Question

Prove
$3 + 2 \sqrt{5} \: \: is \: irrational \: number$
​

Answer 1

Answer:

Prove 3+2

5

is irrational.

→ let take that 3+2

5

is rational number

→ so, we can write this answer as

⇒3+2

5

=

b

a

Here a & b use two coprime number and b



=0.

⇒2

5

=

b

a

−3

⇒2

5

=

b

a−3b

∴

5

=

2b

a−3b

Here a and b are integer so

2b

a−3b

is a rational number so

5

should be rational number but

5

is a irrational number so it is contradict

- Hence 3+2

5

is irrational.

Answer 2

$\huge{\dag}{\underline{\boxed{\sf{\red{Answer:- }}}}}$

${\underline{\boxed{\sf{\purple{\mathtt{ Question :-}}}}}}$

$Prove \: that \: \\ \\ 3+2√5 \: is \: irrational \: number.$

$\small\underline\mathcal\pink{Requried \: Answer :-}$

Let 3+2√5 be rational.

3+2√2 = p/q where q≠ 0 & p and q are integers.

$3 + \sqrt{5} = \frac{a}{b} \: \: \: \: \: \:$

a and b are coprime integers

$2 \sqrt{5} = \frac{a}{b} - 3 \\ 2 \sqrt{5} \: \frac{a - 3}{b} \\ \sqrt{5} = \frac{1}{2} ( \frac{a}{b} - 3)$

$Since \: a \: and \: b \: are \: integers \: \frac{1}{2} ( \frac{a}{b} - 3) \: \\ will \: also \: be \: rational. \\ therefore \: \sqrt{5} \: is \: rational$

This contradicts that the fact √5 is irrational.

Hence our assumption that 3+2√5 is irrational number is false

Therefore 3+2√5 is irrational number.