Is where a ring with the usual operations of addition and multiplication?
It needs to be closed under addition and multiplication which it seems like
Addition needs to be commutative and associative which is quite straightforward.
It must have an identity element which I am not sure about
and it must have an additive inverse which I am also unclear about
Multiplication must be associative which it should be and then it must be distributive over addition meaning that a*(b+c)=a*b+a*c which i also think it is.
Aug 18th 2012, 09:03 PM
Ok I figured it out
Aug 18th 2012, 09:11 PM
you need to PROVE it is all those things. saying "it seems like it" isn't good enough.
for example, to PROVE closure under addition, we need to show that:
(a+bi) + (c+di) can be written in the form: k+mi, for some integers k and m. i suggest looking at k = a+c, and m = b+d.
let's see if we can "discover" what 0 and -(a+bi) would have to be (and just maybe, we'll show they do exist along the way).
so we are looking for some z+wi, with z and w integers, with:
(a+bi) + (z+wi) = a+bi, for EVERY integer a and b.
at this point, we have to have some idea of how to "add" two things of the form (a+bi). the actual definition is:
(a+bi) + (c+di) = (a+c) + (b+d)i (as you may have guessed from what i wrote above).
now, let's see what this tells us about our "mystery" additive identity:
(a+bi) + (z+wi) = (a+z) + (b+w)i, and we want this to equal a+bi.
this means: a+z = a, and b+w = b.
now those are just two equations involving only regular ol' integers, so we can solve those.
a+z = a
-a+(a+z) = -a+a
(-a+a)+z = 0
0+z = 0
z = 0
so the "z part" in our additive identity (if it exists) has to be 0. so we're looking for something of the form 0+wi.
now do the same thing for b. what is your conclusion? is there an additive identity?
to find the multiplicative identity, you need to have some idea of what "ordinary" multiplication is.
i will tell you, it is defined like so:
(a+bi)*(c+di) = (ac-bd) + (ad+bc)i
now suppose we have a multiplicative identity, e+fi.
(a+bi)*(e+fi) = a+bi
(ae-bf) + (af+be)i = a+bi, so:
ae-bf = a
af+be = b
let's re-arrange these equations:
ae - a = bf
be - b = af, and again:
a(e-1) = bf
b(e-1) = af. using an old algebraic trick, multply the first one by a, and the second one by b:
a2(e-1) = abf = b2(e-1).
now we can obviously pick a and b so that a2 ≠ b2, so the only way this is *always* going to be true (for every a and b) is if e-1 = 0, so e = 1.
so our identity is of the form 1+fi.
going back to:
ae - a = bf, since we now know e = 1, this becomes:
a - a = bf
0 = bf.
since we can pick some b that isn't 0, the only way for this to be true for ALL b, is if f = 0.
so our identity, if there is one, is 1+0i. now prove that this works.
proving multiplication is associative and commutative is a bit of a pain, but you should do it, straight from the definitions, at least once in your life.
distributivity of multiplication of addition is going to be the hardest. keep careful track of your terms.
Aug 18th 2012, 09:20 PM
Thanks. I partially prove these properties. The easy ones I prove and the annoying ones I just state. Thanks for helping me out. You are very clear