Question

prove that 7+3√2 is an irrational number​

Answer 1

Step-by-step explanation:

We know, √3 is an irrational number

Let 7√3 be rational

so,

7√3 = p/q , q is not 0, HCF(p,q) = 1

√3 = p/7q

A contradiction as LHS is irrational and RHS is a rational. So, 7√3 is not rational number.

Answer 2

Step-by-step explanation:

If possible , let

If possible let √3 be a rational number and its simplest form be

a/b then, a and b are integers having no common factor

other than 1 and b =|0

$\sqrt{3} = \frac{a}{b} = 3 \frac{a ^{2} }{b ^{2} } (on \: squaring \: both \: sides \: )$

$or \: 3 {b}^{2} = a ^{2}$

$3 \: divides \: a ^{2} (3 \: divides \: 3 {b}^{2}$

3 divides a

let a=3c for some integer c

Putting a=3c in (i), we get

$or \: 3 {b}^{2} = 9 {c}^{2} = b ^{2} = 3 {c}^{2}$

$3 \: divides \: b ^{2} (3 \: divides \: 3c ^{2} )$

3 divides a

Thus 3 is a common factor of a and b

This contradicts the fact that a and b have no common factor other than 1.

The contradiction arises by assuming

$\sqrt{3} \: is \: a \: rational$

$hence \: \sqrt{3} \: is \: irrational \:$

2nd part

$if \: possible \: let \: (7 + 2 \sqrt{3} )be \: a \: rational \: number$

$7 - (7 + 2 \sqrt{3} )is \: a \: rational$

$- 2 \sqrt{3} \: is \: rational$

$this \: contradics \: the \: fact \: that - 2 \sqrt{3} \: is \: an \: irrational \: number$

Since, the contradiction arises by assuming

$7 + 3 \sqrt{2} \: is \: a \: rational \:$

$\: hence \: 7 + 3 \sqrt{2} \: is \: an \: irrational \: \\ proved$