Question

prove that 7+2√5 is an irrational number​

Answer 1

Step-by-step explanation:

Let us assume that 7

5

is rational number

Hence 7

5

can be written in the form of

b

a

where a,b(b

=0) are co-prime

⟹7

root5

=

b

a

⟹

root 5

=

7b

a

But here

root 5

is irrational and

7b

a

is rational

as Rational

=Irrational

This is a contradiction

so 7 root

5

is a irrational number

Answer 2

Answer:

$Let \: 7 + 2 \sqrt{5} \: \: is \: not \: irrational, \\ so \: 7 + 2 \sqrt{5} \: \: is \: a \: rational \: number. \\ Now, \: let \: 7 + 2 \sqrt{5} = x, \\ \: a \: rational \: number \\ Therefore \: \: \: \: \: x {}^{2} = (7 + 2 \sqrt{5} ) {}^{2} \\ = 49 + 20 + 2 \times 7 \times 2 \sqrt{5} \\ = 69 + 28 \sqrt{5} \\ = > 28 \sqrt{5} = x {}^{2} - 69 \\ and, \sqrt{5} = \frac{x {}^{2} - 69}{28} \\ Since, it \: is \: assumed \: that \: \\ 7 + 2 \sqrt{5} = x \: i s \: rational \\ = > x {}^{2} \: is \: rational \: \: \: .............(I) \\ x {}^{2} - 69 \: is \: rational \\ and, \: so \: \frac{x {}^{2 } - 69}{28} \: is \: rational \\ = > \frac{x {}^{2} - 69 }{28} = \sqrt{5} \: is \: rational \\ But \: \sqrt{5} \: is \: irrational \: i.e. \: \\ x {}^{2} \: is \: irrational \: \: \: \: \: .............(II) \\ From\: (I), \: x {}^{2} \: is \: rational \: and \\ from\:(II), \: x {}^{2} \: is \: irrational \\ Therefore, \: we \: arrive \: at \: a \: \\ contradiction. \\ So, our \: assumption \: that \\ \: 7 + 2 \sqrt{5} \: \: i s \: a \: rational \: number \: \\ is \: wrong. \\ Therefore, \: 7 + 2 \sqrt{5} \:\: is \: irrational \\ Hence\: Proved........$