Prove their is a unique element y such that for all x, x+y=x

[snes_nerd]

Homework Statement

Prove there is a unique element y such that for all x, x+y=x

Homework Equations

I also have to prove their is a unique element y such that for all x, xy=x

The Attempt at a Solution

x+y=x, x+y-x=x-x, y=0. xy=x, xy(1/x)=x(1/x). y=1. The problem with the way I did it is that I had to assume the additive and multiplicative inverses are not defined yet. So I have to prove these statements another way. I am not really sure how to though.

I was given a hint which states that for all x, x+y prime=x. Then consider y+y prime and prove that y=y prime. It didnt really help but when I worked through it I found myself using the additive and multiplicative inverses again which I cant use. How does one go about proving these.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Deveno]

a unique element of...what?

 

[snes_nerd]

A unique element for yεℝ such that for all xεℝ, x+y=x

 

[Deveno]

suppose we had 2 real numbers, y,y' such that x+y = x = x+y', for all real numbers x.

what can we say about y+y'?

 

[snes_nerd]

Im not quite sure how to prove that. I mean I can look at this problem and figure it out, but to logically prove it is kind of different. Never had to do something like this before.

x+y= x = x+y' implies that y=y' which is what I am trying to prove since that would be the conclusion. However, I have no logical justification for this conclusion, especially since I have to pretend the additive inverse has not been proven yet in this problem.

 

[daveb]

The additive inverse is a property of the reals as a field, so what you are trying to prove is the uniqueness of the zero element for the field. You can use the additive inverse since its existence is required for it to be a field.

 

[HallsofIvy]

Deveno's question "of what" was due to the fact that you are asking about the existenced of an "additive inverse" of some algebraic system but how you prove that depends upon what algebraic system you are talking about. Since you answer that you are dealing with real numbers, I see nothing wrong with just noting that, for any real number, x, x+ 0= x. And, further, that if x+y= x then, subtracting x from both sides, y= 0. That is, "0" is the unique y such that x+ y= x.

 

[Deveno]

Deveno's question "of what" was due to the fact that you are asking about the existenced of an "additive inverse" of some algebraic system but how you prove that depends upon what algebraic system you are talking about. Since you answer that you are dealing with real numbers, I see nothing wrong with just noting that, for any real number, x, x+ 0= x. And, further, that if x+y= x then, subtracting x from both sides, y= 0. That is, "0" is the unique y such that x+ y= x.

except, presumably, all he is given is that R is a field (perhaps not even that, since apparently he isn't allowed to use inverses), and the explicit definitions of 0 and 1 haven't been made yet.

the question will be more difficult if he does not yet have access to commutativity for addition, since all he is explicitly given is the existence of ONE right-identity. in fact, without the existence of right-inverses, i don't think he can prove it at all without commutativity.

if, however, he knows that y+y' = y'+y, it is much easier.

@snes_nerd:

since x+y = x for ALL x, it is true for the particular: x = y'.

so y'+y = y'

since we are assuming that y' also has this property (that x+y' = x for ALL x),

it is true in particular, for x = y:

y+y' = y.

now, do you have as given, in your rules for R, that a+b = b+a for ALL a,b?

because if you do, what do the two equations above yield for a = y, b = y'?