求证：当r不被k整除时，整数kn+r不被k整除及原理解析

Hey there! I get why this might feel like it's "obvious" but hard to wrap your head around formally—let's unpack it step by step, no fancy math jargon required.

First, Let's Refresh the Definition of Divisibility #

By definition, an integer a is divisible by k (where k is a non-zero integer) if there exists some integer m such that a = k * m. In plain terms: a is a perfect multiple of k.

Let's Prove This Using Contradiction #

Sometimes it's easier to show that the opposite of our claim can't possibly be true. Here's how that works:

1. Suppose (for contradiction) that even though r isn't divisible by k, the number kn + r is divisible by k.
2. By the divisibility definition, there must be some integer m where:

   kn + r = k * m
   

3. Let's rearrange this equation to isolate r:

   r = k*m - kn
   r = k*(m - n)
   

4. Now, m - n is just an integer (since m and n are both integers). That means r is equal to k multiplied by an integer—which would make r divisible by k.
5. But wait! The original problem states that r is not divisible by k. This is a direct contradiction—our initial assumption can't be true.

So that's the proof: if r isn't divisible by k, then kn + r can't be divisible by k either.

A Concrete Example to Make It Click #

Let's use real numbers to see this in action:

• Let k = 5, n = 3 (so kn = 15, a perfect multiple of 5)
• Let r = 2 (which isn't divisible by 5)
• kn + r = 15 + 2 = 17—and 17 divided by 5 leaves a remainder of 2, so it's definitely not divisible by 5.

No matter what n you pick, kn will always be a multiple of k. Adding a number r that's not a multiple of k will never result in another multiple of k—it's like adding an odd number to an even number and expecting an even number, just applied to multiples instead of parity.

内容的提问来源于stack exchange，提问作者pavle