Question

prove that 3-2√5 is irrational​

Answer 1

Proof

$\sf Let \ us \ assume \ that \ 3-2\sqrt{5} \ is \ a \ rational \ number.$

$\sf Rational \ numbers \ can\ be\ expressed \ in \ the \ form \ \dfrac{p}{q}$

$\sf Where \ p \ and \ q \ are \ co-prime \ and \ q \neq 0$

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

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

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

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

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

$\sf \dfrac{3q-p}{2q} \ is \ a \ rational \ number$

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

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

$\sf Hence \ our \ assumption \ is \ wrong$

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

                                                                               

Answer 2

Answer:

anwer = is 5 (five)

OK please like me