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$\sf\bold \red{Question :- }$
Prove that the following are irrational

$\sf(i) \large \big \: \frac{1}{ \sqrt{2} }$
$\sf(ii) \large \: 7 \sqrt{5}$
$\sf(iii) \large \: 6 + \sqrt{2}$
$\sf\bold \red{Note :- }$
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Answer 1

Step-by-step explanation:

(i) Prove that 1/√2 is irrational.

Let's say that 1/√2 is rational and can be written in the form a/b

1/√2 = a/b

√2 = a/b

We know that √2 is irrational number & a/b is rational. But irrational number can't be equal to rational number; which means our assumption is wrong. 1/√2 is a irrational number.

How √2 is a irrational number?

Let's say that √2 is rational number and a & b are coprime & b ≠ 0.

So,

√2 = a/b

b√2 = a

On squaring on both sides, we get

2b² = a² (where 2 divides a², 2 divide a)

So, we can write it lik: a = 2c where c is integer. Now, substitute the value of a = 2c in 2b² = a².

→ 2b² = (2c)²

→ 2b² = 4c² (2 divides b², 2 divides b)

{ Let 2 be a prime number, if 2 divides b² or a² then 2 divides b or a, where b and a are positive integers }

Therefore, 'a' and 'b' have 2 as common factor. But they've no common factor other than 1. Which means our assumption was wrong.

Hence, √2 is a irrational number.

(ii) 7√5 is irrational

Let's say that 7√5 is rational number which can be written in the form a/b having a and b as coprime number.

7√5 = a/b

√5 = a/7b

Here, a/7b is rational number but √5 is irrational number. Again rational number can't be equal to irrational number. Hence, our assumption is wrong 7√5 is irrational number.

(iii) 6 +√2 is irrational

Let's say the 6 + √2 is rational number which can be written in the form a/b having a and b as coprime number.

6 + √2 = a/b

√2 = a/b - 6

√2 = (a - 6b)/b

Here, (a - 6b)/b is rational and √2 is irrational number. So, 6 + √2 is irrational number, our assumption was wrong.

Answer 2

✿ Solutions :-

$\: \: \sf\bold{\green{\sf(i) \large \big \: \dfrac{1}{ \sqrt{2} }}}$

$➤ \sf Let \: us \: suppose \: that \: \dfrac{1}{ \sqrt{2} } \: is \: a \: rational \: number$

→ As the rational number are in the form of p/q were p and q are co - prime and q is not equal to zero , so we can write :-

$\sf→ \dfrac{1}{ \sqrt{2} } = \dfrac{ p}{q}$

Squaring on both sides

$\sf→ ( { \dfrac{1}{ \sqrt{2} } )}^{2} = ({\dfrac{p}{q} )}^{2}$

$\sf→ \dfrac{ {1}^{2} }{ \sqrt{2} ^{2} } = \dfrac{ {p}^{2} }{ {q}^{2} }$

$\sf→ \dfrac{ {1}^{2} }{ \cancel {{\sqrt{2} }} \: \cancel{{^{2}}} } = \dfrac{ {p}^{2} }{ {q}^{2} }$

$\sf→ \sf {q}^{2} = 2 \: {p}^{2} - - - - - (1)$

$• \sf \: {q}^{2} \: is \: divisible \: by \: 2$

By the theorem that if p [ a prime number ] divides a² , then p divides a , were a is a positive integer

$• \sf \: q \: is \: also \: divisible \: by \: 2$

Let ,

• q = 2m for some integer m -------- ( 2 )

→ Substitute the value of q from equation (2) in equation (1) , we get

$\sf→( {2m)}^{2} = 2 {p}^{2}$

$→ \sf \: 4 {m}^{2} = 2 {p}^{2}$

${\sf { →{\dfrac{{\cancel{{4 }} ^{2} {m}^{2} }}{{\cancel{{2}}^{1}}}}}} = \sf {p}^{2}$

$\sf →2 {m}^{2} = {p}^{2}$

$• \sf \: {p}^{2} is \: divisible \: by \: 2$

Again , by the theorem that if p [ a prime number ] divides a² , then p divides a , were a is a positive integer

$• \sf \: p \: is \: also \: divisible \: by \: 2$

Thus ,

• 2 is a common factor of p and q

• This is a contradiction that q and q are co - prime

$\sf Hence \: , our \: supposition \: that \: \dfrac{1}{ \sqrt{2} } \: is \: a \: rational \: number \: is \: wrong \:$

$∴ \sf \: \dfrac{1}{ \sqrt{2} } \: is \: an \: irratinal \: number$

$\: \: \sf\bold{\red{\sf{(ii) \large \: 7 \sqrt{5} }}}$

$➤ \sf Let \: us \: suppose \: that \: 7 \sqrt{5} \: is \: a \: rational \: number$

→ As the rational number are in the form of p/q were p and q are co - prime and q is not equal to zero , so we can write :-

$→ \sf7 \sqrt{5} = \dfrac{p}{q}$

$→ \sf \sqrt{5} = \dfrac{ p}{7q}$

$\sf \: This \: shows \: \: \dfrac{p}{7q} is \: rational$

• Therefore , √5 is also rational which contradicts the fact that √5 is irrational

• Therefore 7√5 is irrational

$\: \: \sf\bold{\pink{\sf{ \sf(iii) \large \: 6 + \sqrt{2}}}}$

$➤ \sf Let \: us \: suppose \: that \: 6 + \sqrt{2} \: is \: a \: rational \: number$

→ As the rational number are in the form of p/q were p and q are co - prime and q is not equal to zero , so we can write :-

$→ \sf \: 6 + \sqrt{2} = \dfrac{p}{q}$

$→ \sqrt{2} = \dfrac{p}{q} - 6$

$→ \sqrt{2} = \dfrac{p - 6q}{q}$

$\sf \: This \: shows \: \dfrac{p - 6q}{q} \: is \: rational$

• Therefore , √2 is also rational which contradicts the fact that √2 is irrational

• Therefore 6 + √2 is irrational

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We know :-

• $\: \: \sf\bold{\orange{\sf{ \sf Irrational \: number :-}}}$

A number that cannot be expressed in the form of p/q a p and q are integers and q is not equal to zero is called as a irrational number . The decimal expansion of such numbers are non terminating and non repeating .

• $\: \: \sf\bold{\orange{\sf{ \sf co - prime \: numbers :-}}}$

Numbers which have only one as common factor .