How to prove that $\sqrt 2 + \sqrt 4$ is irrational? [duplicate]

So while doing all sorts of proving and disproving statements regarding irrational numbers, I ran into this one and it quite stumped me:

Prove that $\sqrt{2} + \sqrt{4}$ is irrational.

I tried all the usual suspects like playing with $\sqrt{2} + \sqrt{4} = \frac{a}{b}$ for $a,b\in \mathbb{Z}$ , but got nowhere.

I also figured maybe I should play with it this way:

$2^\frac{1}{3} + 4^\frac{1}{3}=2^\frac{1}{3} + (2^2)^\frac{1}{3}=2^\frac{1}{3} + 2^\frac{2}{3}=2^\frac{1}{3} + 2^\frac{1}{3}\times 2^\frac{1}{3}=2^\frac{1}{3}(1+2^\frac{1}{3})$

But there I got stumped again, because while $1+2^\frac{1}{3}$ is irrational, nothing promises me that $2^\frac{1}{3} \times (1+2^\frac{1}{3})$ is irrational, and I feel like trying to go further down this road is moot.

So what am I missing (other than sleep and food)? What route should I take to prove this? Thanks in advance!

marked as duplicate by Watson, Chinnapparaj R, user10354138, Brahadeesh, KReiserNov 26 '18 at 8:38

• – punctured dusk Apr 7 '15 at 12:22
• – Watson Nov 25 '18 at 19:45
• – Watson Nov 25 '18 at 21:45
• – Watson Jan 29 at 9:06

Note

$$1 + \sqrt{2} + \sqrt{4} = \frac{(\sqrt{2})^3 - 1}{\sqrt{2} - 1} = \frac{1}{\sqrt{2} - 1}.$$

So if $\sqrt{2} + \sqrt{4}$ is rational, then $1/(\sqrt{2} - 1)$ is rational, which implies $\sqrt{2} - 1$ is rational. Then $\sqrt{2}$ is rational, a contradiction.

• Very nice answer. +1 – Timbuc Mar 26 '15 at 18:00
• Thanks for the reply. I must be tired or confused because I'm missing something very basic and don't realize how $1 + \sqrt{2} + \sqrt{4} = \frac{(\sqrt{2})^3 - 1}{\sqrt{2} - 1}$. I apologize for my stupidity - it's been a tough day. – Elad Avron Mar 26 '15 at 18:01
• Let $x = \sqrt{2}$. Use the factorization $(1 + x + x^2)(x - 1) = x^3 - 1$ to obtain the identity. – kobe Mar 26 '15 at 18:04
• Of course, silly me. Thank you so much, this solution is brilliant! – Elad Avron Mar 26 '15 at 18:09
• @Elad One easly proves a much more general result: for irrational cube roots of rationals, said property fails only for $\,\sqrt1,\,$ see my answer. $\ \$ – Bill Dubuque Mar 26 '15 at 20:21

$y = \sqrt{2} + \sqrt{4}$ is a root of the equation $y^3 - 6y - 6 = 0$.

(To see this, let $x = \sqrt{2}$ and $y = \sqrt{2}+\sqrt{4}=x+x^2$. Then $y^3 = x^3 + 3x^4 + 3x^5 + x^6 = 2 + 6x + 6x^2 + 4 = 6 + 6y$.)

By the rational root theorem, we know that any rational roots of $y^3 - 6y - 6 = 0$ would have to be in the set $\{\pm1,\pm2,\pm3,\pm6\}$; we can quickly confirm that none of these is in fact a root, meaning that the equation does not have any rational roots.

Therefore, $y$ must be irrational.

• Short, sweet, and intuitive. Very nice. – Jon Mar 26 '15 at 21:16

An easy approach: If $p=\sqrt{2}+\sqrt{4}$ is rational:

$$p^2 = \sqrt{4}+2\cdot 2+ 2\sqrt{2} = p+4+\sqrt{2}.$$

So $\sqrt{2}=p^2-p-4$ would be rational.

An alternative, more general approach.

Claim: If $a,b$ are integers that are not perfect cubes, and $a\neq -b$, then $\sqrt{a}+\sqrtb$ is irrational.

