Page 41 - DMTH502_LINEAR_ALGEBRA

Unit 1: Vector Space over Fields

Proof: (a) We know that Notes
a0 = a (0 + 0) = a0 + a0 a R (using distributive law)
Since R is a group under addition, applying right cancellation law,
a0 = a0 + a0 0 + a0 = a0 + a0 a0 = 0

Similarly, 0a = (0 + 0) a = 0a + 0a (using distributive law)
0 + 0a = 0a + 0a (because 0 + 0a = 0a)
Applying right cancellation law for addition, we get
0 = 0a i.e., 0a = 0

Thus a0 = 0a = 0.
(b) To prove that a (–b) = –ab we would show that
ab + a (–b) = 0
We know that a [b + (–b)] = a0 [because b + (–b) = 0]

= 0 (with the virtue of result (a) above)
or ab + a (–b) = 0 (by distributive law)
a (–b) = – (ab).
Similarly, to show (–a) b = – ab, we must show that

ab + (–a) b = 0
But ab + (–a) b = [a + (–a)] b = 0b = 0
– (a) b = – (ab)
Hence the result.

(c) Actually to prove (–a) (–b) = ab is a special case of foregoing article. However its proof is
given as under:
(–a) (–b) = – [a (–b)] [by result b]

= [– (ab)] [because a (–b) = –ab]
= ab
because – (–x) = x is a consequence of the fact that in a group inverse of the inverse of an
element is element itself.

Illustrative Examples

Example 12: Prove that the set of all rational numbers is a ring with respect to ordinary
addition and multiplication.
Let Q be the set of all rational numbers.
R (Q, +) is abelian.
1
(R ) Let a, b Q then a + b Q because sum of two rational numbers is a rational number.
11

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