radicals - Using the fact that $sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $sqrt{3} + sqrt{7} + sqrt{21}$ is irrational.

Question:

Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational.

Following from the question, I tried:

Let $N = \sqrt{3} + \sqrt{7} + \sqrt{21}$. Then,

$$
\begin{align}
N+1 &= 1+\sqrt{3} + \sqrt{7} + \sqrt{21}\\
&= 1+\sqrt{3} + \sqrt{7} + \sqrt{3}\sqrt{7}\\

&= (1+\sqrt{3})(1+\sqrt{7}).
\end{align}
$$

Using the above stated fact, $\sqrt{3}$ and $\sqrt{7}$ are irrational. Also, sum of a rational and irrational number is always irrational, so $1+\sqrt{3}$ and $1+\sqrt{7}$ are irrational. Similarly, if we prove that $N+1$ is irrational, $N$ will also be proved to be irrational.

But, how do I prove that product of $1+\sqrt{3}$ and $1+\sqrt{7}$ are irrational.

Answer

If $(1+\sqrt{3})(1+\sqrt{7})$ is rational, then

$$\displaystyle \frac{12}{(1+\sqrt{3})(1+\sqrt{7})}=\frac{12(1-\sqrt{3})(1-\sqrt{7})}{(-2)(-6)}=1-\sqrt{3}-\sqrt{7}+\sqrt{21}$$ is also rational.

So, $\displaystyle \frac{1}{2}[(1+\sqrt{3})(1+\sqrt{7})+1-\sqrt{3}-\sqrt{7}+\sqrt{21}]-1=\sqrt{21}$ is rational.

This leads to a contradiction.