Reading my post ‘only integers behave nicely’, I came up with another proof (and another variation of it) of the following fact:

Let be a pair of real numbers having the property that: for all natural numbers , where denotes the largest integer . Then and are either the same, one of them equals , or they are both integral. So let’s assume that the last 2 possibilites do not hold, and dive right into it.

Dividing by (which is possible for large enough ), we see that the ratio of and is rational, say , with and coprime. Substitute for , dividing by and multiplying by we obtain:

Since the RHS is divisible by , LHS must be a multiple of too. In particular this is true for . Since we may assume that the fractional part of is between and (duh), this implies that the fractional part of is actually contained in the interval . Otherwise we would have that times the integer part of is divisible by (since we assumed for the moment that ), while times the fractional part of is (rounded down) NOT divisible by , while their sum has to be a multiple of , because and are coprime. Let now, with integral. By induction we now see that the fractional part of is contained in the interval , for all ; contradiction.

Another (admittedly only slightly different) route to take, after we showed that , is the following:

Let (as it definitely resembles the in the inductionproof) be the smallest integer such that the fractional part of is larger than . Then (otherwise would have been a smaller integer such that the fractional part of is larger , which contradicts our definition of . But now, letting be the fractional part of , setting and keeping in mind that is a multiple of , we have the following:

Which shows . By symmetry, and once again, Q.E.D.

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