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I need some help with this problem: the task is to determine if the number $\sqrt{\sqrt{5}+3}+\sqrt{\sqrt{5}-2}$ is rational or not. Unfortunately I barely have an idea how to start and hence would appreciate not a solution but rather a way to get there.

PS. I am a new guy here, could you explain to me how to type the number itself so that it would appear in the title? I have no idea of typing in formulas in MathStack.

Thanks!

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  • $\begingroup$ By the rational root theorem you only have to prove that the number is an algebraic integer but not an integer. For the first one, you find a monic equation with integer coefficients. For the second one, you compute an approximation. $\endgroup$ – lhf Sep 19 '14 at 22:07
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    $\begingroup$ A start on typing math is here $\endgroup$ – Ross Millikan Sep 20 '14 at 0:35
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First notice $\sqrt{\sqrt{5}+3}$ is not rational. For otherwise, its square would be rational, but its square is $\sqrt{5}+3$, where $\sqrt{5}$ is irrational. And if $\sqrt{5}+3=a$ is a rational, then $\sqrt{5}=a-3$ is also a rational, contradiction, hence $\sqrt{5}+3$ is irrational and so is $\sqrt{\sqrt{5}+3}$.

Now, suppose your number $x=\sqrt{\sqrt{5}+3}+\sqrt{\sqrt{5}-2}$ is rational.

Then

$$\left(x-\sqrt{\sqrt{5}+3}\right)^2=x^2-2x\sqrt{\sqrt{5}+3}+\sqrt{5}+3=\sqrt{5}-2$$

Thus

$$x^2+5=2x\sqrt{\sqrt{5}+3}$$

$$\sqrt{\sqrt{5}+3}=\frac{x^2+5}{2x}$$

But if $x$ is rational, then the RHS is rational too. However the LHS is not rational. Contradiction, so $x$ is irrational.

As a side note, the proof is mostly similar to this one.

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  • $\begingroup$ Nice argument, as is the one you provided a link for! $\endgroup$ – user84413 Sep 19 '14 at 22:21
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Let $\alpha=\sqrt{\sqrt{5}+3}+\sqrt{\sqrt{5}-2}$.

Then $\alpha$ is a root of $x^8-4 x^6-26 x^4-100 x^2+625=0$.

If $\alpha$ is rational, then it must be a integer, by the rational root theorem.

From $2 < \sqrt 5 < 3$ and $0< \sqrt{5}-2 < 0.25$, we get that $2 < \alpha < 3.5$, so if $\alpha$ were an integer, it would have to be $3$. But $3$ is not root of the equation above because $3$ does not divide $625$.

Actually, from $2 < \sqrt 5 < 2.25$ we get $2 < \sqrt{5} < \sqrt{\sqrt{5}+3} < \sqrt{5.25} < \sqrt{6} < 2.5$ and so $2 < \alpha < 3$, and thus $\alpha$ is not an integer.

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You have not given a specific number, so I guess what you want is a way to check whether any arbitrary number is rational. Unfortunately, that rdoes not exist at today's level of math knowledge.

The way to prove that a number is rational is of course to demostrate its numerator and denominator.

One typical way to prove that a particular number involve assuming the number is rational, in which case it can be written as $p/q$ with $p$ and $q$ integers and the greatest common divisor of $p$ and $q$ 1 -- and then showing from the properties of the number that some factor must divide both $p$ and $q$. For example, have a look at $x=\sqrt{2}$. If $x$ is rational, write it as $$ x = \frac{p}{q}$$ with g.c.d(p,q) = 1. (For example, $14/10$ would be rejected because you can write that as $7/5$.) Then since $x^2=2$, $$ \frac{p^2}{q^2}=2 \rightarrow p^2 = 2q^2 \rightarrow p^2 \mbox{ is even} \rightarrow p \mbox{ is even} \rightarrow p = 2s $$ for some integer $s$. But then $$ (2s)^2 = 2q^2 \rightarrow 2s^2 = q^2 \rightarrow q^2 \mbox{ is even}\rightarrow q \mbox{ is even} $$ But if $p$ and $q$ are both even, we could have reduced that fraction $p/q$, because the g.c.d of $p$ and $q$ was $2$ (possibly times some other whole number). So $\sqrt{2}$ can't be written as a fraction in reduced form, so it is irrational.

