NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.2#

Since $225 > 135$, we apply Euclid’s division lemma to $225$ and $135$:

$$225 = 135 \times 1 + 90$$

Now apply to $135$ and $90$:

$$135 = 90 \times 1 + 45$$

Now apply to $90$ and $45$:

$$90 = 45 \times 2 + 0$$

The remainder is $0$, so the process stops.

The last non-zero remainder is $\mathbf{45}$.

∴ $\text{HCF}(135, 225) = \mathbf{45}$

Since $38220 > 196$, we apply Euclid’s division lemma to $38220$ and $196$:

$$38220 = 196 \times 195 + 0$$

The remainder is $0$, so the process stops immediately.

The last non-zero remainder is $\mathbf{196}$.

∴ $\text{HCF}(196, 38220) = \mathbf{196}$

Since $867 > 255$, we apply Euclid’s division lemma to $867$ and $255$:

$$867 = 255 \times 3 + 102$$

Now apply to $255$ and $102$:

$$255 = 102 \times 2 + 51$$

Now apply to $102$ and $51$:

$$102 = 51 \times 2 + 0$$

The remainder is $0$, so the process stops.

The last non-zero remainder is $\mathbf{51}$.

∴ $\text{HCF}(867, 255) = \mathbf{51}$

Let $a$ be any positive integer and $b = 6$.

By Euclid’s division algorithm, there exist integers $q$ and $r$ such that:

$$a = 6q + r, \quad \text{where } 0 \leq r < 6$$

So the possible values of $r$ are $0, 1, 2, 3, 4, 5$.

This gives us six cases:

Value of $r$	Form of $a$	Even or Odd?
$r = 0$	$a = 6q$	Even (divisible by 2)
$r = 1$	$a = 6q + 1$	Odd
$r = 2$	$a = 6q + 2$	Even
$r = 3$	$a = 6q + 3$	Odd
$r = 4$	$a = 6q + 4$	Even
$r = 5$	$a = 6q + 5$	Odd

Since $a$ is an odd integer, it cannot be of the form $6q$, $6q+2$, or $6q+4$ (as these are all even).

Therefore, any positive odd integer must be of the form $6q + 1$, $6q + 3$, or $6q + 5$, for some non-negative integer $q$. $\hspace{1cm}$ [Hence Proved]

The maximum number of columns is the HCF of $616$ and $32$.

We apply Euclid’s division algorithm:

$$616 = 32 \times 19 + 8$$$$32 = 8 \times 4 + 0$$

The remainder is $0$, so the process stops.

$\text{HCF}(616, 32) = \mathbf{8}$

∴ The maximum number of columns in which they can march is 8.

Let $a$ be any positive integer and $b = 3$.

By Euclid’s division algorithm:

$$a = 3q + r, \quad \text{where } 0 \leq r < 3$$

So $r$ can be $0$, $1$, or $2$.

Case 1: $r = 0$

$$a = 3q$$

$$a^2 = 9q^2 = 3(3q^2) = 3m, \quad \text{where } m = 3q^2$$

Case 2: $r = 1$

$$a = 3q + 1$$

$$a^2 = (3q+1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1 = 3m + 1, \quad \text{where } m = 3q^2 + 2q$$

Case 3: $r = 2$

$$a = 3q + 2$$

$$a^2 = (3q+2)^2 = 9q^2 + 12q + 4 = 9q^2 + 12q + 3 + 1 = 3(3q^2 + 4q + 1) + 1 = 3m + 1$$

where $m = 3q^2 + 4q + 1$

In all cases, $a^2$ is either of the form $3m$ or $3m + 1$.

∴ The square of any positive integer is either of the form $3m$ or $3m + 1$. $\hspace{1cm}$ [Hence Proved]

Let $a$ be any positive integer and $b = 3$.

By Euclid’s division algorithm:

$$a = 3q + r, \quad \text{where } 0 \leq r < 3$$

So $r$ can be $0$, $1$, or $2$.

