# uniquiness

• Aug 15th 2011, 07:12 AM
psolaki
uniquiness
Suppose G is an additive group with the following properties:

1) x+(y+z) =(x+y) +z for all x,y,z in G

2) There exists 0 such that: x+0 =0+x = x for all x in G

3) For all x in G there exists -x such that : x+(-x) = (-x)+x =0

Now do we need to prove that 0 and -x are unique ,before we prove that:

-(x+y) = (-y)+(-x) and -(-x) = x
• Aug 15th 2011, 07:43 AM
Drexel28
Re: uniquiness
Quote:

Originally Posted by psolaki
Suppose G is an additive group with the following properties:

1) x+(y+z) =(x+y) +z for all x,y,z in G

2) There exists 0 such that: x+0 =0+x = x for all x in G

3) For all x in G there exists -x such that : x+(-x) =0

Now do we need to prove that 0 and -x are unique ,before we prove that:

-(x+y) = (-y)+(-x) and -(-x) = x

Additive in terms of abelian? A lot of this is actually unecessary. For example if you have a semigroup $(S,\ast)$ (i.e. just a set with an associative binary operation $\ast:S\times S\to S$) then any identity would have to be unique, for if $i,i'$ were both identities then $i=i\ast i'$ because $i'$ is an identity and $i\ast i'=i'$ since $i$ is an identity. Make sense? Now use this fact to prove that $-x$ is unique, i.e. assume that $y,z$ are both inverses of $x$ and note that $x+y$ and $x+z$ are identities and so $x+y=x+z$...so
• Aug 15th 2011, 10:38 AM
psolaki
Re: uniquiness
Quote:

Originally Posted by Drexel28
Additive in terms of abelian? A lot of this is actually unecessary. For example if you have a semigroup $(S,\ast)$ (i.e. just a set with an associative binary operation $\ast:S\times S\to S$) then any identity would have to be unique, for if $i,i'$ were both identities then $i=i\ast i'$ because $i'$ is an identity and $i\ast i'=i'$ since $i$ is an identity. Make sense? Now use this fact to prove that $-x$ is unique, i.e. assume that $y,z$ are both inverses of $x$ and note that $x+y$ and $x+z$ are identities and so $x+y=x+z$...so

I am sorry i did not ask for a proof for the uniqueness of 0 and -x

I ask if uniqueness is necessary for the proof of :

-(x+y) = (-y)+(-x) and -(-x) =x i.e do we have to prove that 0 and -x are unique before we prove that : -(x+y) = (-y)+(-x) and -(-x) =x.

Because that is what a lot of books do.1st they prove uniqueness of 0 and -x and then they use this fact to prove : -(x+y) =(-y)+(-x) and -(-x) = x

By the way the group i mentioned is not abelian
• Aug 15th 2011, 11:22 AM
Drexel28
Re: uniquiness
Quote:

Originally Posted by psolaki
I am sorry i did not ask for a proof for the uniqueness of 0 and -x

I ask if uniqueness is necessary for the proof of :

-(x+y) = (-y)+(-x) and -(-x) =x i.e do we have to prove that 0 and -x are unique before we prove that : -(x+y) = (-y)+(-x) and -(-x) =x.

Because that is what a lot of books do.1st they prove uniqueness of 0 and -x and then they use this fact to prove : -(x+y) =(-y)+(-x) and -(-x) = x

By the way the group i mentioned is not abelian

Not strictly, no. But that all depends how you are thinking about the problem. If the problem is "If $x+y=0$ and $y+z=0$ then $x=z$ since $x=x+(y+z)=(x+y)+z=0+z=z$. That said, to say that $x=-(-x)$ you need uniqueness for a purely semantical reason. Namely, unless you are doing something strange, one thinks of $-$ as a function $G\to G$, right? So that calling this function $f:G\to G$ then the theorem states that $f$ is an involution, namely that $f\circ f=\text{id}$. The problem is that $f$ isn't a priori a function if we don't know that $-x$ is unique. In much simpler terms, you need uniqueness to say that 'THE inverse of THE inverse of $x$ is $x$". Make sense?
• Aug 15th 2011, 01:18 PM
psolaki
Re: uniquiness
Quote:

Originally Posted by Drexel28
In much simpler terms, you need uniqueness to say that 'THE inverse of THE inverse of $x$ is $x$". Make sense?

But the following proof indicates that we can prove -(-x) =x without using the uniqueness of -x.

for all a :a+(-a) =(-a)+a = 0 by using rule 2 of the above group.

Now put a = -x and we have that: (-x) +[-(-x)] = [-(-x)] + (-x)= 0

Add x to both sides and we have that:{[-(-x)] +(-x)} +x =0+x = x.....................(1) ..........by using rule 2 of the group

But {[-(-x)]+(-x)} +x = [-(-x)] +[ (-x)+x] =................................by using rule 1 of the group

= [-(-x)] +0 =................................................. ......................by using rule 3 of the group

= [-(-x)].............(2) .................................................. ............................by using rule 2 of the group

Hence :[-(-x)] = x................................................. ................By using equality rule : if a=b and b=c ,then a= c
• Aug 15th 2011, 02:47 PM
Drexel28
Re: uniquiness
Quote:

Originally Posted by psolaki

But the following proof indicates that we can prove -(-x) =x without using the uniqueness of -x.

for all a :a+(-a) =(-a)+a = 0 by using rule 2 of the above group.

Now put a = -x and we have that: (-x) +[-(-x)] = [-(-x)] + (-x)= 0

Add x to both sides and we have that:{[-(-x)] +(-x)} +x =0+x = x.....................(1) ..........by using rule 2 of the group

But {[-(-x)]+(-x)} +x = [-(-x)] +[ (-x)+x] =................................by using rule 1 of the group

= [-(-x)] +0 =................................................. ......................by using rule 3 of the group

= [-(-x)].............(2) .................................................. ............................by using rule 2 of the group

Hence :[-(-x)] = x................................................. ................By using equality rule : if a=b and b=c ,then a= c

I think you missed the point entirely. Let me ask you this, what is $-(-x)$? Define it.
• Aug 15th 2011, 05:50 PM
psolaki
Re: uniquiness
The inverse of the inverse
• Aug 15th 2011, 05:54 PM
Drexel28
Re: uniquiness
Quote:

Originally Posted by psolaki
The inverse of the inverse

Why not a inverse of a inverse?
• Aug 15th 2011, 07:06 PM
psolaki
Re: uniquiness
you mean an inverse of an inverse?If so what is the difference??
• Aug 16th 2011, 04:57 AM
HallsofIvy
Re: uniquiness
Quote:

Originally Posted by psolaki
you mean an inverse of an inverse?If so what is the difference??

The difference is in the words "a" and "the". When you say "the inverse of the inverse" you are assuming that the inverse is unique.

Similarly, showing that -x+(-(-x))= 0 and that -(-x)+ (-x)= 0 does not show that -(-x)= x unless you have shown that the inverse is unique. Otherwise, -(-x) and x might be two different inverses for -x.
• Aug 16th 2011, 07:50 PM
psolaki
Re: uniquiness
before i answer to your post,surely you mean x +(-x)=0 and (-x) +[-(-x)]=0 and not

-x +(-(-x)) =0 and -(-x) + (-x) =0