Question

prove that 5+2√2 is irrational
Please help me with this! ​

Answer 1

Let us assume that 5+2√2 be a  rational number.

So,

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

[In which 'p' and 'q' are integers and q is not equal to 0, also HCF(p,q) = 1]

$\implies2\sqrt{2} = \frac{p}{q}-5$

[By transporting 5 to RHS]

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

[By taking LCM as q in RHS]

$\implies \sqrt{2} = \frac{p-5q}{q} \div 2$

[By transporting 2 to RHS]

$\implies \sqrt{2} = \frac{p-5q}{q}\times \frac{1}{2}$

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

Now,

Since, (p), (q), (-5), and (2) are integers and q is not equal to 0 so,

$\dfrac{p-5q}{2q}$ is n rational number

But √2 is not a rational number (i.e. it is an irrational number)

Hence,

$\boxed{ \sqrt{2} \ne \frac{p-5q}{2q}}$

Thus,

5+2√2 is irrational number.

Answer 2

Answer:

Let assume that 5+2√2 is a rational number.

So , By defination of rational no.

$= > \: 5 + 2 \sqrt{2} = \frac{p}{q}$

Transporting 5 to R.H.S. , we get :-

$= > \: 2 \sqrt{2} = \frac{p}{q} - 5$

By Taking L.CM. , we get:-

$= > \: 2 \sqrt{2} = \frac{p - 5q}{q}$

$= > \: \sqrt{2} = \frac{p - 5q}{q} \div 2$

$= > \: \sqrt{2} = \frac{p - 5q}{q} \times \frac{1}{2}$

$= > \: \sqrt{2} = \frac{p - 5q}{2q}$

Now

Since , (p) , ( q) , (2) ,(-5) are integers and q is not equal to 0.

THEREFORE,

$\frac{p - 5q}{2q} \: is \: a \: rational \: number.$

But √2 is not a rational number

or, √2 is a irrational number.

Therefore , our assumption was wrong

_____________________________

So, 5+2√2 is a irrational number

_____________________________