(-1)(-1)=1

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>>8449781
Reference:
http://mathworld.wolfram.com/FieldAxioms.html.

>>8449781
[math](-1)=(-1)1[/math] (3)
[math]=(-1)(-1)(1/(-1))[/math] ([math]1=(-1)(1/(-1))[/math]) (4)
[math]=(-1)(1/(-1))[/math] ([math](-1)(-1)=-1[/math])
[math]=1[/math]
What do now?

(-1)^2 = -1 is true in characteristic two but not in any other characteristic

what do you mean by 'fixing the axioms'? you would no longer be working in a general field if you forced (-1)^2=-1.

>>8449795
Thank you.
I was reviewing the axioms for [math]\mathbb{R}[/math] from my Analysis book and though about whether that would make sense.
I will keep your post as a reference.

– 1 = 0 – 1
...so (– 1)^2 is
(– 1)(– 1) = (0 – 1)(0 – 1) = 0^2 – 2*0 + 1 = 1
EZ–PZ

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>>8450968
Hold on John Nash.
[math]-1=0-1[/math]
[math](-1)(-1)[/math]
[math]=(0-1)(0-1)[/math]
[math]=0(0-1)+(-1)(0-1)[/math]
[math]=00+0(-1)+(-1)0+(-1)(-1)[/math]
[math](-1)(-1)[/math]

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>>8451035
I hate when this happens.

>>8451035
Can someone explain the purpose of this to me

>>8451062
The purpose of this
>>8451035
post?
>>8450968
wanted to show that [math](-1)(-1)=1[/math].
The post
>>8451035
contains all the steps of
>>8450968
and we see that
>>8450968
failed.