Schaum's Outline of Discrete Mathematics, Third Edition (Schaum's Outlines)

450 ALGEBRAIC SYSTEMS [APP. B

EXAMPLE B.17 Letf(t)=t^4 − 3 t^3 + 6 t^2 + 25 t−39. Find all the roots off(t)given thatt= 2 + 3 iis a
root.

Since 2+ 3 iis a root, then 2− 3 iis a root andc(t)=t^2 − 4 t+13 is a factor off(t). Dividingf(t)byc(t)
we get
f(t)=(t^2 − 4 t+ 13 )(t^2 +t− 3 )

The quadratic formula witht^2 +t−3 gives us the other roots off(t). That is, the four roots off(t)are as
follows:
t= 2 + 3 i, t= 2 − 3 i, t=(− 1 +

√

13 )/ 2 ,t=(− 1 −

√

13 )/ 2

SolvedProblems

OPERATIONS AND SEMIGROUPS

B.1.Consider the setQof rational numbers, and let∗be the operation onQdefined by

a∗b=a+b−ab

(a) Find: (i) 3∗4; (ii) 2∗(−5); (iii) 7∗(1/2).

(b)Is(Q,∗) a semigroup? Is it commutative?

(c) Find the identity element for∗.

(d) Do any of the elements inQhave an inverse? What is it?

(a) (i) 3∗ 4 = 3 + 4 − 3 ( 4 )= 3 + 4 − 12 =− 5

(ii) 2∗(− 5 )= 2 +(− 5 )+ 2 (− 5 )= 2 − 5 + 10 = 7

(iii) 7∗( 1 / 2 )= 7 +( 1 / 2 )− 7 ( 1 / 2 )= 4

(b) We have:

(a∗b)∗c=(a+b−ab)∗c=(a+b−ab)+c−(a+b−ab)c

=a+b−ab+c−ac−bc+abc=a+b+c−ab−ac−bc+abc

a∗(b∗c)=a∗(b+c−bc)=a+(b+c−bc)−a(b+c−bc)

=a+b+c−bc−ab−ac+abc

Hence∗is associative and (Q,∗) is a semigroup. Also

a∗b=a+b−ab=b+a−ba=b∗a

Hence (Q,∗) is a commutative semigroup.

(c) Anelementeis an identity element ifa∗e=afor everya∈Q. Compute as follows:

a∗e=a, a+e−ae=a, e−ea= 0 ,e( 1 −a)= 0 ,e= 0

Accordingly, 0 is the identity element.

(d) In order forato have an inversex, we must havea∗x=0 since 0 is the identity element by Part (c).

Compute as follows:

a∗x= 0 ,a+x−ax= 0 ,a=ax−x, a=x(a−l), x=a/(a−l)

Thus ifa=1, thenahas an inverse and it isa/(a− 1 ).

B.2.LetSbe a semigroup with identitye, and letbandb′be inverses ofa. Show thatb=b′, that is, that

inverses are unique if they exist.

We have:

b∗(a∗b′)=b∗e=b and (b∗a)∗b′=e∗b′=b′

SinceSis associative, (b∗a)∗b′=b∗(a∗b′); henceb=b′.