2$. By the lemma (with prime $p = 2$), $2 \mid p$. So $p = 2m$ for some integer $m$. Substituting: $(2m)^2 = 2 q^2 \implies 4 m^2 = 2 q^2 \implies q^2 = 2 m^2.$ So $2 \mid q^2$, and again by the lemma, $2 \mid q$.

But then both $p$ and $q$ are divisible by $2$, contradicting $\gcd(p,q) = 1$. The contradiction means our assumption was false. Hence $\sqrt{2}$ is irrational. $\blacksquare$

The same script proves that $\sqrt{3}$, $\sqrt{5}$, indeed $\sqrt{p}$ for any prime $p$, is irrational. Just replace $2$ by $p$ throughout. The lemma needs $p$ to be prime; for composites it can fail ($4 \mid 4 \cdot 1 = 2 \cdot 2$ but $4 \nmid 2$).

Operations: rationals and irrationals

A few useful facts:

• (rational) $\pm$ (rational) $=$ rational.
• (rational) $\pm$ (irrational) $=$ irrational. Proof. If $r + s = t$ with $r, t$ rational and $s$ irrational, then $s = t - r$ is rational , contradiction.
• (non-zero rational) $\times$ (irrational) $=$ irrational.
• (irrational) $\pm$ (irrational) may be rational or irrational. Example: $\sqrt{2} + (-\sqrt{2}) = 0$ is rational, but $\sqrt{2} + \sqrt{3}$ is irrational.

These let us conclude things like "$3 + 2\sqrt{5}$ is irrational" without redoing the full contradiction script.

Worked examples

Example 1. Prove that $\sqrt{5}$ is irrational.

Suppose $\sqrt{5} = p/q$ with $\gcd(p, q) = 1$. Then $p^2 = 5 q^2$, so $5 \mid p^2$, hence (lemma) $5 \mid p$. Write $p = 5m$. Substituting, $25 m^2 = 5 q^2$, so $q^2 = 5 m^2$, hence $5 \mid q$. But then $5 \mid \gcd(p, q) = 1$ , contradiction. So $\sqrt{5}$ is irrational. $\blacksquare$

Example 2. Show that $3 + 2 \sqrt{5}$ is irrational.

Suppose $3 + 2\sqrt{5} = r$ with $r$ rational. Then $2 \sqrt{5} = r - 3$, so $\sqrt{5} = \dfrac{r - 3}{2}$. The right side is rational (rationals are closed under subtraction and division by a non-zero rational). But $\sqrt{5}$ is irrational (Example 1). Contradiction. So $3 + 2\sqrt{5}$ is irrational. $\blacksquare$

Example 3. Show that $7 \sqrt{3}$ is irrational.

If $7 \sqrt{3}$ were rational, say $7 \sqrt{3} = r$, then $\sqrt{3} = r / 7$, which is rational. But $\sqrt{3}$ is irrational , contradiction.

Example 4. Prove that $\sqrt{2} + \sqrt{3}$ is irrational.

Suppose $\sqrt{2} + \sqrt{3} = r$ with $r$ rational. Then $\sqrt{3} = r - \sqrt{2}$. Squaring, $3 = r^2 - 2 r \sqrt{2} + 2 \implies 2 r \sqrt{2} = r^2 - 1 \implies \sqrt{2} = \frac{r^2 - 1}{2 r}$ (assuming $r \ne 0$; if $r = 0$ then $\sqrt{2} + \sqrt{3} = 0$ which is false). The right side is rational, but $\sqrt{2}$ is irrational. Contradiction.

Example 5. Show that $\dfrac{1}{\sqrt{2}}$ is irrational.

$\dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2} = \dfrac{1}{2} \cdot \sqrt{2}$. This is (non-zero rational) $\times$ (irrational), hence irrational.

Try it yourself

1. Prove that $\sqrt{3}$ is irrational.
2. Prove that $\sqrt{7}$ is irrational.
3. Prove that $2 + \sqrt{2}$ is irrational.
4. Prove that $5 - 2\sqrt{3}$ is irrational.
5. Prove that $\dfrac{1}{\sqrt{5}}$ is irrational.
6. If $a$ and $b$ are rationals with $a + b\sqrt{2} = 0$, show that $a = 0 = b$.
7. Can the sum of two irrational numbers be rational? Give one example.
8. Prove that $\sqrt{2} \cdot \sqrt{8}$ is rational. Reconcile with "irrational $\times$ irrational".
9. Show that $3 \sqrt{7} - 2$ is irrational.
10. Prove that $\sqrt{6}$ is irrational. (Hint: assume rational, square, use FTA on the prime $2$ , or $3$.)

Pitfalls / Insight

• Always start "suppose, on the contrary, $\sqrt{n} = p/q$ with $\gcd(p, q) = 1$". Without coprimality the proof collapses.
• The lemma needs $p$ to be prime. Trying it for $p = 4$ or $p = 6$ is wrong.
• Irrational $+$ irrational is not automatically irrational. Always reduce to a known irrational like $\sqrt{2}$ or $\sqrt{p}$.

Insight. Proof by contradiction has a beautiful shape: assume the opposite of what you want, follow the logic strictly, and let the universe tear itself apart. The contradiction proves your original claim. This pattern returns again and again in geometry, algebra, and beyond.