This might be a silly question after all. But I guess that assuming associativity and commutativity for $+$ in a ring, one could reorder the operations and obtain the commutativity for $\cdot$.
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5No. Matrices. ;; – Aloizio Macedo May 14 '15 at 17:57
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2Dear Jesus: I am disappointed; I thought Jesus Christ knew all such things! In any event, consider Aloizio Macedo's comment! Barachot! – Robert Lewis May 14 '15 at 18:02
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@RobertLewis No. See my profile for explanation. – Red Banana May 14 '15 at 18:08
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1NIce! I always had faith that Jesus uses a flat-screen monitor, and dresses for business! – Robert Lewis May 14 '15 at 18:13
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In short, the answer is no. Matrix multiplication is not commutative while matrix addition is, for instance.
Further, I've never come across a ring where the addition was not commutative. It is inbuilt in the definition I know of rings. In part this might be because having a $1$ forces addition to be commutative, and typically we investigate rings (not rngs). This is talked about more in this question.
davidlowryduda
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Of, it seems I confused one thing. Here. He says: "In addition, if $ab=ba$, then it is a commutative ring". I thought he was talking about addition. Silly me. – Red Banana May 14 '15 at 18:06
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1It's not every day I get to leave a comment for @JesusChrist, so I'll try to make it good. A ring, by definition, includes commutativity of addition. If a ring is labelled as commutative or non-commutative, it is referring the multiplication operation. – Sean Henderson May 14 '15 at 18:07
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Note that $(x+y)+(x+y) = 2(x+y) = 2x + 2y = x+x+y+y$, whence you get that $x+y=y+x$ i.e. addition is always commutative in a ring.
Alex M.
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