Question

show that under root 3 + 2 under root 5 is irrational

Answer 1

Hello,.

Let's do it by the method of contradiction.

Let
$\sqrt{3} + 2 \sqrt{5}$
Be an Irrational number.

So,. It can be expressed in the form of p/q where p and q are integers and Co primes and q ≠0.

So,
$\sqrt{3} + 2 \sqrt{5} = \frac{p}{q}$
Squaring both the sides we get that

${( \sqrt{3} + 2 \sqrt{5} ) }^{2} = ( \frac{ {p}^{2} }{ {q}^{2} } ) \\ \\ 3 + 20 + 4 \sqrt{15} = \frac{ {p}^{2} }{ {q}^{2} } \\ \\ 4 \sqrt{15} = \frac{ {p}^{2} }{ {q}^{2} } - 23 \\ \\ 4 \sqrt{15} = \frac{ {p}^{2} - 23 {q}^{2} }{ {q}^{2} } \\ \\ \sqrt{15} = \frac{ {p}^{2} - 23 {q}^{2} }{4 {q}^{2} }$
Here,


Root 15 is an Irrational number
But the RHS is a rational number.

An Irrational number can never be equal to a rational number.
Hence, it's a contradiction.
Our assumption was wrong that the given number is rational.

Hence, roo3 + 2 root5 is an Irrational number.


Hope this will be helping you....