Not only is the symbol daunting, but the words are too

[flyingpig]

For this field, -1 = 1 . This is because the thing you need to add to 1 to give you the identity element for addition is 1.

OK, for A4, you need to use that -1=1 and -0=0. You define this to be so.

I overlooked something. since WHEN DID I DEFINE -1 = 1?

 

[micromass]

Well, you need to find for each x, an element y such that

x+y=0

By definition, this element is -x.

So, for x=0, you need an element such that 0+y=0. When you look at the axioms, you see that y=0 is the only possibility. So -0=0.

Now, what is -1?

 

[flyingpig]

Oh is it from 1 + 1 = 0, that 1 = -1? oh okay

 

[micromass]

Oh is it from 1 + 1 = 0, that 1 = -1? oh okay

Yes.

 

[flyingpig]

I am still stuck on M4...

 

[micromass]

Take x in your field. How would you define x^{-1}??

 

[flyingpig]

xx^{-1} = 1

x^{-1} = 1/x

That doesn't change the original definition...

I still can't use 0.

 

[flyingpig]

Oh wait, M4 actually states for all x in R, without 0, the axiom holds, I don't even need to test x = 0 right?

Ah shoot for M3, I put 0 * 1 = 0, that's wrong isn't it? Because it says x without 0

 

[micromass]

Sigh...

For each x nonzero, you need to find a y such that xy=1.
If x=1. What can you take as y?

 

[micromass]

Oh wait, M4 actually states for all x in R, without 0, the axiom holds, I don't even need to test x = 0 right?

Right.

 

[flyingpig]

This is probably another stupid question.

Since M3 states the same condition, it would be wrong/unnecessary to even do 0*1 = 0 because it contradicts M4 \ {0} right?

 

[micromass]

M3 does not state the same thing as M4...

 

[flyingpig]

M3 does not state the same thing as M4...

But they have a similar condition...

 

[flyingpig]

Oh wait you know what, never mind. I was looking at 2.49 and I thought 0^{-1} = 0 and I had to test 0*1 = 0.

So yes I do need to test 0*1 = 0 for x =0