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Michael Spivak in "Calculus" asserts that $\sqrt2$ cannot be proven to exist, and that such a proof is impossible. What does he mean by "exist"? How are you to prove that any number "exists"? Why can't we define $\sqrt2$ as a number that fits under some arbitrary definition of existence, while asserting that its most concise expression is with a functional root?

I'm sorry if these questions seem a bit sophomoric; in some ways it resembles an 8 year old repeatedly asking "why". But given that his prose is very concise and technical, his usage of "exist" was out of the ordinary.

(I used two tags representing the book's field of study; and one representing the actual relevant tag.)

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Oh, I'm sorry. I misquoted. My question still stands, though; how has he defined existence such that $\sqrt2$ might possible not be within it.

Direct quote: "We have not proved that any such number exists..." in reference to $\sqrt2$.

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Where in the book, exactly? –  lhf Jun 10 '12 at 18:53
    
Page 26 in the Prologue. –  whoops Jun 10 '12 at 18:55
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I don't think Spivak says anything like that. What I read in his wonderful book is that "at present a proof [of its existence] is impossible for us...unless you can give an exact quote (with the page number, say) where he says otherwise. –  DonAntonio Jun 10 '12 at 18:56
    
Oh, I'm sorry. I misquoted. My question still stands, though; how has he defined existence such that $\sqrt2$ might possible not be within it. Let me edit the original post. –  whoops Jun 10 '12 at 18:57
    
Exactly: in page 26 Spivaks does not say that, but only that at that moment the square root of 2 can't be proved to exist (in some sense). –  DonAntonio Jun 10 '12 at 18:58

6 Answers 6

The point Spivak is making is that the properties of numbers that have been studied till that point in the book are not enough to prove that there is a number whose square is 2. This follows from the proof that no rational number will do for this task. Since the rational numbers satisfy all properties till that point, it is clear that some other property is needed (and that the rationals cannot have this property). That property is completeness and is a what characterizes real numbers. Using completeness, one can and does prove that there is a positive number whose square is 2; we call it $\sqrt2$.

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Spivak's point is: one can't prove that $\rm\:x^2\!-\!2\:$ has a root in $\rm\:\mathbb R\:$ using only the axioms $\rm\: P1\!-\!P12,\:$ i.e. axioms for an ordered field, since he has just proved that it has no root in the ordered field $\rm\mathbb Q.\:$

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Spivak says that $\,\sqrt{2}\,$ exists means that there's a real number $\,x\,\,s.t.\,\,x^2=2\,$ , and this cannot be proved with the knowledge assumed in page 26 of his book.

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So "exist" essentially refers to any proposition that can be made following from the 12 assumed propositions. Got it. –  whoops Jun 10 '12 at 19:03
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@whoops: No, the point is not whether it exists or not, but whether it can be proved to exist. Those are two very different things, and you're doing yourself and the discussion a disservice by continuing to pretend that the author is saying "does not exist" when all he actually says is "cannot be proved to exist". –  Henning Makholm Jun 10 '12 at 23:03

The construction of the irrational numbers can be undertaken in several ways. You can think of them as the "completion" of the rational numbers in the sense that after you add them to the mix, every Cauchy sequence will be convergent. You can also construct the irrational numbers through Dedekind cuts. If I remember well, this is done towards the end of the book.

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Actually upto that point, Least upper bound property was no known. And it is impossible to prove(existence of roots) without using least upper bound property. Read chapter $7$ from the same book, its explained well in the end

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I believe that the existence of √2 was proven in 300 B.C. give or take.

Just draw a right triangle, with the sides being exactly 1 unit in length, then the hypotenuse of the triangle - using the Pythagorean theorem - will be √2 in length.

I fully concede that I probably missed that whole point of this conversation, but that's how you prove the existence of √2 anyway :)

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In the context of the real numbers, $\sqrt 2$ can be proved to exist using the least upper bound property: $\sqrt 2=\sup\{t:t^2<2\}$. –  Jonas Meyer Jul 1 '12 at 10:06

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