Proof:

Assume $\sqrt{a}+\sqrt{b}$ is rational. Cube it and get:

$$(\sqrt{a}+\sqrt{b})^3 = a+ 3\sqrt{ab}(\sqrt{a}+\sqrt{b}) + b$$

Now, since $a,b$ are rational and $\sqrt{a}+\sqrt{b}$ is non-zero and rational, this means that $\sqrt{ab}$ is rational.

Letting $p=\sqrt{a}+\sqrt{b}$ and $q=\sqrt{ab}$, this means that

$$(x-\sqrt{a})(x-\sqrt{b}) = x^2-px+q$$ is a rational polynomial. It shares at least one root with $x^3-b$, But the GCD of these two polynomials has to be a rational polynomial, so the GCD cannot be linear (since it would be $x-\sqrtb$, which is not a rational polynomial.)

This means that $x^2-px+q$ has to divide $x^3-b$. That means that $\sqrt{a}$ is a root of $x^3-b$, which means that $a=b$. But there are no repeated roots of $x^3-b$, which yields a contradiction.

Corollary: If $a,b$ are rationals such that $a\neq -b$ and $\sqrt{a}$ and $\sqrt{b}$ are irrational, then $\sqrt{a}+\sqrt{b}$ is irrational.

Proof: Rationalize the denominators and revert to the above theorem for integers.

It holds true for any  irrational $\,x=\sqrtn,\,$ except $\,n=1\,$ (so $\,x^2+x=-1\in\Bbb Q)$

More generally:  if $\ r\in\Bbb Q\$ and $\,x=\sqrtr\not\in\Bbb Q\,$ then $\,x^2+x = q\in\Bbb Q\iff r = 1.$

Proof $\,\ qx = x^3+x^2 = r+x^2\$ so $\ qx-r = x^2 = q-x,\$ so $\,(\color{#c00}{q\!+\!1})\,x = r+q.$

Therefore $\,\ x\not\in\Bbb Q\,\Rightarrow\,\color{#c00}{q = -1}\,\Rightarrow\, 0 = (\color{#c00}{q\!+\!1})x = r+q = r-1,\$ thus $\,\ r = 1.$

• Um, if $n=1$, isn't it $x^2+x=2$? And is $x=\sqrt{1}$ irrational? I'm confused. – Thomas Andrews Mar 29 '15 at 4:54
• @Thomas The hypothesis is that $\,x\,$ is an irrational cube root of $\,n.\ \$ – Bill Dubuque Mar 29 '15 at 6:41
• Well, $\sqrt{n}$ is a single-valued function, by convention. You could say any irrational $x$ with $x^3=n$, I suppose. – Thomas Andrews Mar 29 '15 at 15:25
• @Thomas Algebraists not too infrequently overload the notation to denote any root. I will change it to avoid possible confusion. Thanks for pointing that out. – Bill Dubuque Mar 29 '15 at 16:25

Here is another approach which generalises to many similar examples, and which involves no complicated simplifications of surds. (In fact, no easy simplifications of surds either.)

First, $\sqrt2$ is an algebraic integer, that is, it is a root of $$x^3-2$$ which is a monic polynomial (leading coefficient $1$) with integer coefficients. Similarly, $\sqrt4$ is an algebraic integer.

It is known that the sum of two algebraic integers is an algebraic integer. Thus, $x=\sqrt2+\sqrt4$ is an algebraic integer.

It is also known that if an algebraic integer is rational, then it is a rational integer - that is, an ordinary integer, $0,1,-1,2,-2$ etc. However, it is not hard to find the estimates $$1<\sqrt2<\frac43 ,\quad \frac43<\sqrt4<\frac53\ ;$$ so $\frac73<x<3$, hence $x$ is not an integer and must be irrational.

If $a$ is rational, $a^2-a-4$ is rational. What is $a^2-a-4$ when $a=\sqrt2+\sqrt4$?

(After you work that out, I bet you'll wonder where I got $a^2-a-4$ from. Or, try to figure it out yourself. Now that you have a sort of idea on how to deal with this sort of problem, I leave you with another, similar problem: What polynomial can I use to prove that $\sqrt3+\sqrt9$ is irrational?)

• I was going to suggest proving $\sqrt2+\sqrt3$ irrational, but then I realized that the polynomial involved is an eight-degree one. And I'm not sure how hard the one I gave is; I didn't try it yet. – Akiva Weinberger Mar 26 '15 at 20:49