prove that $2\sqrt5 +\sqrt{11}$ is irrational

Solution 1:

Here's a more advanced approach, with some common details to the above.

A monic polynomial with integer coefficients:

$$ (x^2 - (2\sqrt 5 + \sqrt{11})^2)(x^2 - (2\sqrt 5 - \sqrt{11})^2) = x^4 - 62x^2 +81 $$

Now $x = 2\sqrt 5 + \sqrt{11}$ is a root of this polynomial, but by the Rational Roots Thm., any rational root would be an integer divisor of $81$.

Thus one only needs to verify that $2\sqrt 5 + \sqrt{11}$ is not an integer. A simple hand computation (or calculator computation) shows this positive number lies strictly between $7$ and $8$.

Solution 2:

You simply need to show that $\sqrt{55}$ is irrational since $a+bc$ is irrational whenever $a,b$ rational (and $b\ne 0)$ and $c$ irrational. This produces a contradiction. To show $\sqrt{55}$ is irrational, there is a famous proof based on the fundamental theorem of arithmetic that the square root of any non-perfect-square natural number is irrational (or you can do an ad-hoc proof, similar to the proof that $\sqrt{2}$ is irrational).

Solution 3:

If $2\sqrt{5}+ \sqrt{11}=r$ is rational then $4*5 + 4*\sqrt{55} + 11 = r^2$ is rational. Then $\sqrt{55} = \frac {r^2 - 31}4$ is rational.

NOW do the proof by contradiction

Let $\sqrt{55} = \frac ab$ so $55b^2 = a^2$ so $5|a$ and so $25|a^2$ so $5|b^2$ so $5|b$ so $a$ and $b$ aren't in lowest terms.... yadda, yadda, yadda....