1. the first 5 conditons you have listed amount to showing (R,+,0) is an abelian group. there are some important rules you did not list:

(R,*) is an associative semi-group:

a*b (often writtten just ab) must be in R (closure of multiplication)

(a*b)*c = a*(b*c) (associativity of multplication)

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other important rules a ring must satisfy:

a*(b+c) = a*b + a*c (distributivity of left-multiplication)

(a+b)*c = a*c + b*c (distributivity of right multiplication). we have to check these separately because it may not be the case that a*b = b*a (such rings are commutative).

if a ring IS commutative, checking one distributive axiom will suffice.

a ring may also possess a multiplicative element, called unity, and wriiten 1, as a multiplicative identity (a*1 = 1*a for all a in R).

if a commutative ring with unity possesses multiplicative inverses for all non-zero a in R, it is a field

(this can be said more succintly by stating R is a ring with (R-{0},*,1) an abelian group).

now for Z[x], pick 3 "arbitrary polynomials":

and

and show the axioms hold. you may find it helpful to make all 3 polynomials the same degree by setting t = max({deg(p),deg(q),deg(r)}), and using coefficients of 0, for powers of x between m and t, n and t, and s and t.

a caveat: since you are dealing with integer coefficients, "dividing" is something you should stay away from.

3. is the set of all real numbers of the form a+b√2, where a,b are rational. addition will be easy....multiplication is messy, but not too hard. to find an inverse, look at what (a + b√2)(a - b√2) is. if this is rational (and non-zero), say it's q, then what would (1/q)(a + b√2)(a - b√2) be?

be sure to state your final answer only in terms of a and b.

for 6: the small c's mean: "complement". if your "universal set" is U, then Ac = U - A = {everything in U, except the stuff in A}. for example, if U = {1,2,3,4,5}, and A = {2,4}, then Ac = {1,3,5}.

in a venn diagram, if A is drawn as a circle (colored in), then Ac would be everything colored in outside the circle, and the circle A left blank (clear).