Case 1: $r = 0$, i.e., $a = 3q$

$$a^3 = (3q)^3 = 27q^3 = 9(3q^3) = 9m, \quad \text{where } m = 3q^3$$

Case 2: $r = 1$, i.e., $a = 3q + 1$

$$a^3 = (3q+1)^3 = 27q^3 + 27q^2 + 9q + 1$$

$$= 9(3q^3 + 3q^2 + q) + 1 = 9m + 1, \quad \text{where } m = 3q^3 + 3q^2 + q$$

Case 3: $r = 2$, i.e., $a = 3q + 2$

$$a^3 = (3q+2)^3 = 27q^3 + 54q^2 + 36q + 8$$

$$= 9(3q^3 + 6q^2 + 4q) + 8 = 9m + 8, \quad \text{where } m = 3q^3 + 6q^2 + 4q$$

In all cases, $a^3$ is of the form $9m$, $9m+1$, or $9m+8$.

∴ The cube of any positive integer is of the form $9m$, $9m+1$, or $9m+8$. $\hspace{1cm}$ [Hence Proved]

Proof by contradiction.

Assume $\sqrt{2}$ is rational. Then there exist integers $a$ and $b$ (with $b \neq 0$) such that:

$$\sqrt{2} = \dfrac{a}{b}$$

where $a$ and $b$ are co-prime (i.e., $\text{HCF}(a,b) = 1$).

Squaring both sides:

$$2 = \dfrac{a^2}{b^2} \implies a^2 = 2b^2$$

This means $a^2$ is even, so $a$ must be even (since if $a$ were odd, $a^2$ would be odd).

Let $a = 2c$ for some integer $c$. Then:

$$a^2 = 4c^2 \implies 2b^2 = 4c^2 \implies b^2 = 2c^2$$

This means $b^2$ is even, so $b$ must also be even.

But if both $a$ and $b$ are even, they have a common factor of $2$, which contradicts our assumption that $\text{HCF}(a,b) = 1$.

∴ Our assumption is wrong. Hence, $\sqrt{2}$ is irrational.

Proof by contradiction.

Assume $\sqrt{3}$ is rational. Then there exist co-prime integers $a$ and $b$ ($b \neq 0$) such that:

$$\sqrt{3} = \dfrac{a}{b}$$

Squaring both sides:

$$3b^2 = a^2$$

This means $3$ divides $a^2$. Since $3$ is prime, $3$ divides $a$.

Let $a = 3c$. Then:

$$3b^2 = 9c^2 \implies b^2 = 3c^2$$

This means $3$ divides $b^2$, so $3$ divides $b$.

But then $3$ divides both $a$ and $b$, contradicting $\text{HCF}(a,b) = 1$.

∴ $\sqrt{3}$ is irrational.

Proof by contradiction.

Assume $\sqrt{5}$ is rational. Then there exist co-prime integers $a$ and $b$ ($b \neq 0$) such that:

$$\sqrt{5} = \dfrac{a}{b}$$

Squaring both sides:

$$5b^2 = a^2$$

This means $5$ divides $a^2$. Since $5$ is prime, $5$ divides $a$.

Let $a = 5c$. Then:

$$5b^2 = 25c^2 \implies b^2 = 5c^2$$

This means $5$ divides $b^2$, so $5$ divides $b$.

But then $5$ divides both $a$ and $b$, contradicting $\text{HCF}(a,b) = 1$.

∴ $\sqrt{5}$ is irrational.

We have already proved that $\sqrt{5}$ is irrational.

Proof by contradiction.

Assume $3 + 2\sqrt{5}$ is rational. Then there exist integers $a$ and $b$ ($b \neq 0$) such that:

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

Rearranging:

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

Since $a$, $b$ are integers, $\dfrac{a-3b}{2b}$ is rational, which means $\sqrt{5}$ is rational.

But this contradicts the fact that $\sqrt{5}$ is irrational.

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