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!
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.
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.
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:
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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.
