# Intervals with no integers (second version)

While studying the theme “Integral and fractional part of a number” I've come across the following problem: Let [α , β ] be an interval which contains no integers with 0 < α < β < 1 and β - α < 1/6. One needs to prove that there is a positive integer n such that [nα , nβ ] still contains no integers but has length at least 1/6. I' ve managed to prove that any such interval containing ½ satisfies the assertion. Also I've come to conclusion that the assertion is equivalent to the following: [αn]=[βn] & {αn} is not zero. Here I'm stuck. I do need your help. Thanks in advance.

• Did you mean to say that they have length at most $1/6$? That's what $\beta-\alpha < 1/6$ reads as to me. – Jack Crawford Aug 14 at 6:03
• We're seeking intervals [nα , nβ ] with no integers and with lengths more or equal to 1/6. – Игорь Ж Aug 14 at 6:12

## 1 Answer

Fun fact: for given n, if [a,b] doesn't contain any rational of the form m/n, then [na, nb] isn't going to contain an integer.

Let x be the greatest integer such that [xa, xb] doesn't contain an integer. For the purposes of contradiction, assume that the length of [xa, xb] is less than 1/6. That interval cannot contain a number of the form m/2 and a number of the form m/3 (because [1/3, 1/2] and [1/2, 2/3] both have length 1/6). Therefore, either [2xa, 2xb] or [3xa, 3xb] would have no integer in them. That violates our assumption that x was the greatest multiplier. Therefore, our assumption that the length of the multiplied interval was less than 1/6 was incorrect.

• Remarkable! Thank you, Matthew. – Игорь Ж Aug 14 at 6:36
• @Игорь Ж It was a fun problem. I wish I could upvote the question more than once. ^_^ – Matthew Daly Aug 14 at 6:37