how would you prove that $2\sqrt5 +\sqrt{11}$ is irrational? I started with a proof by contradiction that assumes that $2\sqrt5 +\sqrt{11}$ is rational and therefore there exist integers $a$ and $b$ such that $\frac{a}{b}=2\sqrt5 +\sqrt{11}$ and squaring both sides yields $\frac{a^2}{b^2}=31 +4\sqrt5\sqrt{11}$ and from this point im stuck as i dont know how to continue to arrive at a contradiction.
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$\begingroup$ That might be the best approach (with suitable technique) unless you know some higher algebra. Can you add a few words about your math background? $\endgroup$ – hardmath Nov 23 '17 at 0:56
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1$\begingroup$ $31 + 4\sqrt{5}\sqrt{11} = 31 + 4\sqrt{55}$. If $\sqrt{55} $ is not rational, neither is $31 + 4\sqrt{55}$. So do $\sqrt{55} =\frac { \frac {a^2}{b^2} - 31}4 = \frac cd$ leads to a contradiction. $\endgroup$ – fleablood Nov 23 '17 at 1:02
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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 $8$.
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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).
answered Nov 23 '17 at 0:55
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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....
