Question

prove that root 5 minus 3 root 2
is an irrational number by contradiction method​

Answer 1

Proof :-

Assume that $\sqrt{5}-3 \sqrt{2}$ be a Rational number.

So, It can be expressed in the form of p/q where p and q are integers and q is not equal to 0. Also, P and Q are Co-primes.

Thus,

$\sqrt{5} - 3 \sqrt{2} = \frac{p}{q} \\ \\ squaring \: both \: sides \\ \\ {( \sqrt{5} - 3 \sqrt{2})}^{2} = \frac{ {p}^{2}}{ {q}^{2} } \\ \\ \\5 + 18 - 6 \sqrt{10} = \frac{ {p}^{2}}{ {q}^{2} } \\ \\ \\23 - 6 \sqrt{10} = \frac{ {p}^{2}}{ {q}^{2} } \\ \\ \\6 \sqrt{10} = 23 - \frac{ {p}^{2}}{ {q}^{2} } \\ \\ \\6 \sqrt{10} = \frac{23 {q}^{2} - {p}^{2}}{ {q}^{2} } \\ \\ \\ \sqrt{10} = \frac{23 {q}^{2} - {p}^{2}}{ 6{q}^{2} }$

$\rule{200}{2}$

We already know that $\sqrt{10}$ is an irrational Number as it cannot be expressed in the form of p /q where p and q are integers and q is not equal to 0.

But,

We took p and q as integers in the first step where q was not equal to 0.

Hence :-

$\dfrac{23q^{2}-p^{2}}{6q^{2}}$ is also a Rational number.

$\rule{200}{2}$

We already know that a rational number can never be equal to an irrational number.

Thus, what we got is never possible.

Hence, It is a contradiction.

This has arose because we took $\sqrt{5}-3 \sqrt{2}$ as a rational number.

Therefore,

$\sqrt{5}-3 \sqrt{2}$ is an Irrational Number.

$\rule{200}{2}$