How to Prove x(y − z) = xy − xz in a Ring

To prove $x(y-z)=xy-xz$ in a ring, use the ring’s distributive law and the definition of subtraction. The key detail is that multiplying an additive inverse by $x$ gives the additive inverse of the product.

The proof works whether or not the ring is commutative and whether or not it has a multiplicative identity.

The ring properties you need

The elements of a ring form an abelian group under addition. In particular, each element $a$ has an additive inverse $-a$, and $a+(-a)=0$. A ring also satisfies the distributive law. The left distributive law says

$$x(a+b)=xa+xb.$$

Subtraction is defined using the additive inverse: $y-z$ means $y+(-z)$. These properties are enough to establish the identity.

Proof

First, show that $x(-z)=-(xz)$. Distribute $x$ over the sum $z+(-z)$:

$$\begin{aligned} xz+x(-z) &= x\bigl(z+(-z)\bigr) \\ &= x0 \\ &= 0. \end{aligned}$$

Here $x0=0$: by distributivity, $x0=x(0+0)=x0+x0$, and cancellation in the additive group gives $x0=0$. Thus $x(-z)$ is an additive inverse of $xz$, so $x(-z)=-(xz)$.

Now apply distributivity and substitute that fact:

$$\begin{aligned} x(y-z) &= x\bigl(y+(-z)\bigr) \\ &= xy+x(-z) \\ &= xy+(-(xz)) \\ &= xy-xz. \end{aligned}$$

That proves the identity.

Worked example with matrices

The identity also holds in noncommutative rings, such as square matrices under ordinary addition and multiplication. Take

$$A=\begin{pmatrix}1&2\\0&1\end{pmatrix},\quad B=\begin{pmatrix}2&1\\3&0\end{pmatrix},\quad C=\begin{pmatrix}1&2\\1&1\end{pmatrix}.$$

First subtract the matrices:

$$B-C=\begin{pmatrix}1&-1\\2&-1\end{pmatrix}.$$

Then multiply on the left by $A$:

$$A(B-C)=\begin{pmatrix}5&-3\\2&-1\end{pmatrix}.$$

Multiplying separately gives

$$AB=\begin{pmatrix}8&1\\3&0\end{pmatrix},\qquad AC=\begin{pmatrix}3&4\\1&1\end{pmatrix},$$

so

$$AB-AC=\begin{pmatrix}5&-3\\2&-1\end{pmatrix}=A(B-C).$$

This is a concrete instance of the same ring law. The matrices need not commute; the proof only multiplies the sum on the left by $A$.

Common mistakes

• Assuming subtraction distributes by definition. The identity follows from distributivity over addition plus the additive-inverse rule; it is useful to show that rule when writing a proof.
• Changing the order of factors. In a noncommutative ring, $xy$ need not equal $yx$. The proof gives $x(y-z)=xy-xz$, not $(y-z)x=yx-zx$ by rearranging factors. The latter identity follows from right distributivity, with the order preserved.
• Assuming the ring has a unit. No multiplicative identity is used. The proof relies on addition, additive inverses, and distributivity.
• Confusing this with a product of differences. The rule concerns multiplication by one fixed element $x$. It does not say that $(x-y)(u-v)$ equals a single difference; expanding such a product creates four terms.

The general rule to remember is that multiplication distributes over subtraction on either side, provided you preserve the order of the factors and use the corresponding distributive law.