Example 1.6. Given 101 integers fro

Example 1.6. Given 101 integers from 1, 2, . . . , 200, there are at least two integers such that one of them is divisible
by the other.
Proof. By factoring out as many 2’s as possible, we see that any integer can be written in the form 2k
· a, where
k ≥ 0 and a is odd. The number a can be one of the 100 numbers 1, 3, 5, . . . , 199. Thus among the 101 integers
chosen, two of them must have the same a’s when they are written in the form, say, 2r
· a and 2s
· a with r 6= s. if
r < s, then the first one divides the second. If r > s, then the second one divides the first.

Results (English) 3:

实例1.6。从1的101个整数，2，。..，200，至少有两个整数，其中一个是可分的
由其他
证明。通过分解出多达2的可能，我们看到，任何整数可以以书面的形式2K
·一，在
K≥0和是奇数。一个可数的100个数字1，3，5，。..199。因此，在101的整数
选择，他们两个都有相同的一个时，以书面的形式，说，
·2R和2S
·与R 6 =如果
R＜s，然后fiRST一分二。如果R >的，然后第二个分fiRST。