On Proofs Concerning Irrational Numbers and Their Sums

关于无理数及其和的证明 苏剑林 · 2009-07-31 · Mathematics proof summation irrational number

] In middle school, rational numbers are defined as the set of integers and fractions—put uniformly, numbers that can be written as the ratio of two integers. Correspondingly, irrational numbers are naturally those that cannot be written as the ratio of two integers, i.e., infinite non-repeating decimals, such as $\pi,\sqrt{2}$ and so on. Historically, the discovery of irrational numbers triggered the first crisis in the foundations of mathematics, and gave birth to a "golden egg"—though its discoverer lost his life for it. Let us forever remember him: Hippasus.

History:

http://baike.baidu.com/view/1167.htm#2]

I won't say much more about irrational numbers here; the main point is to discuss the related proofs.

Let me note upfront that what follows is my own method of proof. I believe there is a general method that covers all these cases, but I haven't managed to find it. more

1. $\sqrt{2}$ is irrational.

Proof: Suppose $\sqrt{2}$ is rational, and let $\sqrt{2}=p/q$, where $p/q$ is in lowest terms. Since $\sqrt{2}$ is not an integer, we have q > 1.
Squaring both sides gives $2={p^2}/{q^2}$. Since $p/q$ is in lowest terms, ${p^2}/{q^2}$ is also in lowest terms. But the only way for 2 to appear as a fraction in lowest terms is $2/1$, which forces q = 1—contradicting q > 1. So the assumption fails, and $\sqrt{2}$ is irrational.

2. $\sqrt{2}+\sqrt{3}$ is irrational.

Proof: This one is simple, and I'm sure everyone already knows it.
Suppose $\sqrt{2}+\sqrt{3}=p$ is rational. Squaring both sides gives $5+2\sqrt{6}=p^2\Rightarrow \sqrt{6}={p^2-5}/2$, so $\sqrt{6}$ is rational—a contradiction. Hence the assumption fails.

3. $\sqrt{2}+\sqrt{3}+\sqrt{5}$ is irrational.

Proof: This one isn't so simple—it took me quite a while to work out.

(1) Suppose $\sqrt{2}+\sqrt{3}+\sqrt{5}=p$ is rational. Squaring both sides gives
$10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}=p^2\Rightarrow \sqrt{6}+\sqrt{10}+\sqrt{15}={p^2-10}/2$,
so $\sqrt{6}+\sqrt{10}+\sqrt{15}$ is rational. Squaring again gives:
$31+10\sqrt{6}+6\sqrt{10}+4\sqrt{15}={(p^2-10)^2}/4$,
so $5\sqrt{6}+3\sqrt{10}+2\sqrt{15}={{(p^2-10)^2}/4-31}/2$.
Since $\sqrt{6}+\sqrt{10}+\sqrt{15}$ is rational, then
$$3\sqrt{6}+\sqrt{10}={{(p^2-10)^2}/4-31}/2-2(\sqrt{6}+\sqrt{10}+\sqrt{15})$$
has a rational number on its left-hand side. But by the method used in item 2, $3\sqrt{6}+\sqrt{10}$ can be shown to be irrational—a contradiction. So the assumption fails.

(2) Likewise, suppose $\sqrt{2}+\sqrt{3}+\sqrt{5}=p$ is rational. Rewrite the equation as
$\sqrt{2}+\sqrt{3}=p-\sqrt{5}$, and square both sides:
$$\begin{aligned}(\sqrt{2}+\sqrt{3})^2=(p-\sqrt{5})^2 \\ \Rightarrow 5+2\sqrt{6}=p^2+5-2p\sqrt{5}\end{aligned}$$
which shows that $\sqrt{6}+p\sqrt{5}$ is rational—again a contradiction.

4. $\sqrt{2}+\sqrt{3}+\sqrt{5}+\sqrt{7}$ is irrational.

Proof: I spent a whole day scribbling on newspaper before finally seeing a way through.
Again, suppose $\sqrt{2}+\sqrt{3}+\sqrt{5}+\sqrt{7}=p$ is rational. Rewrite the equation as
$\sqrt{2}+\sqrt{3}+\sqrt{5}=p-\sqrt{7}$, and square both sides to get:
$$10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}=p^2+7-2p\sqrt{7}$$
$\Rightarrow \sqrt{6}+\sqrt{10}+\sqrt{15}+p\sqrt{7}=(p^2-3)/2$ is a rational number.
Squaring both sides again:
$$\begin{aligned}(\sqrt{6}+\sqrt{10}+\sqrt{15})^2=((p^2-3)/2-p\sqrt{7})^2 \\ \Rightarrow 31+10\sqrt{6}+6\sqrt{10}+4\sqrt{15}=((p^2-3)/2)^2+7p^2-p(p^2-3)\sqrt{7} \\ \Rightarrow 2\sqrt{10}+4\sqrt{6}+p(p^2-7)\sqrt{7}=((p^2-3)/2)^2+7p^2-4(\sqrt{6}+\sqrt{10}+\sqrt{15}+p\sqrt{7})\end{aligned}$$
Since $\sqrt{6}+\sqrt{10}+\sqrt{15}+p\sqrt{7}=(p^2-3)/2$ is rational, $2\sqrt{10}+4\sqrt{6}+p(p^2-7)\sqrt{7}$ is rational—a contradiction.

The above are proofs concerning sums of square roots. The proofs are somewhat crude and roundabout, though the underlying principle is quite easy to grasp. The method has limited reach—it works at most up to sums of four square roots—and a general method is still being sought...