Question

Prove that the 3-5 rootover 3 is irrational​

Answer 1

Answer:

we know that root3 is irrational

and rational number+ irrational number= irrational

so

we can say 3-5 rootover 3 is irrational

but for more information

prefer dear sir's channel

Answer 2

$\huge\boxed{\fcolorbox{red}{ink}{SOLUTION:}}$

Let consider

$a = \sqrt{3} - \sqrt{5}$

is rational

Then

$\frac{1}{a} = \frac{1}{ \sqrt{3 - \sqrt{5} } }$

will also be rational

Now ,

$\frac{1}{ \sqrt{3 - \sqrt{5} } }$

$\frac{1}{ \sqrt{3 - \sqrt{5} } } \times \frac{ \sqrt{3 + \sqrt{5} } }{ \sqrt{3} + \sqrt{5} }$

$\frac{ \sqrt{3 + \sqrt{5} } }{ - 2}$

Sum of rational number is again rational

a is rational as assumed ,

$2 \times \frac{1}{a}$

is also rational

Now so

$a - 2 \times \frac{1}{a}$

$= ( \sqrt{3} - \sqrt{5} ) + (\sqrt{3} + \sqrt{5} )$

$= 2 \times \sqrt{3}$

• But we know √3 is irrational , hence a contradiction, hence

$( \sqrt{3} - \sqrt{5})$

is irrational.

$\huge\boxed{\dag\sf\red{Thanks}\dag}$