Revisiting Irrational Numbers

What Makes a Number Irrational?

Recall from Section 1: a number is irrational if it cannot be written as $\dfrac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Its decimal expansion goes on forever without ever settling into a repeating block.

Familiar examples are $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$, and $\pi$. In this section we don't just believe these are irrational — we prove it, using a powerful technique called proof by contradiction.

Key Point: To prove a number is irrational, we assume the opposite (that it is rational), and then show this assumption leads to something impossible. Since the assumption breaks, the number must be irrational.

[Board Important] 'Prove that $\sqrt{2}$ is irrational' (and similar) is one of the most repeated board questions. Learn the standard proof structure thoroughly — it is worth 2 to 3 marks and the steps are always the same.

The Key Theorem We Will Use

The irrationality proofs rest on one important theorem that follows from the Fundamental Theorem of Arithmetic.

Theorem: Let $p$ be a prime number. If $p$ divides $a^2$ (where $a$ is a positive integer), then $p$ divides $a$.

Why this is true (the idea)

If $p$ divides $a^2$, then $p$ must appear in the prime factorisation of $a^2$. But the primes of $a^2$ are exactly the primes of $a$, each appearing twice as often. So if $p$ is a factor of $a^2$, it must already be a factor of $a$.

Example

Since $2$ divides $a^2$, it follows that $2$ divides $a$. Likewise, if $3 \mid a^2$ then $3 \mid a$, and if $5 \mid a^2$ then $5 \mid a$.

Key Point: This theorem is the engine behind every irrationality proof in this chapter. State it before you use it in the exam.

[Board Important] The theorem requires $p$ to be prime. It can fail for non-primes — for instance $4 \mid 6^2 = 36$, but $4 \nmid 6$.

Proof that $\sqrt{2}$ is Irrational

This is the model proof. Learn its structure and you can adapt it to $\sqrt{3}$, $\sqrt{5}$, and many others.

To prove: $\sqrt{2}$ is irrational.

Proof (by contradiction):

1. Assume the opposite: Suppose $\sqrt{2}$ is rational. Then we can write $\sqrt{2} = \dfrac{a}{b}$, where $a$ and $b$ are co-prime integers (no common factor other than 1) and $b \neq 0$.
2. Rearrange and square: $\sqrt{2}\, b = a \Rightarrow 2b^2 = a^2$.
3. Apply the theorem: Since $2$ divides $a^2$, by the theorem $2$ divides $a$. So write $a = 2c$ for some integer $c$.
4. Substitute back: $2b^2 = (2c)^2 = 4c^2 \Rightarrow b^2 = 2c^2$. So $2$ divides $b^2$, and hence $2$ divides $b$.
5. The contradiction: Now $2$ divides both $a$ and $b$. But we assumed $a$ and $b$ are co-prime (no common factor). This is impossible.
6. Conclusion: Our assumption was wrong, so $\sqrt{2}$ is irrational. $\blacksquare$

Exam Tip: The contradiction always comes from assuming co-primeness and then finding a shared factor. Always begin by stating '$a$ and $b$ are co-prime'.

Irrationality of Sums and Products

Once we know $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$ are irrational, we can prove that combinations like $5 - \sqrt{3}$ or $3\sqrt{2}$ are also irrational. These proofs are shorter and use two facts you already met in Section 1:

• (rational) ± (irrational) = irrational
• (non-zero rational) × (irrational) = irrational

Method (by contradiction)

To prove, say, $5 - \sqrt{3}$ is irrational:

1. Assume $5 - \sqrt{3}$ is rational, say $= r$.
2. Then $\sqrt{3} = 5 - r$. The right side is a difference of two rationals → rational.
3. So $\sqrt{3}$ would be rational — but we know $\sqrt{3}$ is irrational. Contradiction.
4. Hence $5 - \sqrt{3}$ is irrational.

Key Point: The trick is always to isolate the known irrational (like $\sqrt{3}$) on one side. If the other side is forced to be rational, you have your contradiction.

[Board Important] State clearly which known result you are using (e.g. '$\sqrt{3}$ is irrational'). Examiners expect you to name the fact, not just assume it.

Solved Examples

Example 1: Prove $\sqrt{3}$ is irrational

Solution:

1. Assume $\sqrt{3} = \dfrac{a}{b}$ with $a, b$ co-prime, $b \neq 0$.
2. Then $3b^2 = a^2$, so $3$ divides $a^2$, hence (theorem, $p=3$) $3$ divides $a$. Write $a = 3c$.
3. Substitute: $3b^2 = 9c^2 \Rightarrow b^2 = 3c^2$, so $3$ divides $b^2$, hence $3$ divides $b$.
4. Then 3 divides both $a$ and $b$ — contradicting co-primeness.

Final Answer: $\sqrt{3}$ is irrational.

Takeaway: Identical structure to the $\sqrt{2}$ proof, with 2 replaced by 3.