If you have had pre-calculus, you may be familar with the number $$ e = \sum_{k=0}^\infty\frac{1}{k!} $$. We can prove that $e$ is irrational in a slightly different way: If $e$ is rational, write it as $p/q$ with $p$ and $q$ whole numbers. Then consider the real number $$ r = q!\left( e - \sum_{k=0}^q\frac{1}{k!} \right) $$ Since $q!e$ would be $(q-1)!p$ and thus an integer, and every term of the form $\frac{q!}{k!}$ for $k \leq q$ is obviously an integer, $r$ would be an integer.

But let's estimate $r$ in another way.
$$ r = q!\left(\sum_{k=0}^\infty\frac{1}{k!} - \sum_{k=0}^q\frac{1}{k!}\right) = \sum_{k=q+1}^\infty\frac{q!}{k!} $$ This is obviously positive.
But for $k > q$, $$ \frac{q\,!}{k!} = \frac{q\,!}{q\,!} \frac{1}{q+1} \frac{1}{q+2} \cdots \frac{1}{k} < \frac{1}{q+1} \frac{1}{q+1} \cdots \frac{1}{q+1} $$ with $k-q$ terms in that product. So $$ \frac{q\,!}{k!} < \frac{1}{(q+1)^k-q}$$ and $$ r < \sum_{k=q+1}^\infty\frac{1}{(q+1)^{k-q}}= \frac{1}{(q+1)^{q+1}}\frac{1}{1-\frac{1}{q+1}} =\frac{1}{(q+1)^{q+1}} \frac{q+1}{q} < \frac{1}{q(q+1)^q} < 1 $$ So if $e$ were rational, $r$ (defined in that way) would be a positive integer but less than $1$; this can't be so, so $e$ is irrational.

In general, any n-th root of an interger that is not a perfect n-th power is always irrational (for $n>1$). Other than those, you probably won't be able to prove any number is irrational.

For example, consider $$ \sum_{n=1}^{\infty}\frac{1}{n^3} $$ If proving that is irrational comes naturally to you, then I can safely say you have the talent to be a professional mathematician. At least.

For your P.S:

Get a book on the LaTeX document preparation system, created by Knuth. Expressions in Stack Exchange written in the form they would be written in LaTeX are typeset as if they were LaTeX. For example, the text expression \int x^\alpha dx, if enclosed by dollar signs at the beginning and end, comes out as $\int x^\alpha dx$.

A little trick: you can see the source of an answer by trying to edit that answer. It won;t let you post your edits, but you will see the LaTeX forms used.

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  • $\begingroup$ I hate when bad formatting in the initial question makes me answer a harder questions than was asked... I could not see the number you wanted to determine rationallity of! $\endgroup$ – Mark Fischler Sep 19 '14 at 23:03
  • $\begingroup$ But still +1 when I'll be able (vote limit reached by now), because, even though you don't answer the question, this is a very nice intro to irrational numbers! $\endgroup$ – Jean-Claude Arbaut Sep 19 '14 at 23:05
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A nice trick is to consider that over $\mathbb{F}_{41}$ both $5$ and $\pm\sqrt{5}+3=\pm 13+3$ are quadratic residues, while $\pm\sqrt{5}-2$ are not, so the splitting field of the minimal polynomial of $\alpha=\sqrt{\sqrt{5}+3}+\sqrt{\sqrt{5}+2}$ over $\mathbb{F}_{41}$ is $\mathbb{F}_{41^2}$ and $\alpha$ cannot be a rational number.

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