# Is 4+sqrt7 rational?

Mar 10, 2018

$\sqrt{7}$ is irrational.

Therefore $4 + \sqrt{7}$ will also be irrational.

#### Explanation:

$\sqrt{7}$ is an an irrational number, meaning that it is an infinite non-recurring decimal which cannot be written as a common fraction.

$\sqrt{7} = 2.64575131 \ldots \ldots$

If you use an irrational number in an operation, the answer will be irrational.

Note that $\left({\sqrt{7}}^{2} = 7\right)$

Mar 10, 2018

No

#### Explanation:

If $4 + \sqrt{7} = \frac{p}{q}$ for some integers $p , q$ then $\sqrt{7} = \frac{p - 4 q}{q}$ would also be rational.

To see that $\sqrt{7}$ is irrational, we can proceed as follows...

Suppose $x > 0$ satisfies:

$x = 2 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{2 + x}}}}$

Then:

$x = 2 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{2 + x}}}}$

$\textcolor{w h i t e}{x} = 2 + \frac{1}{1 + \frac{1}{1 + \frac{2 + x}{3 + x}}}$

$\textcolor{w h i t e}{x} = 2 + \frac{1}{1 + \frac{3 + x}{5 + 2 x}}$

$\textcolor{w h i t e}{x} = 2 + \frac{5 + 2 x}{8 + 3 x}$

$\textcolor{w h i t e}{x} = \frac{21 + 8 x}{8 + 3 x}$

Multiplying both ends by $\left(8 + 3 x\right)$ we find:

$3 {x}^{2} + 8 x = 21 + 8 x$

Subtracting $8 x$ from both sides we find:

$3 {x}^{2} = 21$

Hence:

${x}^{2} = 7$

So:

$x = \sqrt{7}$

We have found:

$\sqrt{7} = 2 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{2 + \sqrt{7}}}}}$

$\textcolor{w h i t e}{\sqrt{7}} = 2 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{4 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{4 + \ldots}}}}}}}}$

Since this continued fraction does not terminate, it does not represent a rational number.

So $\sqrt{7}$ is irrational and so is $4 + \sqrt{7}$