Example 2: Prove $\sqrt{5}$ is irrational

Solution:

1. Assume $\sqrt{5} = \dfrac{a}{b}$, $a, b$ co-prime, $b \neq 0$.
2. $5b^2 = a^2 \Rightarrow 5 \mid a^2 \Rightarrow 5 \mid a$. Write $a = 5c$.
3. $5b^2 = 25c^2 \Rightarrow b^2 = 5c^2 \Rightarrow 5 \mid b^2 \Rightarrow 5 \mid b$.
4. So 5 divides both $a$ and $b$ — contradiction.

Final Answer: $\sqrt{5}$ is irrational.

Takeaway: Works for the square root of any prime number.

Example 3: Prove $5 - \sqrt{3}$ is irrational

Solution:

1. Assume $5 - \sqrt{3}$ is rational, say $= r$.
2. Then $\sqrt{3} = 5 - r$. Since 5 and $r$ are rational, $5 - r$ is rational.
3. So $\sqrt{3}$ would be rational — contradicting the known fact that $\sqrt{3}$ is irrational.

Final Answer: $5 - \sqrt{3}$ is irrational.

Takeaway: Isolate $\sqrt{3}$; the rest must be rational, giving the contradiction.

Example 4: Prove $3\sqrt{2}$ is irrational

Solution:

1. Assume $3\sqrt{2}$ is rational, say $= r$.
2. Then $\sqrt{2} = \dfrac{r}{3}$. The right side is a rational divided by a non-zero rational → rational.
3. So $\sqrt{2}$ would be rational — contradicting that $\sqrt{2}$ is irrational.

Final Answer: $3\sqrt{2}$ is irrational.

Takeaway: A non-zero rational times an irrational is always irrational.

Example 5: Prove $\dfrac{1}{\sqrt{2}}$ is irrational

Solution:

1. Note $\dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$ (rationalising).
2. Assume $\dfrac{\sqrt{2}}{2}$ is rational, say $= r$. Then $\sqrt{2} = 2r$, which is rational.
3. This contradicts $\sqrt{2}$ being irrational.

Final Answer: $\dfrac{1}{\sqrt{2}}$ is irrational.

Takeaway: Rationalise first if it helps, then isolate the known irrational.

Example 6: Prove $\sqrt{2} + \sqrt{3}$ is irrational

Solution:

1. Assume $\sqrt{2} + \sqrt{3} = r$ (rational).
2. Square both sides: $2 + 2\sqrt{6} + 3 = r^2 \Rightarrow 2\sqrt{6} = r^2 - 5$.
3. So $\sqrt{6} = \dfrac{r^2 - 5}{2}$, which would be rational.
4. But $\sqrt{6}$ is irrational (6 is not a perfect square). Contradiction.

Final Answer: $\sqrt{2} + \sqrt{3}$ is irrational.

Takeaway: Squaring can convert a sum of surds into a single surd, which we then show is irrational.

Example 7: Prove $7\sqrt{5}$ is irrational

Solution:

1. Assume $7\sqrt{5} = r$ (rational).
2. Then $\sqrt{5} = \dfrac{r}{7}$, a rational divided by a non-zero rational → rational.
3. This contradicts $\sqrt{5}$ being irrational.

Final Answer: $7\sqrt{5}$ is irrational.

Takeaway: Same one-line idea as $3\sqrt{2}$ — divide out the rational coefficient.

Example 8: Prove $2 + 3\sqrt{5}$ is irrational

Solution:

1. Assume $2 + 3\sqrt{5} = r$ (rational).
2. Then $3\sqrt{5} = r - 2$, so $\sqrt{5} = \dfrac{r - 2}{3}$, which is rational.
3. This contradicts $\sqrt{5}$ being irrational.

Final Answer: $2 + 3\sqrt{5}$ is irrational.

Takeaway: First subtract the rational, then divide the coefficient, to isolate $\sqrt{5}$.

Example 9: Is $\sqrt{6}$ irrational?

Solution:

1. Assume $\sqrt{6} = \dfrac{a}{b}$, $a, b$ co-prime.
2. $6b^2 = a^2$, so $6 \mid a^2$, which means $2 \mid a^2$ and $3 \mid a^2$; hence $2 \mid a$ and $3 \mid a$, so $6 \mid a$. Write $a = 6c$.
3. $6b^2 = 36c^2 \Rightarrow b^2 = 6c^2$, so $6 \mid b$ too.
4. Then 6 divides both $a$ and $b$ — contradiction.

Final Answer: $\sqrt{6}$ is irrational.

Takeaway: The square root of any non-perfect-square positive integer is irrational.

Example 10: Sum/difference of two irrationals can be rational

Show by an example that the sum of two irrational numbers can be rational.

Solution:

1. Take the two irrational numbers $\sqrt{2}$ and $-\sqrt{2}$ (the second is irrational since it is $-1 \times$ an irrational).
2. Their sum is $\sqrt{2} + (-\sqrt{2}) = 0$, which is rational.

Final Answer: $\sqrt{2} + (-\sqrt{2}) = 0$ is rational.

Takeaway: Unlike (rational)+(irrational), the sum of two irrationals is NOT always irrational — it depends on the numbers.