Real Numbers — NCERT Solutions, AI Tutor & Practice

We use repeated division by the smallest primes.

(i) 140

$140 = 2 \times 70 = 2 \times 2 \times 35 = 2 \times 2 \times 5 \times 7$

So $140 = 2^2 \times 5 \times 7$.

(ii) 156

$156 = 2 \times 78 = 2 \times 2 \times 39 = 2 \times 2 \times 3 \times 13$

So $156 = 2^2 \times 3 \times 13$.

(iii) 3825

$3825 = 3 \times 1275 = 3 \times 3 \times 425 = 9 \times 5 \times 85 = 9 \times 5 \times 5 \times 17$

So $3825 = 3^2 \times 5^2 \times 17$.

Prime factorisation:

$96 = 2^5 \times 3$

$404 = 2^2 \times 101$

HCF = product of lowest powers of common primes $= 2^2 = 4$.

LCM = product of highest powers of all primes $= 2^5 \times 3 \times 101 = 32 \times 303 = 9696$.

Verification:

$\text{HCF} \times \text{LCM} = 4 \times 9696 = 38784$

$96 \times 404 = 38784$

Since both are equal, the result is verified.

We use the identity for any two positive integers $a$ and $b$:

$\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b$

Here $a = 306$, $b = 657$, HCF $= 9$.

$\text{LCM} = \frac{a \times b}{\text{HCF}} = \frac{306 \times 657}{9}$

$= \frac{201042}{9} = 22338$

So the LCM of 306 and 657 is $22338$.

If a number ends in the digit $0$, it must be divisible by $10$, i.e. by $2 \times 5$. So its prime factorisation must contain both $2$ and $5$.

Now consider $6^n$:

$6^n = (2 \times 3)^n = 2^n \times 3^n$

The only prime factors of $6^n$ are $2$ and $3$. There is no factor of $5$.

By the Fundamental Theorem of Arithmetic, the prime factorisation of a number is unique, so $5$ can never appear in $6^n$.

Therefore $6^n$ can never end with the digit $0$ for any natural number $n$.

A composite number has at least one factor other than 1 and itself.

First number:

$7 \times 11 \times 13 + 13 = 13(7 \times 11 + 1) = 13 \times (77 + 1) = 13 \times 78$

Since it is a product of $13$ and $78$ (both greater than 1), it is a composite number.

Second number:

$7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5(7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1)$

$= 5 \times (1008 + 1) = 5 \times 1009$

Since it is a product of $5$ and $1009$ (both greater than 1), it too is a composite number.

Prime factorisations:

$12 = 2^2 \times 3$

$15 = 3 \times 5$

$21 = 3 \times 7$

HCF = product of the lowest powers of primes common to all three.

The only prime common to all three is $3$ (each has $3^1$).

$\text{HCF} = 3$

LCM = product of the highest powers of every prime that appears.

$\text{LCM} = 2^2 \times 3 \times 5 \times 7 = 4 \times 3 \times 5 \times 7 = 420$

So HCF $= 3$ and LCM $= 420$.

The bells toll together again after a time that is a multiple of all three intervals, i.e. the LCM of 9, 12 and 15.

Prime factorisations:

$9 = 3^2$

$12 = 2^2 \times 3$

$15 = 3 \times 5$

LCM = highest powers of all primes

$\text{LCM} = 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180 \text{ minutes}$

$180$ minutes $= 3$ hours.

Starting from 8:00 a.m., adding 3 hours gives 11:00 a.m.

So the bells will next toll together at 11:00 a.m.

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.