# 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!

• – Bart Michels Apr 7 '15 at 12:22
• – Watson Nov 25 '18 at 19:45
• – Watson Nov 25 '18 at 21:45
• – Watson Jan 29 '19 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.

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.

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

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