In middle school, the definition of a rational number is the set of integers and fractions; more formally, it is any number that can be written as the ratio of two integers. Correspondingly, an irrational number is naturally a number that cannot be written as the ratio of two integers—that is, an infinite non-repeating decimal, such as \pi, \sqrt{2}, etc. Historically, the discovery of irrational numbers brought about the first mathematical crisis and laid a "golden egg," though the discoverer lost his life because of it. Let us forever remember—Hippasus.
History:
I won’t say much more about irrational numbers themselves here; I mainly want to discuss the related proofs.
First, let me clarify that the following are my own proof methods. While I believe a universal method exists, I have not yet identified it.
1. \sqrt{2} is an irrational number.
Proof: Suppose \sqrt{2} is a rational number, and let \sqrt{2} = p/q, where p/q is a fraction in its simplest form. Since \sqrt{2} is not an integer, q > 1.
Squaring both sides, we get 2 = p^2/q^2. Since p/q is in simplest form, p^2/q^2 is also in simplest form. The simplest fraction form of 2 can only be 2/1, which implies q = 1. This contradicts q > 1. Therefore, the assumption is false, and \sqrt{2} is an irrational number.
2. \sqrt{2} + \sqrt{3} is an irrational number.
Proof: This one is simple; I believe everyone knows how to do it.
Suppose \sqrt{2} + \sqrt{3} = p is a rational number. Squaring both sides gives 5 + 2\sqrt{6} = p^2 \Rightarrow \sqrt{6} = (p^2 - 5)/2. Thus \sqrt{6} would be a rational number, which is a contradiction. Therefore, the assumption is false.
3. \sqrt{2} + \sqrt{3} + \sqrt{5} is an irrational number.
Proof: This is not quite so simple; I thought about it for a long time.
(1) Suppose \sqrt{2} + \sqrt{3} + \sqrt{5} = p is a rational number. 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. Thus \sqrt{6} + \sqrt{10} + \sqrt{15} is a rational number. Squaring both sides again: 31 + 10\sqrt{6} + 6\sqrt{10} + 4\sqrt{15} = ((p^2 - 10)^2)/4, which implies 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}) The left side of this equation must be a rational number. However, using Method 2, it can be proven that 3\sqrt{6} + \sqrt{10} is irrational. This is a contradiction, so the assumption is false.
(2) Similarly, suppose \sqrt{2} + \sqrt{3} + \sqrt{5} = p is a rational number. We rearrange the equation as: \sqrt{2} + \sqrt{3} = p - \sqrt{5}. Squaring 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} This implies \sqrt{6} + p\sqrt{5} is a rational number, which is also a contradiction.
4. \sqrt{2} + \sqrt{3} + \sqrt{5} + \sqrt{7} is an irrational number.
Proof: After sketching on paper all day, I finally have a lead.
Similarly, suppose \sqrt{2} + \sqrt{3} + \sqrt{5} + \sqrt{7} = p is a rational number. We rearrange the equation as: \sqrt{2} + \sqrt{3} + \sqrt{5} = p - \sqrt{7}. Squaring both sides gives: 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, which 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, it follows that 2\sqrt{10} + 4\sqrt{6} + p(p^2 - 7)\sqrt{7} is rational, which is a contradiction.
The above are proofs regarding the sums of square roots. The proofs are somewhat rough and tedious, but the principles are extremely easy to understand. The scope of this method is limited, supporting up to the sum of four square roots; I am currently searching for a universal method...
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