Algebra Know-It-ALL

Chapter 5

1. If you start with a positive integer p and multiply by −3, you get a negative integer
   whose absolute value is 3 |p|. If you start with a negative integer n and multiply by −3,
   you get a positive integer whose absolute value is 3 |n|.


2. If you start with any integer and keep multiplying by −3 over and over, the polarity
   (that is, the “positivity” or “negativity”) of the end product keeps alternating, and the
   absolute value increases by a factor of 3 every time. So with any integer q, whether
   positive or negative, you get

|− 3 q|= 3 |q|

|− 3 × (− 3 q)|= 3 × 3 |q|

|− 3 × (−3)× (− 3 q)|= 3 × 3 × 3 |q|

↓

and so on, forever

As you can see, the absolute value grows rapidly as you keep on multiplying by −3, unless

the starting integer q happens to be 0.

3. Evaluating this requires close attention and patience. Here is the initial expression for
   reference:

4 + 32 / 8 × (−2)+ 20 / 5 / 2 − 8

Table A-4 is an S/R breakdown of the evaluation process.

Table A-3. Solution to the second part of Prob. 10 in Chap. 4.

As you read down the left-hand column, each statement is equal

to every statement above it.

Statements Reasons

a+ (b−c)+ (a−b)+c Begin here

a+ [b+ (−c)]+ [a+ (−b)]+c Change subtractions into negative additions;

replace original parentheses with brackets

for temporary clarification

a+b+ (−c)+a+ (−b)+c Get rid of brackets; they served a good

purpose but are not necessary in straight sums

a+a+b+ (−b)+c+ (−c) Commutative law for addition, generalized

a+a b+ (−b) = 0

and

c+ (−c)= 0

so the b’s and c’s “cancel out”

Chapter 5 599