# Let $R$ a ring prove that $x(y-z)=xy - xz$

1,462

## Solution 1

1. $a-b=a+(-b)$ by the definition of subtraction.

2. $x(-z)+xz={?}$.

## Solution 2

From Wikipedia:

For each $a$ in $R$ there exists $−a$ in $R$ such that $a + (−a) = (−a) + a = 0$ ($−a$ is the inverse element of $a$)

$$x(y-z)=x(y+(-z))$$

\begin{align} x(y+(-z))&=xy+x(-z)\\ &=xy-xz,\qquad\text{ as above.} \end{align}

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### 90intuition

Updated on August 01, 2022

• 90intuition over 1 year

Let $R$ a ring (not necessarily commutative) prove that $x(y-z)=xy - xz$.

\begin{align*} x(y-z)&=x(y+(-z)) \\ &=xy +x(-z) \\ &=xy+-(xz) \\ &=xy-xz \end{align*}

I think all my steps are valid, but I however don't see why:

1. $a-b=a+(-b)$
2. $x(-z)=-(xz)$

How can I show this rigoursly ? I was thinking that 1) may be a definition.

• I know the definition of a ring. @DietrichBurde
• The first one is a definition. We define $a-b$ as the sum $a+(-b)$.
• @DietrichBurde I don't see anywhere defined that $a-b=a+(-b)$
• @DietrichBurde I see that answer. And I'm aware of that definition. But doesn't define an binary operation $-:R×R→R: (a,b) ↦ a-b = a + (-b)$ Right ?
• The Chaz 2.0 about 10 years
"Some basic properties of a ring follow immediately from the axioms. The additive identity and the additive inverse are unique."
• The Chaz 2.0 about 10 years
^That's from wikipedia. Once we have established that the additive identity is unique, then we may (well-)define "subtraction"...
• Suppose you wrote the axioms for a group using "addition" as the operation suppressed instead of "multiplication". That is, every group has an "addition" operation. Or equivalently xy means (x+y). Then, the inverse axiom for groups says "for all x, there exists a -x such that x-x=0." Again, I've suppressed the addition operation, so x-x=0 implicitly best gets read as meaning (x+-x)=0. So, no, you don't need to define a binary operation of subtraction for this sort of problem. You just need to recognize that x(y-z) has multiplication suppressed first and addition suppressed second.
• • $x(-z)+xz=x(-z+z)=x⋅0=0$. aaaaah ! thanks
• @90intuition you mean $=x\cdot0=0$
• That definition doesn't define an binary operation "substraction", right ? Something like: $-:R×R→R: (a,b) ↦ a-b = a + (-b)$
• 