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Define $\oplus$ on $\mathbb{R} \times \mathbb{R}$ by setting $$(a,b) \oplus (c,d)=(ac-bd, ad+bc).$$ How to show that $(\mathbb{R} \times \mathbb{R}, \oplus)$ is an algebraic system. I don't understand the difference between algebraic structure andalgebraic system I read this article but I didnt understand it properly. Can anybody help me me understand by giving hints?

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It's generally considered rude here to ask questions in the imperative. You might get better responses if you phrase them in the form "I have a question about this problem: ..." or something similar. – Calvin McPhail-Snyder Aug 16 '12 at 21:29
What do you mean by "algebraic system"? Do you have a specific definition in mind? What have you tried so far? – Calvin McPhail-Snyder Aug 16 '12 at 21:30
continued from here – Santosh Linkha Aug 16 '12 at 21:31
@CalvinMcPhail-Snyder I am new to this site and not sure what is required and what are the requirements. Well the only algebraic structure that I am familiar with is a group. So I am assuming we can show that it is a group – math101 Aug 16 '12 at 21:37
@math101 No problem! Just letting you know for the future. – Calvin McPhail-Snyder Aug 16 '12 at 21:39
up vote 3 down vote accepted

$\oplus$ on $\mathbb{R} \times \mathbb{R}$ is a just multiplication of the complex numbers on $\mathbb{C}$. It is not a group since $(0,0) \in \mathbb{R} \times \mathbb{R}$ does not have a inverse under $\oplus$.

However $\mathbb{R} \times \mathbb{R} - \{0\}$ is a group under $\oplus$.

$(\mathbb{R} \times \mathbb{R}, \oplus)$ however is a commutative monoid. It is associative, has an identity element, but not every element has an inverse.

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Are algebraic structure and algebraic system same? – Santosh Linkha Aug 16 '12 at 22:00
@experimentX Both term algebraic system and algebraic structure are not generally used. In logic and model theory, I have heard the term algebraic language as a language with no relation symbols, and a algebraic structure as a structure in an algebraic language. Groups, vector spaces, rings are usually given an algebraic language. They are hence algebraic structures. – William Aug 16 '12 at 22:03
Oh, this is just simple undergraduate algebra. anyway thanks for response +1 – Santosh Linkha Aug 16 '12 at 22:06
Thanks for helping me out – math101 Aug 16 '12 at 22:27

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