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+11=1 1 x (1+11) = (1x1) (1x1) + (1) + (1x1) = (1x1) 1 + (1x1) = 0 (1x1) = 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 $11=0$ and do the same derivation. I would suggest that you state which axiom you are using at each stage. 
August 7th, 2016, 04:25 AM  #3 
Member Joined: Aug 2016 From: Used to be Earth Posts: 64 Thanks: 14  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 > 11=0 subtract 1 from both sides > 1+11=1 multiply by 1 on both sides > 1 x (1+11) = (1x1) distributivity > (1x1) + (1) + (1x1) = (1x1) subtract (1x1) from both sides > 1 + (1x1) = 0 add 1 to both sides > (1x1) = 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 $11=0$, I meant you to multiply that by $1$. 
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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