Question

prove that 3+2 rute.7
us an irrational​

Answer 1

Step-by-step explanation:

it is not irrational as it has a denominator not having root...

Answer 2

Answer:

Let us assume that $3 + 2\sqrt{7}$ is an rational

            such that  $3 + 2\sqrt{7}$ = $\frac{p}{q}$ ( p and q are integers )

                                   $2\sqrt{7}$ = $\frac{p}{q} - 3$

                                    $2\sqrt{7}$= $\frac{p-3q}{q}$

                                    $\sqrt{7}$=  $\frac{p-3q}{2q}$

∵ p and q are integers. So,   $\frac{p-3q}{2q}$ is a rational no. and  $\sqrt{7}$ is also a rational no. But this contradict the fact that  $\sqrt{7}$ is an irrational no. This contradiction has arisen due to incorrect assumption that $3 + 2\sqrt{7}$ is a rational no. Hence, $3 + 2\sqrt{7}$ is an irrational no.