Real number happens to be an integer - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T19:24:35Z http://mathoverflow.net/feeds/question/31517 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/31517/real-number-happens-to-be-an-integer Real number happens to be an integer Martin Brandenburg 2010-07-12T09:19:58Z 2010-07-12T13:45:03Z <p>Sometimes you have a real number (with a rather complicated definition), and with some effort you can show that</p> <p>1) this real number is, actually, an integer;</p> <p>2) the distance of this real number to an integer, say $0$, is less than $1/2$.</p> <p>Thus you can conclude that this real number <em>is</em> $0$! 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.</p> <p>So my question is: Can you give nice, explicit instances of this trick?</p> http://mathoverflow.net/questions/31517/real-number-happens-to-be-an-integer/31531#31531 Answer by lhf for Real number happens to be an integer lhf 2010-07-12T10:58:44Z 2010-07-12T10:58:44Z <p>One prime example of a similar trick is the <a href="http://en.wikipedia.org/wiki/Proof_that_e_is_irrational" rel="nofollow">proof</a> that <em>e</em> is irrational. The books <em>Irrational Numbers</em> by Ivan Niven and <em>Making Transcendence Transparent</em> by Burger and Tubbs contain other examples.</p> http://mathoverflow.net/questions/31517/real-number-happens-to-be-an-integer/31533#31533 Answer by Douglas S. Stones for Real number happens to be an integer Douglas S. Stones 2010-07-12T11:07:08Z 2010-07-12T11:07:08Z <p>I'm not sure how related this particular example is to your question (but I think it's interesting). <a href="http://www.math.upenn.edu/~wilf/DownldGF.html" rel="nofollow">generatingfunctionology</a> 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<sub>n</sub> is the nearest integer to the above approximation.</p> http://mathoverflow.net/questions/31517/real-number-happens-to-be-an-integer/31540#31540 Answer by SJR for Real number happens to be an integer SJR 2010-07-12T12:16:47Z 2010-07-12T12:16:47Z <p>Your question has a logical formulation that leads to an open problem.</p> <p>A Discretely Ordered Ring is an ordered ring in which the inequality <code>$x&lt;y&lt;x+1$</code> 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:</p> <p>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? </p> <p>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.</p> <p>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 </p> http://mathoverflow.net/questions/31517/real-number-happens-to-be-an-integer/31541#31541 Answer by Roland Bacher for Real number happens to be an integer Roland Bacher 2010-07-12T12:19:56Z 2010-07-12T12:19:56Z <p>I guess the series expansion for the partition number $p(n)$ by Rademacher (see <a href="http://en.wikipedia.org/wiki/Partition_(number_theory" rel="nofollow">http://en.wikipedia.org/wiki/Partition_(number_theory</a>) ) is a nice illustration. The number $p(n)$ is obviously an integer which is uniquely defined by enough terms of Rademacher's formula.</p> http://mathoverflow.net/questions/31517/real-number-happens-to-be-an-integer/31546#31546 Answer by Wadim Zudilin for Real number happens to be an integer Wadim Zudilin 2010-07-12T13:14:30Z 2010-07-12T13:14:30Z <p>I could not manage to find a reasonable link to Runge's method for diophantine equations (the wiki page on <a href="http://en.wikipedia.org/wiki/Carl_David_Tolm%C3%A9_Runge" rel="nofollow">Runge</a> 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 $1/2$.</p> <p>Everybody who can help with a link is welcome!</p> http://mathoverflow.net/questions/31517/real-number-happens-to-be-an-integer/31550#31550 Answer by Timothy Chow for Real number happens to be an integer Timothy Chow 2010-07-12T13:45:03Z 2010-07-12T13:45:03Z <p>See <i>Making Transcendence Transparent: An Intuitive Approach to Classical Transcendental Number Theory</i> by Burger and Tubbs. It is only a slight exaggeration to say that the <i>entire book</i> is devoted to showing that the whole subject of diophantine approximation is based on the paradigm you describe.</p>