Question

If R is a ring then prove that a(-b)=(-a)b=(-ab) for all a,b€R​

Answer 1

Answer:

3.1.1. The following subsets of Z (with ordinary addition and multiplication) satisfy all but one of the

axioms for a ring. In each case, which axiom fails.

(a) The set S of odd integers.

• The sum of two odd integers is a even integer. Therefore, the set S is not closed under addition.

Hence, Axiom 1 is violated.

(b) The set of nonnegative integers.

• If a is a positive integer, then there is no solution of a + x = 0 that is also positive. Hence, Axiom

5 is violated.

3.1.2

(a) Show that the set R of all multiples of 3 is a subring of Z.

(b) Let k be a fixed integer. Show that the set of all multiples of k is a subring of Z.

• Clearly, (b) implies (a); so let us just prove (b). Let

S = {z ∈ Z | z = nk for somen ∈ Z} .

In general, to show that a subset S of a ring R, is a subring of R, it is sufficient to show that

(i) S is closed under addition in R

(ii) S is closed under multiplication in R;

(iii) 0R ∈ S;

(iv) when a ∈ S, the equation a + x = 0R has a solution in S.

Let a, b, c ∈ S ⊂ Z with a = rk, b = sk, c = tk.

(i) a + b = rk + sk = (r + s)k ∈ S

(ii) ab = (rk)(sk) = (rsk)k ∈ S

(iii) 0Z = 0 = 0 · k ∈ S

(iv) a = rs ∈ S ⇒ x = −rs ∈ S is a solution ofa + x = 0S

Thus, S is a subring of Z.

3.1.3. Let R = {0, e, b, c} with addition and multiplication defined by the tables below:

+ 0 e b c · 0 e b c

0 0 e b c 0 0 0 0 0

e e 0 c b e 0 e b c

b b c 0 e b 0 b e c

c c b e 0 c 0 c c 0

Assume distributivity and associativity and show that R is a ring with identity. Is R commutative?

Axioms (1) and (6) are satisfied by virtue of the tables above. We are also allowed to assume that Axioms (2),

(7) and (8) hold. Axiom (3), commutatively of addition, is also evident from the symmetry of the addition

table. Similarly, the symmetry of the multiplication table implies that multiplication is commutative for

1

2

this set. From the addition table it is also clear that 0 + a = a for any a ∈

Explanation:

If R is a ring then prove that a(-b)=(-a)b=(-ab) for all a,b€R