Prove by contradiction that $\sqrt{2}$ is an irrational number. - AQA - A-Level Maths: Mechanics - Question 10 - 2018 - Paper 3

Squaring both sides of the equation, we have:

$2 = \frac{a^2}{b^2}$

Multiplying both sides by $b^2$, we conclude:

$a^2 = 2b^2$

From the equation $a^2 = 2b^2$, we see that $a^2$ is even, since it is equal to twice an integer ($2b^2$). If $a^2$ is even, it follows that $a$ must also be even (as the square of an odd number is odd). Thus, we can express $a$ as:

$a = 2k$

for some integer $k$.

Substituting $a = 2k$ back into the equation $a^2 = 2b^2$, we get:

$(2k)^2 = 2b^2$ $4k^2 = 2b^2$ $2k^2 = b^2$

This implies that $b^2$ is also even, which means that $b$ is also even.

Since both $a$ and $b$ are even, it means that they have a common factor of 2. This contradicts our initial assumption that $a$ and $b$ have no common factors.

Therefore, our assumption that $\sqrt{2}$ is rational must be incorrect. Hence, we conclude that $\sqrt{2}$ is indeed an irrational number.