Exercise 1.2: Real Numbers — NCERT Solutions

We use proof by contradiction.

Suppose, on the contrary, that $\sqrt{2}$ is rational. Then we can write

$\sqrt{2} = \frac{p}{q}$

where $p$ and $q$ are integers, $q \neq 0$, and $p, q$ have no common factor other than 1 (the fraction is in lowest terms).

Squaring both sides:

$2 = \frac{p^2}{q^2} \implies p^2 = 2q^2 \quad (1)$

So $p^2$ is even, which means $p$ is even. Let $p = 2m$ for some integer $m$.

Substitute into (1):

$(2m)^2 = 2q^2 \implies 4m^2 = 2q^2 \implies q^2 = 2m^2$

So $q^2$ is even, which means $q$ is even.

But now both $p$ and $q$ are even, so they have a common factor 2. This contradicts our assumption that they had no common factor other than 1.

Hence our assumption is wrong, and $\sqrt{2}$ is irrational.

We argue by contradiction.

Assume $\sqrt{5}$ is rational. Then

$\sqrt{5} = \frac{p}{q}$

where $p, q$ are integers, $q \neq 0$, and $p, q$ have no common factor other than 1.

Squaring:

$5 = \frac{p^2}{q^2} \implies p^2 = 5q^2 \quad (1)$

So $5$ divides $p^2$. Since $5$ is prime, $5$ divides $p$. Write $p = 5m$.

Substitute into (1):

$(5m)^2 = 5q^2 \implies 25m^2 = 5q^2 \implies q^2 = 5m^2$

So $5$ divides $q^2$, and therefore $5$ divides $q$.

Thus both $p$ and $q$ are divisible by 5, contradicting the fact that they have no common factor other than 1.

Hence our assumption is false, and $\sqrt{5}$ is irrational.

We use contradiction, taking as given that $\sqrt{5}$ is irrational.

Suppose $3 + 2\sqrt{5}$ is rational. Then we can write

$3 + 2\sqrt{5} = \frac{a}{b}$

where $a, b$ are integers and $b \neq 0$.

Rearranging to isolate $\sqrt{5}$:

$2\sqrt{5} = \frac{a}{b} - 3 = \frac{a - 3b}{b}$

$\sqrt{5} = \frac{a - 3b}{2b}$

Since $a$ and $b$ are integers, the right-hand side $\dfrac{a - 3b}{2b}$ is a rational number.

This would make $\sqrt{5}$ rational — but that contradicts the given fact that $\sqrt{5}$ is irrational.

Hence our assumption is wrong, and $3 + 2\sqrt{5}$ is irrational.