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August 6th, 2016, 09:39 PM   #1
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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.
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August 6th, 2016, 10:32 PM   #2
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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.
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August 7th, 2016, 04:25 AM   #3
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Quote:
Originally Posted by v8archie View Post
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
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August 7th, 2016, 05:25 AM   #4
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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 $-1$.
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August 7th, 2016, 03:06 PM   #5
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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.
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August 7th, 2016, 05:12 PM   #6
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Fair enough.
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