Real number happens to be an integer - MathOverflow

Real number happens to be an integer

2010-07-12T09:19:58Z

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?

Answer by lhf for Real number happens to be an integer

2010-07-12T10:58:44Z

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.

Answer by Douglas S. Stones for Real number happens to be an integer

2010-07-12T11:07:08Z

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 F_n is the nearest integer to the above approximation.

Answer by SJR for Real number happens to be an integer

2010-07-12T12:16:47Z

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

Answer by Roland Bacher for Real number happens to be an integer

2010-07-12T12:19:56Z

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.

Answer by Wadim Zudilin for Real number happens to be an integer

2010-07-12T13:14:30Z

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!

Answer by Timothy Chow for Real number happens to be an integer

2010-07-12T13:45:03Z

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.