Question

if √5 is an irrational number then prove that 3-2√5 is an irrational number.​

Answer 1

Proof

$\sf Let \ 3 - 2\sqrt{5} \ be \ a \ rational \ number.$

$\sf Rational \ numbers \ are \ written \ in \ the \ form \ \dfrac{p}{q} \ \\where \ p \ and \ q \ are \ co-prime \ and \ q \neq 0$

$\implies 3-2\sqrt{5} = \dfrac{p}{q}$

$\implies -2\sqrt{5} = \dfrac{p}{q}-3$

$\implies -2\sqrt{5} = \dfrac{p-3q}{q}$

$\implies \sqrt{5} = \dfrac{p-3q}{-2q}$

$\implies \sqrt{5} = \dfrac{3q-p}{2q}$

${\sf{\bold{{Now, \implies \dfrac{3q-p}{2q} \ is \ a \ rational \ number}}}$

$\sf{\bold{\sqrt{5} \ is \ a \ rational \ number}$

$\sf{\bold{But \ this \ contradicts \ to \ the \ fact \ that \ \sqrt{5} \ is \ an \ irrational \ number}}}}}$

$\sf{\bold{Hence \ our \ assumption \ is \ wrong}}}$

$\boxed{\boxed{\bold{3-2\sqrt{5} \ is \ an \ irrational \ number}}}}$

                                                                         

Answer 2

Answer:

Let assume that 3-2√5 is a rational numbers

so, it can be written in the form of a/b where a and b is Coprime and b not is equal to 0

$3 - 2 \sqrt{5 } = \frac{a}{ b}$

$\sqrt{5 } = \frac{a}{2b} + 3$

$\sqrt{5} = \frac{a + 6b}{2b}$

Therefore, a and b is a integer, a/2b+3 is rational

so, √5 will be rational

But, our contradiction is wrong √5 is a irrational

hence, it is proved that 3-2√5 is a irrational.

I hope this help you