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We all know the square root of 2 is a irrational number. So I think we cannot use a ruler and draw a line that is the square root of 2 long. I think anybody agree with that.

But we can draw a right angle triangle with two sides of 1 unit, then you can confidently draw a hypotenuse. That hypotenuse is the square root of 2 unit! How to explain that?

peter
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    Non sequitur: Impossibility to construct a length (ratio) of $\sqrt 2$ does not follow from its irrationality. There is nothing to explain. – Hagen von Eitzen Feb 28 '13 at 10:19
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    We can also draw a circle. That doesn't make $\pi$ rational... – xavierm02 Feb 28 '13 at 10:22
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    we can only draw approximations of such perfect shapes, so there is indeed nothing to explain. – Ittay Weiss Feb 28 '13 at 10:36
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    We can't even use a ruler and draw a line that is 1 unit long! –  Feb 28 '13 at 10:49
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    @Hurkyl, well, in compass and straightedge constructions, a segment of length $1$ is a given. – Andreas Caranti Feb 28 '13 at 10:55
  • Constructible numbers(on the x-y plane) are actually real. And any irrational number can be constructed. You can actually talk about imaginary numbers when its about constructible numbers. – Inceptio Feb 28 '13 at 11:39
  • @Inceptio Might I know how to prove that any irrational number is constructible? For example, how to construct $2^{1/3}$? Thanks in any case. – awllower Feb 28 '13 at 17:48
  • “I think anybody agree with that.” I’m a body, and I don’t agree. Further: @Inceptio, it is not true that every irrational number can be constructed: the constructible numbers are very special among algebraic numbers. – Lubin Mar 01 '13 at 23:16
  • @HagenvonEitzen: It would have to be transcendental to be non-construtible, right? – Adam Dec 05 '13 at 18:00
  • @Adam Transcendental is sufficient, but not necessary. For example $\sqrt[3]2$ is not constructible either. More precisely, a number is constructible in this sense if the degree of its minimal polynomial over the rationals is a power of two. – Hagen von Eitzen Dec 05 '13 at 19:00
  • http://math.stackexchange.com/questions/917983/the-proof-of-sqrt2-is-not-rational-number-via-fundamental-theorem-of-arithm –  Nov 20 '15 at 00:34

2 Answers2

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You're mixing up things. You are probably thinking of constructible numbers, and of course they are not confined to be rational.

And then I may be wrong, but to draw a right-angle triangle you need a compass alongside the straight-edge. (It depends on the rules of the game, but in the usual interpretation you do.)

Andreas Caranti
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  • I am not well trained in Math. I just know the term of constructible number. The right angle triangle don't has to be drawn by physical tools. It can be just drawn in our mind. As to contructible number, maybe we need think why it is not fully compatible with the irrational number square root of 2. – peter Feb 28 '13 at 23:54
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The ancient Greeks thought some explanation was needed but their reasoning was:

  • $\sqrt2$ is constructible with ruler and compass

  • $\sqrt2$ is irrational

How can something that is clearly a number be irrational? Thus began a major crisis in Greek mathematics.

Of course, like others have said, nowadays we see nothing wrong with constructible numbers being irrational.

lhf
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  • I believe the starting point of the crisis was the proof of the irrationality of $\varphi$, the golden ratio appearing in the regular pentagon. But $\sqrt{2}$ is easier to grasp, that why it's used universally. – vonbrand Feb 28 '13 at 17:22
  • In Mathematics and its History, Stillwell suggests that this crisis was not resolved until the invention of the real numbers starting in the 1600s. – MJD Dec 15 '14 at 05:46
  • I don't have an answer to my question. The question has stayed in my mind for many years before I asked this question. The question is not simple as it seems on the surface. – peter Nov 22 '15 at 09:57