NCERT Class 10 Mathematics — Real Numbers, Exercise 1.2. All 3 questions solved.

Every question here is a proof that some number is irrational — that it cannot be written as $\dfrac{p}{q}$ with $p$ and $q$ integers and $q \ne 0$. The method is proof by contradiction: assume the number is rational, follow the consequences, and arrive at something impossible. Two facts from the chapter do the work:

• Theorem 1.2. If a prime $p$ divides $a^2$, then $p$ divides $a$ (for a positive integer $a$).
• Theorem 1.3. $\sqrt2$ is irrational — and, by the same argument, so is $\sqrt p$ for any prime $p$.

A third fact is used silently throughout: adding, subtracting, multiplying or dividing (by a non-zero number) two rationals always gives a rational.

Key insight. Only question 1 needs the full argument from scratch. Every other number in the exercise is a rational number combined with $\sqrt5$ or $\sqrt2$. Assume it is rational, rearrange to get $\sqrt5$ or $\sqrt2$ on its own, and the other side is built from rationals — so $\sqrt5$ or $\sqrt2$ would be rational, which is already known to be false.

Question 1

Prove that $\sqrt5$ is irrational.

Solution. Suppose, to the contrary, that $\sqrt5$ is rational. Then we can write

$$\sqrt5 = \frac{a}{b},$$

where $a$ and $b$ are integers, $b \ne 0$, and $a$ and $b$ are coprime (have no common factor other than $1$). We may insist on coprime because any fraction can be reduced to its lowest terms by cancelling common factors — and that insistence is what the contradiction will break.

Multiply both sides by $b$ and square:

$$b\sqrt5 = a \quad\Longrightarrow\quad 5b^2 = a^2.$$

So $5$ divides $a^2$. Since $5$ is prime, Theorem 1.2 says $5$ divides $a$. Write $a = 5c$ for some integer $c$ and substitute:

$$5b^2 = 25c^2 \quad\Longrightarrow\quad b^2 = 5c^2.$$

Now $5$ divides $b^2$, and so, by Theorem 1.2 again, $5$ divides $b$.

So $5$ divides both $a$ and $b$ — but $a$ and $b$ were chosen with no common factor other than $1$. This contradiction came from assuming $\sqrt5$ is rational, so that assumption is false.

Question 2

Prove that $3 + 2\sqrt5$ is irrational.

Solution. Suppose, to the contrary, that $3 + 2\sqrt5$ is rational. Then there are coprime integers $a$ and $b$, with $b \ne 0$, such that

$$3 + 2\sqrt5 = \frac{a}{b}.$$

Isolate the root. Subtract $3$ and divide by $2$:

$$2\sqrt5 = \frac{a}{b} – 3 = \frac{a – 3b}{b} \quad\Longrightarrow\quad \sqrt5 = \frac{a – 3b}{2b}.$$

Since $a$ and $b$ are integers, $a – 3b$ and $2b$ are integers, and $2b \ne 0$. So the right-hand side is rational, which would make $\sqrt5$ rational.

But question 1 proved that $\sqrt5$ is irrational. The contradiction shows the assumption was false.

Question 3

Prove that the following are irrationals:

(i) $\dfrac{1}{\sqrt2}$    (ii) $7\sqrt5$    (iii) $6 + \sqrt2$

Solution. The same pattern three times: assume the number equals $\dfrac{a}{b}$ with $a, b$ coprime integers and $b \ne 0$, then solve for the root.

(i) Suppose $\dfrac{1}{\sqrt2} = \dfrac{a}{b}$. The number $\dfrac{1}{\sqrt2}$ is not zero, so $a \ne 0$ and we may take reciprocals:

$$\sqrt2 = \frac{b}{a}.$$

With $b$ and $a$ integers and $a \ne 0$, this says $\sqrt2$ is rational, contradicting Theorem 1.3. So $\dfrac{1}{\sqrt2}$ is irrational.

(ii) Suppose $7\sqrt5 = \dfrac{a}{b}$. Dividing by $7$,

$$\sqrt5 = \frac{a}{7b},$$

and $a$ and $7b$ are integers with $7b \ne 0$. So $\sqrt5$ would be rational, contradicting question 1. Hence $7\sqrt5$ is irrational.

(iii) Suppose $6 + \sqrt2 = \dfrac{a}{b}$. Subtracting $6$,

$$\sqrt2 = \frac{a}{b} – 6 = \frac{a – 6b}{b},$$

and $a – 6b$ and $b$ are integers with $b \ne 0$. So $\sqrt2$ would be rational, contradicting Theorem 1.3. Hence $6 + \sqrt2$ is irrational.

Common mistakes

• Question 1, not assuming $a$ and $b$ coprime. Without that assumption, finding that $5$ divides both $a$ and $b$ is no contradiction at all — it just means the fraction was not in lowest terms. The whole proof hangs on this one phrase.
• Question 1, using Theorem 1.2 without $5$ being prime. “$p$ divides $a^2$, so $p$ divides $a$” is false for composite $p$: $4$ divides $6^2 = 36$ but not $6$. Say explicitly that $5$ is prime when you use it.
• Question 1, stopping at “$5$ divides $a$”. That alone contradicts nothing. The contradiction appears only after substituting $a = 5c$ and showing that $5$ divides $b$ as well.
• Question 2, quoting the answer instead of proving it. Writing “rational + irrational is irrational, so $3 + 2\sqrt5$ is irrational” skips exactly what the question asks you to show. The proof is the rearrangement to $\sqrt5 = \dfrac{a – 3b}{2b}$.
• Question 3(i), dividing by $a$ without checking it is non-zero. Taking reciprocals needs $a \ne 0$, which holds because $\dfrac{1}{\sqrt2} \ne 0$. It is one short phrase, and leaving it out leaves a gap in the logic.
• Any question, arguing from decimals. “$\sqrt5 = 2.2360679\ldots$ never ends” is not a proof: a calculator shows only finitely many digits, and some rationals such as $\dfrac17$ also have long decimal expansions. Irrationality is about whether the number can be a fraction, which only an argument like the ones above can settle.

Practise next

• Exercise 1.1 — prime factorisation, HCF and LCM. The uniqueness of prime factorisation practised there is what Theorem 1.2 rests on.
• Chapter 2, Exercise 2.1 — the next chapter, on polynomials and their zeroes.