Htykuut

I'm reading R. Courant & H. Robbins' "What is Mathematics: An Elementary Approach to Ideas and Methods" for fun. I'm on page $60$ and $61$ of the second addition. There are three exercises on proving numbers irrational spanning these pages, the last is as follows.

Exercise $3$: Prove that $phi=sqrt{2}+sqrt{3}+sqrt{5}$ is irrational. Try to make up similar and more general examples.

My Attempt:

Lemma: The number $sqrt{2}+sqrt{3}$ is irrational. (This is part of Exercise 2.)

Proof: Suppose $sqrt{2}+sqrt{3}=r$ is rational. Then
$$begin{align}
2&=(r-sqrt{3})^2 \
&=r^2-2sqrt{3}+3
end{align}$$ is rational, so that
$$sqrt{3}=frac{r^2+1}{2r}$$ is rational, a contradiction. $square$

Let $psi=sqrt{2}+sqrt{3}$. Then, considering $phi$,
$$begin{align}
5&=(phi-psi)^2 \
&=phi^2-psiphi+5+2sqrt{6}.
end{align}$$

I don't know what else to do from here.
My plan is/was to use the Lemma above as the focus for a contradiction, showing $psi$ is rational somehow.

Please help :)

Thoughts:

The "try to make up similar and more general examples" bit is a little vague.

The question is not answered here as far as I can tell.

Hint:

Assume by contradiction that $psi=sqrt{2}+sqrt{3}+sqrt{5}$ is rational. Then
$$(psi-sqrt{5})^2=(sqrt{2}+sqrt{3})^2 \
psi^2-2sqrt{5}psi+5=5+2sqrt{6} \
psi^2=2sqrt{6} +2sqrt{5}psi\
$$

Square it one more time, and you reach the contradiction.

For the general case, here is a nice trivial solution from Kvant: prove the following lemma:

Lemma If $a_1,a_2,..,a_k$ are distinct integers $geq 2$, none of which is divisible by a square, and $b_1,..,b_n$ are integers such that
$$b_1sqrt{a_1}+...+b_nsqrt{a_n} in mathbb Q$$
then $b_1=...=b_n=0$.

Proof: Do induction by the number $m$ of primes which divide $a_1...a_n$.

The inductive step $P(m)Rightarrow P(m+1)$ is done by contradiction: move all terms for which $a_k$ is divisible by the $p_{m+1}$ on one side, everything else on the other side and square, to end in the case $P_m$.

Okay, if $sqrt{2}+sqrt{3}+sqrt{5}=rinmathbb Q$, then:
$$(sqrt{2}+sqrt{3})^2=(r-sqrt{5})^2$$
$$2+2sqrt6+3=r^2-2sqrt5 r+5$$
$$2sqrt6=r^2-2sqrt5 r$$
Square this once again and you obtain that $sqrt5$ is rational, which is a contradiction.

We assume we have
$$sqrt{2}+sqrt{3}+sqrt{5}=frac{m}{n}$$ with $gcd(m,n)=1$. We write
$$sqrt{2}+sqrt{3}=frac{m}{n}-sqrt{5};$$
squaring, we have
$$2sqrt{6}=frac{m^2}{n^2}-2frac{m}{n}sqrt{5};$$
and squaring once again then simplifying, we arrive at
$$5frac{n}{m}+frac{m}{4n}-6frac{n^3}{m^3}=sqrt{5}.$$

This is a contradiction, since we have on the left only rational numbers and on the right side an irrational number.

An alternative solution. Assume that $alpha=sqrt{2}+sqrt{3}+sqrt{5}=frac{a}{b}inmathbb{Q}$. By quadratic reciprocity, there is some prime $p>b$ such that $3$ and $5$ are quadratic residues $!!pmod{p}$ while $2$ is not. That implies that $alpha$ is an algebraic number over $mathbb{K}=mathbb{F}_p$ with degree $2$, since $sqrt{2}$ does not belong to $mathbb{K}$ but belongs to a quadratic extension of $mathbb{K}$. On the other hand $b<p$ ensures that $b$ is an invertible element $!!pmod{p}$ and $alphainmathbb{K}$, contradiction.

This proof has a straightforward generalization: the sum of the square roots of some prime numbers is never a rational number.

Use these properties of rational numbers:

1. If $x$ is rational then $x^2$ is also rational.


2. Sum of a rational number and an irrational number is irrational.

Let's suppose $x = sqrt 2 + sqrt 3$ is rational. Then $x^2 = 5 + 2 sqrt 6$ (which is irrational using the second property).

Using the first property, $x$ also becomes irrational.

Now using the second property we can say that since $x(sqrt 2 + sqrt 3)$ is irrational, irrespective of whether $sqrt 5$ is rational or irrational using the second property, the total sum will be irrational. Thus $sqrt 2 + sqrt 3 + sqrt 5$ is irrational.