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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.

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I know the definition of a ring. @DietrichBurde –  90intuition Oct 15 '13 at 21:42
1  
The first one is a definition. We define $a-b$ as the sum $a+(-b)$. –  Stefan Hamcke Oct 15 '13 at 21:42
    
@DietrichBurde I don't see anywhere defined that $a-b=a+(-b)$ –  90intuition Oct 15 '13 at 21:47
    
@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 ? –  90intuition Oct 15 '13 at 21:51
    
"Some basic properties of a ring follow immediately from the axioms. The additive identity and the additive inverse are unique." –  The Chaz 2.0 Oct 15 '13 at 21:51

2 Answers 2

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

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

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$x(-z)+xz=x(-z+z)=x⋅0=0$. aaaaah ! thanks –  90intuition Oct 15 '13 at 21:45
    
@90intuition you mean $=x\cdot0=0$ –  dfeuer Oct 15 '13 at 21:47
    
I edited it now. –  90intuition Oct 15 '13 at 21:53

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 $a$)

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

Multiplication distributes over addition:

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

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That definition doesn't define an binary operation "substraction", right ? Something like: $-:R×R→R: (a,b) ↦ a-b = a + (-b)$ –  90intuition Oct 15 '13 at 21:53

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