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

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

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 :)


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.

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    $\begingroup$ @Wolfram Unlikely, from the chosen reference ... $\endgroup$ – Michael Burr Feb 2 '17 at 15:14
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    $\begingroup$ @Wolfram A little bit, yes, from Galois Theory and Algebraic Number Theory, but I'm approaching this from Elementary Number Theory, as the title of the book suggests. $\endgroup$ – Shaun Feb 2 '17 at 15:15
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    $\begingroup$ You can work out the minimal polynomial of your number, it turns out to be $x^8 - 40 x^6 + 352 x^4 - 960 x^2 + 576$ (I did it with Wolfram Alpha, could be done by hand with some effort). $\endgroup$ – lulu Feb 2 '17 at 15:17
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    $\begingroup$ Using some well known results from field theory one sees that $1, \sqrt{2}, \sqrt{3}$ and $\sqrt{5}$ are members of some basis of $\Bbb{Q}[\sqrt{2},\sqrt{3},\sqrt{5}]$ over $\Bbb{Q}$. $\sqrt{2}+\sqrt{3}+\sqrt{5}$ can be uniquely expressed as a linear combination of elements of this basis, so it is irrational. $\endgroup$ – Vik78 Feb 2 '17 at 15:31
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    $\begingroup$ You write "The "try to make up similar and more general examples" bit is a little vague." The question is quite interesting. You may find the following conclusion: "The sum of an arbitrary nimber square roots of prime numbers is irrational." Try to imagine a few other theorems. The lemma in @N.S.'s answer seems to be the most general. $\endgroup$ – Bernard Massé Feb 2 '17 at 17:08


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 $P_m$.

  • $\begingroup$ It should be +2 sqrt 5 psi $\endgroup$ – user236182 Feb 2 '17 at 15:35

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.

  • $\begingroup$ Ah, of course! Thank you! $\ddot\smile$ $\endgroup$ – Shaun Feb 2 '17 at 15:19

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.


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.


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.

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    $\begingroup$ Here's a MathJax tutorial. $\endgroup$ – Shaun Feb 2 '17 at 15:24
  • $\begingroup$ Hi. Your post could use some layout, you can read on the mathjax tutorial how to layout the math. $\endgroup$ – mathreadler Feb 2 '17 at 17:30
  • $\begingroup$ Cut him some slack, he's new. But he'll be expected to start doing it himself as time goes on. $\endgroup$ – Mr. Brooks Feb 4 '17 at 22:34
  • $\begingroup$ Thanks Sir for the edit.I am new and I am trying to learn stuffs $\endgroup$ – Abhishek Kumar Feb 5 '17 at 5:24

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