Math, asked by Anonymous, 6 hours ago

3.Prove that the following are irrationals:

(i) 1/√2

(ii) 7√5

(iii) 6 + √2

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subject : maths

class 10th​

Answers

Answered by ruchikasingh287287
2

 \sf \: ( \: i \: ) \:  -  \:  \frac{1}{ \sqrt{2} }  \\  \sf \:  Let \:  us \:  assume \:  \frac{1}{ \sqrt{2} }  \: is \: rational. \\  \sf \: So  \: we \:  can  \: write \:  this \:  number \:  as \:  \\ \sf \:  \frac{1}{ \sqrt{2} }  =  \frac{a}{b}  -  1 \\  \sf \: Here \: , a \:  and  \: b  \: are \:  two  \\  \sf co-prime \:  numbers \:  and \:  \\ \sf  b  \: is  \: not  \: equal \:  to  \: zero \\ \\   \sf \: Simplify  \: the  \: equation \:  (1)  \\  \sf \: multiply \:  by \:  \sqrt{2}  \: both \:  sides, \:  we \:  get \\  \sf \: 1 =  \frac{a \:   \sqrt{2} }{b}  \\  \sf \: Now,  \: divide  \: by \:  b, we  \: get \:  \\ \sf \: b = a \sqrt{2}  \: or \:  \frac{b}{a}  =  \sqrt{2}

 \sf \: Here, \:  a  \: and \:  b  \: are \:  integers \:  so,  \:  \\  \sf \: is  \: a \:  rational \:  a  \: number, \\ \sf \: so \:  \sqrt{2} \:  should  \: be  \: a  \: rational \:  number. \\ \sf \: but \:  \sqrt{2}  \: is  \: a  \: irrational \:  number \: ,  \\   \sf \:  so \:  it  \: is \:  contradictory. \\ \sf \: therefore \:  \frac{1}{ \sqrt{2} } is \:  irrational \:  number. \\

Answered by Anonymous
11

Answer:

 \sf{(i) \: Irrational}

 \sf{(ii) \: Irrational}

 \sf{(iii) \: Irrational}

Step-by-step explanation:

Solution :

Let us assume 1/√2 is a rational number

Let us assume 1/√2 = r where r is a rational number

On further calculation we get

1/r = √2

Since r is a rational number, 1/r = √2 is also a rational number

But we know that √2 is an irrational number

So our supposition is wrong.

Hence, 1/√2 is an irrational number.

(ii) 7√5

Solution :

Let’s assume on the contrary that 7√5 is a rational number. Then, there exist positive integers a and b such that

7√5 = a/b where, a and b, are co-primes

⇒ √5 = a/7b

⇒ √5 is rational [∵ 7, a and b are integers ∴ a/7b is a rational number]

This contradicts the fact that √5 is irrational. So, our assumption is incorrect.

Hence, 7√5 is an irrational number.

(iii) 6 + √2

Solution :

Let’s assume on the contrary that 6+√2 is a rational number. Then, there exist co prime positive integers a and b such that

6 + √2 = a/b

⇒ √2 = a/b – 6

⇒ √2 = (a – 6b)/b

⇒ √2 is rational [∵ a and b are integers ∴ (a-6b)/b is a rational number]

This contradicts the fact that √2 is irrational. So, our assumption is incorrect.

Hence, 6 + √2 is an irrational number.

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