0 + a = a ; 0*a + a*a = a*a ; 0*a = a*a - a*a = 0 . WFL On 9/12/13, Bernie Cosell <bernie@fantasyfarm.com> wrote:
My apologies for asking such a dumb/simple question [I feel like I should have to forfeit my MS for even *needing* to ask], but:
I'm working through some texts trying to (a) remember what I once knew about surreal numbers, and (b) get a handle on quaternions (which I never knew). Instead of jumping in and just reading the relevant sections, I thought I'd start at the beginning and try to get some of my very old, not much used, math muscles going again. I was reading the first chapter, on fields, and it was all trivial, easy-going..... until I got to thinking about 0 * a = 0 for all a. That isn't actually mentioned as a theorem in the text [but you needed that for a theorem a few pages in and that's what got me thinking about it]
The field axioms it uses are [I mention them only because I think there are different sets of axioms that equivalently define fields but these are also the ones I remember from forever ago]: associativity: (a+b)+c = a+(b+c) (a*b)*c = a*(b*c) commutivity: a+b=b+a; a*b=b*a distributivity: a*(b+c) = (a*b)+(a*c) additive identity 0 + a = a multiplicitive id 1 * a = a additive inverse all b, there exists "-b" such that b + "-b" = 0 multipiciticative inverse: all nonzero b. there exists c such that b*c = 1
And so I've been trying to go from there to prove that 0 * a = 0. I started with: 0 * a = (1 + (-1)) * a = 1 * a + (-1) * a = a + (-1)*a
Then I run into an odd roadblock: I don't see exactly how to prove: (-1) * a = -a I even have trouble with something similarly simple/obvious like: -b * -a = -(b*a)
I figure I must be missing something unbelievably obvious or maybe this is just the wrong approach [or else I'm approaching senility a LOT faster and scarier than I feared]. I appreciate your forbearance...:o)
/Bernie\
-- Bernie Cosell Fantasy Farm Fibers mailto:bernie@fantasyfarm.com Pearisburg, VA --> Too many people, too few sheep <--
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