Why does dividing by a fraction mean multiplying by its reciprocal?

Computing $a\div\frac{b}{c}$ means finding the number $x$ that satisfies $x\cdot\frac{b}{c}=a.$ Multiply both sides by $\frac{c}{b}$: $x\cdot\frac{b}{c}\cdot\frac{c}{b}=a\cdot\frac{c}{b}.$ The factors on the left simplify to $1$, so $x=a\cdot\frac{c}{b}$. The reciprocal is not merely a memory trick; it is the number that turns the divisor into $1$.

A division is a ratio. Multiplying both the dividend and divisor by the same nonzero number leaves that ratio unchanged. Therefore, $a\div\frac{b}{c}=\left(a\cdot\frac{c}{b}\right)\div\left(\frac{b}{c}\cdot\frac{c}{b}\right)=\left(a\cdot\frac{c}{b}\right)\div1.$ We choose $c/b$ precisely because it makes the divisor equal to $1$.

Consider $6\div\frac{3}{4}=6\cdot\frac{4}{3}=\frac{24}{3}=8.$ Concretely, eight groups of size $3/4$ fit into $6$. Check with the inverse operation: $8\cdot\frac{3}{4}=6.$ Since the product recovers the original dividend, the quotient is correct.

The rule also works with negative fractions. For example, $5\div\left(-\frac{2}{3}\right)=5\cdot\left(-\frac{3}{2}\right)=-\frac{15}{2},$ and $(-15/2)(-2/3)=5$. The divisor cannot be zero: for $b/c$, we need $c\ne0$ so the fraction exists and $b\ne0$ so we are not dividing by zero. A common mistake is to flip the dividend; only the divisor is replaced by its reciprocal.