Question

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

Answer 1

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

Explanation:

Consider the octic polynomial with zeros:

`sigma_1sqrt(2)+sigma_2sqrt(17)+sigma_3sqrt(19)`

where `sigma_1, sigma_2, sigma_3 in { 1, -1 }`

We find:

`(x+sigma_1sqrt(2)+sigma_2sqrt(17)-sqrt(19))(x+sigma_1sqrt(2)+sigma_2sqrt(17)+sqrt(19))`

`=(x+sigma_1sqrt(2)+sigma_2sqrt(17))^2-19`

`=(x^2+2sigma_1sqrt(2)x)+sigma_2(2sqrt(17)x+2sigma_1 sqrt(34))`

Then:

`((x^2+2sigma_1sqrt(2)x)-(2sqrt(17)x+2sigma_1 sqrt(34)))((x^2+2sigma_1sqrt(2)x)+(2sqrt(17)x+2sigma_1 sqrt(34)))`

`=(x^2+2sigma_1sqrt(2)x)^2-(2sqrt(17)x+2sigma_1 sqrt(34))^2`

`=x^4+4sigma_1sqrt(2)x^3+8x^2-68x^2-136sigma_1sqrt(2)x-136`

`=(x^4-60x^2-136)+sigma_1(4sqrt(2)x^3-136sqrt(2)x)`

Then:

`((x^4-60x^2-136)-(4sqrt(2)x^3-136sqrt(2)x))((x^4-60x^2-136)+(4sqrt(2)x^3-136sqrt(2)x))`

`=(x^4-60x^2-136)^2-(4sqrt(2)x^3-136sqrt(2)x)^2`

`=x^8-152x^6+5504x^4-20672x^2+18496`

By the rational roots theorem, any rational zero of this octic is expressible in the form `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) ~~ 1.414+4.123-4.359 ~~ 1.178`

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

Answer 2

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+2sqrt(34)-2csqrt(2)-2csqrt(17)+c^2 = 19`

Subtract `19` from both sides to get:

`2sqrt(34)-2csqrt(2)-2csqrt(17)+c^2 = 0`

That is:

`(2sqrt(2)-2c)sqrt(17) = 2csqrt(2)-c^2`

So:

`sqrt(17) = (2csqrt(2)-c^2)/(2sqrt(2)-2c)`

Square both side to get:

`17 = (8c^2-4sqrt(2)c^3+c^4)/(8-8sqrt(2)c+4c^2)`

Multiply both sides by `(8-8sqrt(2)c+4c^2)` to get:

`136-136sqrt(2)c+68c^2 = 8c^2-4sqrt(2)c^3+c^4`

So:

`(4c^3-136c)sqrt(2) = c^4-60c^2-136`

So:

`sqrt(2) = (c^4-60c^2-136)/(4c^3-136c)`

Now we know that `sqrt(2)` is irrational, so `c` must be irrational too.