This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 2k, 2k + 1, 2k + 2,
and 3k, 3k + 1, 3k + 2, 3k + 3, or in general jk, jk + 1, . . . jk + j . For j k 1
these blocks overlap and no integers are left out. In the last example 24, 25 2 I0
implies 48, 49, 50 2 I0 which implies 72, 73, 74, 75 2 I0 and 96, 97, 98, 99, 100 2
I0 , so we know the result holds for n0 = 96. In fact it actually holds for n0 = 34
but it is not important to get a precise bound.
To show that there are two consecutive integers, we cheat and use a fact
from number theory: if the greatest common divisor of a set Ix is 1 then there
are integers i1 , . . . im 2 Ix and (positive or negative) integer coe cients ci so
that c1 i1 + · · · + cm im = 1. Let ai = c+ and bi = ( ci )+ . In words the ai
i
are the positive coe cients and the bi are 1 times the negative coe cients.
Rearranging the last equation gives
a1 i1 + · · · + am im = (b1 i1 + · · · + bm im ) + 1 and using Lemma 1.15 we have found our two consecutive integers in Ix . 25 1.5. LIMIT BEHAVIOR While periodicity is a the...
View Full
Document
 Spring '10
 DURRETT
 The Land

Click to edit the document details