Why Does ac Divide bc Exactly When a Divides b?

The statement follows directly from the definition of divisibility and cancellation of a nonzero factor. The key condition is that the same positive integer c multiplies both a and b; this does not mean you can cancel factors in every divisibility statement.

Start from the definition

For integers $x$ and $y$, the notation $x\mid y$ means that $y=kx$ for some integer $k$. Thus, to prove an assertion of the form $x\mid y$, you need to produce an integer quotient.

Because $a$, $b$, and $c$ are positive integers, $c\ne 0$. That allows cancellation by $c$ in an equality. We can now prove the two implications separately.

If $ac$ divides $bc$, then $a$ divides $b$

Assume $ac\mid bc$. By the definition of divisibility, there is an integer $k$ such that

$$bc=k(ac).$$

Rearrange the right-hand side and cancel $c$ from both sides:

$$bc=(ka)c \quad\Longrightarrow\quad b=ka.$$

Since $k$ is an integer, this equation says exactly that $a\mid b$.

If $a$ divides $b$, then $ac$ divides $bc$

Now assume $a\mid b$. There is an integer $k$ such that $b=ka$. Multiply both sides by $c$:

$$bc=(ka)c=k(ac).$$

This expresses $bc$ as an integer multiple of $ac$, so $ac\mid bc$.

Together, these implications prove

$$ac\mid bc \quad\Longleftrightarrow\quad a\mid b.$$

A worked example

Take $a=6$, $b=30$, and $c=4$. Since $30=5\cdot 6$, we have $6\mid 30$. Multiplying that quotient equation by $4$ gives

$$30\cdot 4=5(6\cdot 4),$$

so $24\mid 120$. In the other direction, if you start with $24\mid 120$, the quotient is $5$, and cancelling the common factor $4$ in $120=5\cdot 24$ gives $30=5\cdot 6$. Hence $6\mid 30$.

Why the positivity condition matters

The proof uses $c\ne 0$, not that $c$ is positive specifically. It therefore also works for any nonzero integer $c$. If $c=0$, the equivalence can fail: for example, let $a=2$ and $b=1$. Then $ac=bc=0$, so $0\mid 0$ under the definition (because $0=k\cdot 0$ for any integer $k$), but $2\nmid 1$.

Positivity is a convenient way to ensure $c$ is nonzero. In the usual divisibility convention, a nonzero integer divides $0$; $0$ does not divide a nonzero integer.

Common mistakes

• Cancelling in the wrong statement. From $a\mid bc$ alone, you cannot generally conclude $a\mid b$. For example, $2\mid 2\cdot 3$, but $2\nmid 3$. Here, the hypothesis is instead $ac\mid bc$: the entire factor $ac$ divides the entire product $bc$.
• Forgetting what divisibility means. The quotient must be an integer. In the forward direction, the proof uses the integer $k$ supplied by $ac\mid bc$; after cancellation, the same $k$ shows $a\mid b$.
• Cancelling zero. Cancellation is valid here because $c$ is positive and therefore nonzero. If a version of the claim allows $c=0$, it needs a separate check and is not true in general.