# Prove that the number (sqrt(2)+sqrt(17)-sqrt(19)) is irrational?

Feb 23, 2018

Here's a proof based on the rational roots theorem...

#### Explanation:

Consider the octic polynomial with zeros:

${\sigma}_{1} \sqrt{2} + {\sigma}_{2} \sqrt{17} + {\sigma}_{3} \sqrt{19}$

where ${\sigma}_{1} , {\sigma}_{2} , {\sigma}_{3} \in \left\{1 , - 1\right\}$

We find:

$\left(x + {\sigma}_{1} \sqrt{2} + {\sigma}_{2} \sqrt{17} - \sqrt{19}\right) \left(x + {\sigma}_{1} \sqrt{2} + {\sigma}_{2} \sqrt{17} + \sqrt{19}\right)$

$= {\left(x + {\sigma}_{1} \sqrt{2} + {\sigma}_{2} \sqrt{17}\right)}^{2} - 19$

$= \left({x}^{2} + 2 {\sigma}_{1} \sqrt{2} x\right) + {\sigma}_{2} \left(2 \sqrt{17} x + 2 {\sigma}_{1} \sqrt{34}\right)$

Then:

$\left(\left({x}^{2} + 2 {\sigma}_{1} \sqrt{2} x\right) - \left(2 \sqrt{17} x + 2 {\sigma}_{1} \sqrt{34}\right)\right) \left(\left({x}^{2} + 2 {\sigma}_{1} \sqrt{2} x\right) + \left(2 \sqrt{17} x + 2 {\sigma}_{1} \sqrt{34}\right)\right)$

$= {\left({x}^{2} + 2 {\sigma}_{1} \sqrt{2} x\right)}^{2} - {\left(2 \sqrt{17} x + 2 {\sigma}_{1} \sqrt{34}\right)}^{2}$

$= {x}^{4} + 4 {\sigma}_{1} \sqrt{2} {x}^{3} + 8 {x}^{2} - 68 {x}^{2} - 136 {\sigma}_{1} \sqrt{2} x - 136$

$= \left({x}^{4} - 60 {x}^{2} - 136\right) + {\sigma}_{1} \left(4 \sqrt{2} {x}^{3} - 136 \sqrt{2} x\right)$

Then:

$\left(\left({x}^{4} - 60 {x}^{2} - 136\right) - \left(4 \sqrt{2} {x}^{3} - 136 \sqrt{2} x\right)\right) \left(\left({x}^{4} - 60 {x}^{2} - 136\right) + \left(4 \sqrt{2} {x}^{3} - 136 \sqrt{2} x\right)\right)$

$= {\left({x}^{4} - 60 {x}^{2} - 136\right)}^{2} - {\left(4 \sqrt{2} {x}^{3} - 136 \sqrt{2} x\right)}^{2}$

$= {x}^{8} - 152 {x}^{6} + 5504 {x}^{4} - 20672 {x}^{2} + 18496$

By the rational roots theorem, any rational zero of this octic is expressible in the form $\frac{p}{q}$ for integers $p , q$ where $p$ is a divisor of the constant term $18496$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are the integer factors of $18496$.

Now $\sqrt{2} + \sqrt{17} - \sqrt{19} \approx 1.414 + 4.123 - 4.359 \approx 1.178$

Since this is not an integer, it is not a rational zero of the octic.

Feb 23, 2018

Alternative proof using the irrationality of $\sqrt{2}$...

#### Explanation:

Suppose:

$\sqrt{2} + \sqrt{17} - \sqrt{19} = c$

Then:

$\sqrt{2} + \sqrt{17} - c = \sqrt{19}$

Square both sides to get:

$19 + 2 \sqrt{34} - 2 c \sqrt{2} - 2 c \sqrt{17} + {c}^{2} = 19$

Subtract $19$ from both sides to get:

$2 \sqrt{34} - 2 c \sqrt{2} - 2 c \sqrt{17} + {c}^{2} = 0$

That is:

$\left(2 \sqrt{2} - 2 c\right) \sqrt{17} = 2 c \sqrt{2} - {c}^{2}$

So:

$\sqrt{17} = \frac{2 c \sqrt{2} - {c}^{2}}{2 \sqrt{2} - 2 c}$

Square both side to get:

$17 = \frac{8 {c}^{2} - 4 \sqrt{2} {c}^{3} + {c}^{4}}{8 - 8 \sqrt{2} c + 4 {c}^{2}}$

Multiply both sides by $\left(8 - 8 \sqrt{2} c + 4 {c}^{2}\right)$ to get:

$136 - 136 \sqrt{2} c + 68 {c}^{2} = 8 {c}^{2} - 4 \sqrt{2} {c}^{3} + {c}^{4}$

So:

$\left(4 {c}^{3} - 136 c\right) \sqrt{2} = {c}^{4} - 60 {c}^{2} - 136$

So:

$\sqrt{2} = \frac{{c}^{4} - 60 {c}^{2} - 136}{4 {c}^{3} - 136 c}$

Now we know that $\sqrt{2}$ is irrational, so $c$ must be irrational too.