Question

1. Show that 5-2root3 is an irrational number.​

Answer 1

Solution:

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

$\sf{\implies 5-2\sqrt{3}=\dfrac{p}{q}\;\;\;\;[Where\;p\;and\;q\;are\;co-prime\;numbers]}$

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

$\sf{\implies 5q=p+2\sqrt{3}}$

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

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

We know that $\sf{ \dfrac{5q-p}{2}\;is\;rational\;number\;but\;\sqrt{3}\;is\;irrational.}$

And, Rational ≠ Irrational

So, our assumption is wrong.

$\sf{Hence,\;5-2\sqrt{3}\;is\;a\;irrational\;number.}$

Answer 2

Answer:

Step-by-step explanation:

To Prove :-

5+2√3 is rational number.

Solution:-

Let us assume that 5+2√3 is rational number

5+2√3 = p/q ( where p and q are co prime)

2√3 = p/q-5

2√3 = p-5q/q

√3 = p-5q/2q

now p , 5 , 2 and q are integers 

p-5q/2q is rational number.

√3 is rational number.

Hence, 5+2√3 is irrational number.