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### IB Maths HL Option Sets Abelian Group

Posted: Thu Mar 07, 2013 7:45 pm
IB Mathematics HL Option: Sets, Relations and groups (Abelian Group)

How can we show that the set $\mathbb{R}$ under the addition is an Abelian group?

Thanks

### Re: IB Maths HL Option Sets Abelian Group

Posted: Thu Mar 07, 2013 7:59 pm
IB Mathematics HL Option Sets, Relations and Groups - Abelian Group

The basic principles about Abelian Groups are:
First of all I remind you the definition of Group (G,*)
Let G be a non-empty set on which a binary operation * is defined . We say G is a Group under this operation if each of the following four properties hold:

1. G is closed under * , i.e., $x*y \in G$ for every $x,y \in G$

2. The operation * is associative on G,
i.e., $(x*y)*z=x*(y*z)$ for every $x,y \in G$

3. The operation * has an identity element in G,
i.e., $x*e=e*x=x$ for every $x \in G$

4. Each element of G has an inverse under *,
i.e., $x*x^{-1}=x^{-1}*x=e$ for every $x \in G$

A group {G,*} is called Abelian if $x*y=y*x$ for every $x,y \in G$

Also Given a group {G,*} we have the following properties:
Left cancellation law: If x*y=x*z then y=z
Right cancellation law: If y*x=z*x then y=z

The set $\mathbb{R}$ under the addition is an Abelian group since

Closure
: The addition on $\mathbb{R}$ is closed
i.e.$(x+y)\in \mathbb{R}$ for every $x,y \in \mathbb{R}$

Associative: The addition on $\mathbb{R}$ is Associative
i.e. x+(y+z)=(x+y)+z for every $x,y \in \mathbb{R}$

Identity: For every element $x \in \mathbb{R}$ there is an element, the zero, which is the identity element in $\mathbb{R}$ under addition.
i.e. x+0=0+x=0

Inverse
: For every element $x \in \mathbb{R}$ there is an element, which is its opposite in $\mathbb{R}$.
i.e. x+(-x)=(-x)+x=0

Commutative
: The addition on $\mathbb{R}$ is Commutative
i.e. x+y=y+x for every $x,y \in \mathbb{R}$

Hope these help