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Sometimes you have a real number (with a rather complicated definition), and with some effort you can show that

1) this real number is, actually, an integer;

2) the distance of this real number to an integer, say $0$, is less than $1/2$.

Thus you can conclude that this real number is $0$! I think this is a very nice trick. Especially when the argument for 1) is so involved that you don't really see this a priori. However, I don't remember in which context I have seen this. But I guess that this trick works in various situations.

So my question is: Can you give nice, explicit instances of this trick?

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    $\begingroup$ Any irrationality/transcendence proof use this trick (sometimes implicitly). You may get too many examples of this type. In any case, I can't understand why don't you make your question community wiki. $\endgroup$
    – Wadim Zudilin
    Commented Jul 12, 2010 at 11:06
  • $\begingroup$ Ok, I've made it CW. $\endgroup$
    – Martin Brandenburg
    Commented Jul 12, 2010 at 11:21
  • $\begingroup$ Hm, Martin, you don't like proofs by contradiction. In that case your question is somehow contradictory! :-) The only family of examples that seems to be useful is related to a PV number, say $v$: the sum of the $N$th powers of all conjugates of $v$ is an integer which is the nearest to $v^N$. This trick (a la Lobachevsky) is indeed useful in several results on PV numbers (as far as I remember in Cassels' book on diophantine approximations). You can also do a similar thing for the reciprocal of $j((1+\sqrt{-163})/2)$... $\endgroup$
    – Wadim Zudilin
    Commented Jul 12, 2010 at 12:04
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    $\begingroup$ Real number happy to be an integer - I need more coffee! $\endgroup$
    – Franz Lemmermeyer
    Commented Jul 12, 2010 at 14:39

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Your question has a logical formulation that leads to an open problem.

A Discretely Ordered Ring is an ordered ring in which the inequality $x<y<x+1$ has no solutions, or equivalently, an ordered ring containing no element between 0 and 1. There is a simple finite set of axioms for the class of discretely ordered rings. Hilbert's Tenth Problem for discretely ordered rings asks:

Is it decidable whether given a system of polynomial equations with integer coefficients, there is some (at least one) discretely ordered ring in which that system of equations is solvable?

This is equivalent to asking if the set of unsolvable polynomial systems whose unsolvability follows from the axioms for discretely ordered rings can be effectively listed.

The best result so far, due to van den Dries, is that the answer is positive for a single polynomial equation in two variables. The proof uses basic facts about algebraic function fields, especially the Riemann-Roch Theorem. See "Which Curves over Z have Points in a Discretely Ordered Ring?", Transactions of the AMS. March 1981

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I'm not sure how related this particular example is to your question (but I think it's interesting). generatingfunctionology contains a proof that the Fibonacci numbers are approximated by \[F_n \sim \frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n.\] It is followed by the remark that the error term is never more then 1/2. Hence Fn is the nearest integer to the above approximation.

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  • $\begingroup$ This is a special case of the general phenomenon involving PV numbers (Pisot-Vijayaraghavan) mentioned by Wadim. $\endgroup$
    – Victor Protsak
    Commented Jul 18, 2010 at 13:32
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I guess the series expansion for the partition number $p(n)$ by Rademacher (see http://en.wikipedia.org/wiki/Partition_(number_theory) ) is a nice illustration. The number $p(n)$ is obviously an integer which is uniquely defined by enough terms of Rademacher's formula.

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  • $\begingroup$ Yes, this is indeed a nice example. $\endgroup$
    – Wadim Zudilin
    Commented Jul 12, 2010 at 12:42
  • $\begingroup$ Was Rademacher's formula proven by the trick above? $\endgroup$
    – Martin Brandenburg
    Commented Jul 12, 2010 at 22:56
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See Making Transcendence Transparent: An Intuitive Approach to Classical Transcendental Number Theory by Burger and Tubbs. It is only a slight exaggeration to say that the entire book is devoted to showing that the whole subject of diophantine approximation is based on the paradigm you describe.

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One prime example of a similar trick is the proof that e is irrational. The books Irrational Numbers by Ivan Niven and Making Transcendence Transparent by Burger and Tubbs contain other examples.

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    $\begingroup$ Hm, since this is a proof by contradiction, actually there is no such real number. $\endgroup$
    – Martin Brandenburg
    Commented Jul 12, 2010 at 11:21
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I could not manage to find a reasonable link to Runge's method for diophantine equations (the wiki page on Runge contains only a mention); my memory says that it is in Mordell's book (at least in relation with Catalan's equation). The idea of the method for solving the diophantine equation $y^m=F(x)$, where $F(x)$ is a polynomial, is to use the binomial theorem for $\sqrt[m]{F(x)}$ and truncate the tail which is less than $1/2$.

Everybody who can help with a link is welcome!

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    $\begingroup$ Runge's method is explained, with proofs, in "Classical Diophantine Equations" by Sprindzuk. Generalizations due to Schinzel are stated in the Mordell book that you mention. $\endgroup$
    – Sidney Raffer
    Commented Jul 12, 2010 at 16:59

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