Square Root

I will prove that $3\sqrt{7}$ is irrational first. Assume not, i.e. assume $3\sqrt{7}=\frac{p}{q}$ , where $p$ and $q$ are positive integers and $\gcd(p,q)=1$ .

$\begin{aligned} 3\sqrt{7} &= \dfrac {p}{q}\\ 63 &= \dfrac {p^2}{q^2}\\ p^2 &= 63q^2\\ \implies 21&\mid p \end{aligned}$

Let $p=21p_1$ .

$\begin{aligned} 441p_1^2 &= 63q^2\\ 7p_1^2 &= q^2\\ \implies 7&\mid q \end{aligned}$

But we have already said that $\gcd(p, q)=1$ , but we have just shown that both have 7 as a common factor. Thus, we have a contradiction. Therefore, $3\sqrt{7}$ is irrational.

Now, our question in hand asks

$\begin{aligned} \sqrt{2+7} \times \sqrt{7} &= \sqrt{9} \times \sqrt{7}\\ &= 3\sqrt{7} \end{aligned}$

But we have already shown that this number is irrational. Therefore, we have proven this number is irrational.

$\sqrt(2+7)\times\sqrt(7) = 3\sqrt(7)$

We know for any prime number $p$ , the $\sqrt p$ is an irrational number and also as we know product of non-zero rational number and irrational number is an irrational number and hence given number is an $iraational$ number.

$\sqrt{2+7}$ $\cdot$ $\sqrt{7}$

$\sqrt{2+7}$ $\rightarrow$ $\sqrt{2+7=9}$ $\rightarrow$ $\sqrt{9=3}$

$3$ $\times$ $\sqrt{7}$ $\text{}[is ~an~ irrational ~number]$ .

Note: $3$ $\times$ $\sqrt{7}$ $\approx$ $7.9372539332$

$\sqrt{2+7} \times \sqrt{7}$ is a irrational number because $\sqrt{9} \times \sqrt{7} = 3\sqrt{7}$ . 7 is not a perfect square, so it is irrational.

$\sqrt{2+7}×\sqrt7=\sqrt9×\sqrt7=\sqrt{63}$

The square root of any number that is not a perfect square is an irrational number

Because $7^2=49$ and $8^2=64$ , $63$ is not a perfect square $\Rightarrow\color{#D61F06}\boxed{\sqrt{2+7}×\sqrt{7} \ is \ irrational}$