10th Maths 1.3

Exercise 1.3

Q1. Prove that $\sqrt{5}$ $sqrt5$ is irrational.

Sol. :

Let

\sqrt{5}

$sqrt5$ is a rational number. Then we can take

a

$a$ and

b

$b$ two co-prime non-zero integers.

Such that

\sqrt{5} = \frac{a}{b}

$sqrt5=frac(a)(b)$ , where

b \neq 0

$bne0$

b \sqrt{5} = a

$bsqrt5=a$

5 b^{2} = a^{2}

$5b^2=a^2$

Therefore,

5

$5$ divides

a^{2}

$a^2$ so we can take

a = 5 c

$a=5c$ , where

c

$c$ is an integer.

5 b^{2} = {(5 c)}^{2}

$5b^2=(5c)^2$

5 b^{2} = 25 c^{2}

$5b^2=25c^2$

b^{2} = 5 c^{2}

$b^2=5c^2$

b^{2}

$b^2$ , as well as

b

$b$ , also is divisible by

5

$5$ .

a

$a$ and

b

$b$ both have at least one common factor.

But this is contradiction arisen because of our incorrect assumption that

\sqrt{5}

$sqrt5$ is a rational number.

Thus, we conclude that

\sqrt{5}

$sqrt5$ is irrational.

Proved.

Q2. Prove that $3 + 2 \sqrt{5}$ $3+2sqrt5$ is irrational.

Sol. :

Let

3 + 2 \sqrt{5}

$3+2sqrt5$ is a rational number. Then we can take

a

$a$ and

b

$b$ two coprime nonzero integers.

Such that

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

$3+2sqrt5=frac(a)(b)$ , where

b \neq 0

$bne0$

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

$2sqrt5=frac(a)(b)-3$

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

$sqrt5=frac(a)(2b)-frac(3)(2)$

Since

a

$a$ and

b

$b$ are two integers where

b \neq 0

$bne0$ , so

(\frac{a}{2 b} - \frac{3}{2})

$(frac(a)(2b)-frac(3)(2))$ is rational, so

$sqrt5$ is rational.

But this contradiction has arisen because of our incorrect assumption that

$3+2sqrt5$ is rational.

So, we conclude that

$3+2sqrt5$ is irrational.

Proved.

Q3. Prove that following are irrationals :

i) $frac(1)(sqrt2)$ ii) $7sqrt5$ iii) $6+sqrt2$

Sol. :

i) Let

$frac(1)(sqrt2)$ is a rational number.

Then we can take

$a$ and

$b$ two co-prime non-zero integers

such that

$frac(1)(sqrt2)=frac(a)(b)$ , where

$bne0$

$sqrt2=frac(b)(a)$ ,

since

$b$ and

$a$ are non-zero integers

$b/a$ is a rational number and so

$sqrt2$ is also a rational number.

But it is a contradiction to the fact that

$sqrt2$ is an irrational number. It is arisen due to our incorrect assumption.

So, we conclude that

$frac(1)(sqrt2)$ is an irrational number.

Proved.

ii) Let

$7sqrt5$ is a rational number.

Then we can take

$a$ and

$b$ two co-prime non-zero integers

such that

$7sqrt5=frac(a)(b)$ , where

$bne0$

$sqrt5=frac(b)(7a)$ ,

since

$a, b$ and

$7$ are non-zero integers

$b/(7a)$ is a rational number and so

$sqrt5$ is also a rational number.

But it is a contradiction to the fact that

$sqrt5$ is an irrational number. It is arisen due to our incorrect assumption.

So, we conclude that

$7sqrt5$ is an irrational number.

Proved.

iii) Let

$6+sqrt2$ is a rational number.

Then we can take

$a$ and

$b$ two co-prime non-zero integers

such that

$6+sqrt2=frac(a)(b)$ , where

$bne0$

$sqrt2=frac(b)(a)-6$ ,

since

$a, b$ and

$6$ are non-zero integers

$b/a-6$ is a rational number and so

$sqrt2$ is also a rational number.

But it is a contradiction to the fact that

$sqrt2$ is an irrational number. It is arisen due to our incorrect assumption.

So, we conclude that

$6+sqrt2$ is an irrational number.

Proved.

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10th Maths 1.3

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Exercise 1.3

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