I tried to let $$2\sqrt{5} + \sqrt{11} = \frac{a}{b}$$ and find contradictions
(I set $b \ne 0$, $a$ and $b$ are in their simplest form) but I cannot find any
Asked
Active
Viewed 581 times
-1
-
Let's say it is rational, then $a^2/b^2 - n$ is rational for all rational $n$. But I reckon that I can find an $n$, for which we can easily prove this false – mdave16 Nov 25 '17 at 23:26
-
Show $x^2-5$ is irreducible over $\Bbb Q$ and then show $x^2-11$ is irreducible over $\Bbb Q[\sqrt{5}]$. That will do it. – Gregory Grant Nov 25 '17 at 23:30
-
If $r=2\sqrt5+\sqrt{11}$ is rational, then so is $\frac{r^2-141}4=\sqrt{55}.$ Can you prove that $\sqrt{55}$ is irrational? – bof Nov 25 '17 at 23:31
-
@zipirovich Nevertheless my approach differs from these of presented there. – szw1710 Nov 25 '17 at 23:39
-
@szw1710 This is a duplicate, which means it should be closed as a duplicate, especially since the original question was the better of the two questions. You can make a case if you choose, to ask for a merging of the answers to the original, but you cannot deny you answered a question that was asked and answered two days ago. – amWhy Nov 26 '17 at 00:51
-
@amWhy Thanks for your remarks. Maybe the moderators could join all into one piece (too many answers there to write my own in the original question. My best regards. – szw1710 Nov 26 '17 at 09:21
3 Answers
4
If $2\sqrt{5}+\sqrt{11}$ is rational, then also $$2\sqrt{5}-\sqrt{11}=\frac{9}{2\sqrt{5}+\sqrt{11}}$$ would be rational. Subtracting two rationals $(2\sqrt{5}+\sqrt{11})-(2\sqrt{5}-\sqrt{11})=2\sqrt{11}$ we arrive at $2\sqrt{11}$ which must also be rational, hence $\sqrt{11}$ is rational, which is false (proof like $\sqrt{2}$ is irrational).
2
Let $x= 2\sqrt{5}+\sqrt{11}.$ Then
$$x^2 = 4\sqrt{55} + 31.$$
So $4\sqrt{55} = x^2 -31$, and squaring again gives
$$x^4-62x^2 +81 = 0.$$
By the rational root theorem, the only possible rational roots are $\pm 3^k$, where $k=0,1,2,3,4.$ Since $x$ is a root, it must be one of these, but $x$ is between $7$ and $8$, so it can't equal a power of $3$.
B. Goddard
- 33,728
1
Suppose that $2\sqrt{5}+\sqrt{11}=r$ where $r$ is a rational number. Then
\begin{align} r - \sqrt{11} &= 2\sqrt 5\\ r^2 - 2r\sqrt{11}+11 &= 20 \\ r^2 - 9 &= 2r \sqrt{11} \\ r - \dfrac{9}{r} &= 2\sqrt{11} \end{align}
But this is a contradiction since $r - \dfrac{9}{r}$ is, by hypothesis, a rational number and $2\sqrt{11}$ is an irrational number.
Hence $r = 2\sqrt{5}+\sqrt{11}$ is an irrational number.
Steven Alexis Gregory
- 27,619