Let $R$ a ring prove that $x(yz)=xy  xz$
1,462
Solution 1
$ab=a+(b)$ by the definition of subtraction.
$x(z)+xz={?}$.
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 $a$)
$$ x(yz)=x(y+(z)) $$
Multiplication distributes over addition:
$$\begin{align} x(y+(z))&=xy+x(z)\\ &=xyxz,\qquad\text{ as above.} \end{align}$$
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90intuition
Updated on August 01, 2022Comments

90intuition over 1 year
Let $R$ a ring (not necessarily commutative) prove that $x(yz)=xy  xz$.
\begin{align*} x(yz)&=x(y+(z)) \\ &=xy +x(z) \\ &=xy+(xz) \\ &=xyxz \end{align*}
I think all my steps are valid, but I however don't see why:
 $ab=a+(b)$
 $x(z)=(xz)$
How can I show this rigoursly ? I was thinking that 1) may be a definition.

90intuition about 10 yearsI know the definition of a ring. @DietrichBurde

Stefan Hamcke about 10 yearsThe first one is a definition. We define $ab$ as the sum $a+(b)$.

90intuition about 10 years@DietrichBurde I don't see anywhere defined that $ab=a+(b)$

90intuition about 10 years@DietrichBurde I see that answer. And I'm aware of that definition. But doesn't define an binary operation $:R×R→R: (a,b) ↦ ab = 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"...

Doug Spoonwood about 10 yearsSuppose 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 xx=0." Again, I've suppressed the addition operation, so xx=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(yz) has multiplication suppressed first and addition suppressed second.

90intuition about 10 years@DougSpoonwood Ah, that makes sense. Didn't think about it that way. Thanks

90intuition about 10 years$x(z)+xz=x(z+z)=x⋅0=0$. aaaaah ! thanks

dfeuer about 10 years@90intuition you mean $=x\cdot0=0$

90intuition about 10 yearsThat definition doesn't define an binary operation "substraction", right ? Something like: $:R×R→R: (a,b) ↦ ab = a + (b)$

90intuition about 10 yearsI edited it now.