Prove that $\sqrt{2}+\sqrt{3}+\sqrt{5}$ is irrational. Generalise this.

Question

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

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

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.

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

Answer 1

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

Answer 2

Hint:

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

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 $$b_1\sqrt{a_1}+...+b_n\sqrt{a_n} \in \mathbb Q$$ 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 $P_m$.

Answer 3

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 $$2\sqrt{6}=\frac{m^2}{n^2}-2\frac{m}{n}\sqrt{5};$$ and squaring once again then simplifying, we arrive at $$5\frac{n}{m}+\frac{m}{4n}-6\frac{n^3}{m^3}=\sqrt{5}.$$

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

Answer 4

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.

Answer 5

An alternative solution. Assume that $\alpha=\sqrt{2}+\sqrt{3}+\sqrt{5}=\frac{a}{b}\in\mathbb{Q}$. 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 $\alpha\in\mathbb{K}$, contradiction.

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

Answer 6

Using the straightforward fact that the sum of two (or more) algebraic integers is an algebraic integer (see here), one has the following very short argument: since $\sqrt{2},\sqrt{3},\sqrt{5}$ are algebraic integers, their sum $A=\sqrt{2}+\sqrt{3}+\sqrt{5}$ must be also. Hence if $A$ is rational, it must be an integer. However, as a rudimentary estimate of the three square roots shows, we have $5 < A < 6$: contradiction.