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

284 PROPERTIES OF THE INTEGERS [CHAP. 11

11.5. Prove Proposition 11.3: Supposea≤b, andcis any integer. Then: (i)a+c≤b+c,

(ii)ac=bcwhenc>0; butac=bcwhenc<0.

The proposition is certainly true whena=b. Hence we need only consider the case whena<b, that is, when

b−ais positive.

(i) The following difference is positive:(b+c)−(a+c)=b−a. Hencea+c<b+c.

(ii) Suppose c is positive. By property [P 1 ] of the positive integersN, the productc(b−a) is also positive. Thus

ac < bc.

Now supposecis negative. Then−cis positive, and the product(−c)(b−a)=ac−bcis also positive. Accordingly,

bc < ac, whenceac > bc.

11.6. Prove Proposition 11.4 (iii):|ab|=|a||b|.

The proof consists of analysing the following five cases: (a)a=0orb=0; (b)a>0 andb>0; (c)a>0 and

b<0; (d)ba <0 andb>0; (e)ba <0 andb<0. We only prove the third case here. (c) Sincea>0 andb<0,

|a|=aand|b|=−b. Alsoab <0. Hence|ab|=−(ab)=a(−b)=|a||b|.

11.7 Prove Proposition 11.4 (iv):|a±b|≤|a|+|b|.

Nowab≤|ab|=|a||b|, and so 2ab≤ 2 |a||b|. Hence

(a+b)^2 =a^2 + 2 ab+b^2 ≤|a|^2 + 2 |a||b|+|b|^2 =(|a|+|b|)^2

But

√

(a+b)^2 =|a+b|. Thus the square root of the above yields|a+b|≤|a|+|b|. Also,

|a−b|=|a+(−b)|≤|a|+|−b|=|a|+|b|

MATHEMATICAL INDUCTION,WELL-ORDERING PRINCIPLE

11.8. Prove the proposition that the sum of the first n positive integers isn(n+ 1 )/2; that is:

P (n): 1 + 2 +···+n=

1

2

n(n+ 1 )

P( 1 )is true since 1=^12 ( 1 )( 1 + 1 ). AssumingP(k)is true, we addk+1 to both sides ofP(k)obtaining

1 + 2 + 3 +···+k+(k+ 1 ) =

1

2

k(k+ 1 )+(k+ 1 )=

1

2

[k(k+ 1 )+ 2 (k+ 1 )]

=

1

2

[(k+ 1 )(k+ 2 )]

This isP(k+ 1 ). Accordingly,P(k+ 1 )is true wheneverP(k)is true. By the principle of mathematical induction,

Pis true for everyn∈N.

11.9. Supposea=1. ShowPis true for alln≥1 wherePis defined as follows:

P (n): 1 +a+a^2 +···+an=

an+^1 − 1

a− 1

P(1) is true since

1 +a=

a^2 − 1

a− 1

AssumingP(k)is true, we addak+^1 to both sides ofP(k), obtaining

1 +a+a^2 +...+ak+ak+^1 =

ak+^1 − 1

a− 1

+ak+^1 =

ak+^1 − 1 +(a− 1 )ak+^1

a− 1

=

ak+^2 − 1

a− 1

This isP(k+ 1 ). Thus,P(k+ 1 )is true wheneverP(k)is true. By the principle of mathematical induction,Pis true

for everyn∈N.