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 August 6th, 2016, 09:39 PM #1 Member   Joined: Aug 2016 From: Used to be Earth Posts: 64 Thanks: 14 A Quick Proof Check Using the field axioms, prove: (-1 x -1) = 1 -1+0=-1 -1+1-1=-1 -1 x (-1+1-1) = (-1x-1) (-1x-1) + (-1) + (-1x-1) = (-1x-1) -1 + (-1x-1) = 0 (-1x-1) = 1 Is this correct? Thanks. Last edited by sKebess; August 6th, 2016 at 09:42 PM.
 August 6th, 2016, 10:32 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra You could just start with $1-1=0$ and do the same derivation. I would suggest that you state which axiom you are using at each stage. Thanks from sKebess
August 7th, 2016, 04:25 AM   #3
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Quote:
 Originally Posted by v8archie You could just start with $1-1=0$ and do the same derivation.
Yes, quite true. But, since that isn't reducing the number of steps, is it really "better" so to speak? It should give the following, right?

inverses ->
1-1=0

subtract 1 from both sides ->
-1+1-1=-1

multiply by -1 on both sides ->
-1 x (-1+1-1) = (-1x-1)

distributivity ->
(-1x-1) + (-1) + (-1x-1) = (-1x-1)

subtract (-1x-1) from both sides ->
-1 + (-1x-1) = 0

add 1 to both sides ->
(-1x-1) = 1

 August 7th, 2016, 05:25 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra That's OK, but you don't mention "identity" for $-1 \times 1$. Also, in starting with $1-1=0$, I meant you to multiply that by $-1$. Thanks from sKebess
 August 7th, 2016, 03:06 PM #5 Member   Joined: Aug 2016 From: Used to be Earth Posts: 64 Thanks: 14 Hum, but $\displaystyle -1\times0=0$ is not given here though, so the rhs would have complicated things a little. Unless, you had something else in mind.
 August 7th, 2016, 05:12 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra Fair enough.

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