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Problem: Prove that $\sqrt{2}+\sqrt{3}$ is irrational. The book where I encountered this problem had the following hint:

We make a polynomial with integer coefficients called $f(x)$ that $f(\sqrt{2}+\sqrt{3})=0$. (Why?)

Accepting this I solved the problem like this: If $x=\sqrt{2}+\sqrt{3}$ then $x^2=5+2\sqrt{6}$ and so $(x^2-5)^2=24$ thus: $$x^4-10x^2-1=0$$ But I want to know the reason that we should do this.

  • @Arthur: You're absolutely right. I forgot the rational root theorem! – Mathematician 42 Aug 08 '16 at 08:11
  • To show $\sqrt2 + \sqrt3$ is irrational, it is enough to show that your $f(x)$ cannot have rational roots - which is sometimes simpler. Of course, you do not have to do this, there are other ways as well. – Macavity Aug 08 '16 at 08:18

5 Answers5

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An easy way to see this is to notice that $$(\sqrt 2 + \sqrt 3)^2 = 5+2\sqrt 6.$$

Now, if $\sqrt 2 + \sqrt 3$ is rational, then so is its square. This implies that $(\sqrt 2 + \sqrt 3)^2 - 5 = 2\sqrt6$ is rational - a contradiction. Hope this helps.

Zestylemonzi
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The rational root theorem, states that if a rational number solves a given polynomial equation with integer coefficients, then it must be possible to write that number as $\frac pq$ where $p$ is an integer that divides the constant term, and $q$ is an integer that divides the highest-degree coefficient.

In this case we only need to check $\pm 1$, which clearly does not solve the equation. Thus there are no rational numbers that solves $x^4 - 10x^2 - 1 = 0$.

Arthur
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Directly:

If $r = \sqrt{2}+\sqrt{3}$ is rational,

$\begin{array}\\ \dfrac1{r} &=\dfrac1{\sqrt{2}+\sqrt{3}}\\ &=\dfrac1{\sqrt{2}+\sqrt{3}}\dfrac{-\sqrt{2}+\sqrt{3}}{-\sqrt{2}+\sqrt{3}}\\ &=-\sqrt{2}+\sqrt{3}\\ \end{array} $

is also rational.

Adding and subtracting these, $\sqrt{2}$ and $\sqrt{3}$ are rational.

marty cohen
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More generally, suppose $r =\sqrt{a}+\sqrt{b} $ is rational, where $a$ and $b$ are positive integers.

Then $r(\sqrt{a}-\sqrt{b}) =a-b $ so $\sqrt{a}-\sqrt{b} =\dfrac{a-b}{r} $ is also rational.

Adding and subtracting these, $\sqrt{a}$ and $\sqrt{b}$ are rational.

Therefore, if either or both of $\sqrt{a}$ and $\sqrt{b}$ are irrational, then $\sqrt{a}+\sqrt{b}$ is irrational (and similarly for $\sqrt{a}-\sqrt{b}$).

This might form a basis for a proof that $\sum_{k=1}^n \sqrt{a_k} $ is irrational under suitable assumptions on the $a_k$.

marty cohen
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You can do it the way you normally prove $\sqrt2$ is irrational.

Let $x/y = \sqrt2 + \sqrt3$ where x and y are integers with no common factors.

You have

$x^4-10\cdot x^2y^2 + y^4 = 0$

So

$(x^2 - y^2)^2 - 8 \cdot x^2y^2 = 0$

$(x^2 - y^2)^2 = 8 \cdot x^2y^2$

Because the left side is divisible by 4, $x^2 - y^2$ is divisible by $2$. Thus, x and y are the same parity. Since there are no common factors, they are both odd.

Let $x^2-y^2 = 2m$ where m is an integer.

$m^2 = 2\cdot x^2y^2$

m must be even. Let $m=2n$.

$4n^2 = 2\cdot x^2y^2$

$2n^2 = x^2y^2$

So $x^2y^2$ is even. This means one of x and y is even which is a contradiction.

ski Lodge
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