用反证法证明：若a∣b且a∤c则b∤c的思路说明

Let’s walk through this proof using proof by contradiction—a go-to technique for these kinds of number theory statements where a direct proof might feel less straightforward. Here’s the breakdown:

Step 1: Lay out the given conditions #

• By the definition of divisibility, since $a \mid b$, there exists some integer $k \in \mathbb{Z}$ such that:
  $$b = ak$$
• We also know $a \nmid c$, which means there is no integer $l \in \mathbb{Z}$ where $c = al$. This is our key constraint we’ll reference later.

Step 2: Assume the opposite of our target conclusion #

To use contradiction, we start by assuming the statement we want to prove is false. That means we assume $b \mid c$. By divisibility rules, this implies there’s some integer $m \in \mathbb{Z}$ where:
$$c = bm$$

Step 3: Derive a contradiction #

Substitute the expression for $b$ (from Step 1) into the equation for $c$:
$$c = (ak)m = akm$$

Since $k$ and $m$ are both integers, their product $km$ is also an integer (integers are closed under multiplication). Let’s call this product $l = km$—so now we have $c = al$.

But this directly clashes with our given condition that $a \nmid c$, which explicitly states no such integer $l$ exists.

Step 4: Finalize the proof #

Because our assumption ($b \mid c$) led to a clear contradiction with the original premises, the assumption must be false. Therefore, the original statement holds: if $a \mid b$ and $a \nmid c$, then $b \nmid c$.

内容的提问来源于stack exchange，提问作者jitender