Algebra Know-It-ALL

598 Worked-Out Solutions to Exercises: Chapters 1 to 9

We can “zip up” the sum b+c, and call the “package” e. Now our expression looks like

this:

(a+e)+d

The associative law for addition tells us that we can rewrite this as

a+ (e+d)

Now let’s “unzip” e so it turns back into the sum b+c (which it always has been, after

all), to get

a+ (b+c+d)

That’s the result we want, proving that for any four integers a,b,c, and d,

(a+b+c)+d=a+ (b+c+d)

If this seems like a trivial exercise, keep in mind that you’re getting your brain into shape

to do more complicated tricks some day.

10. The first expression can be simplified as shown in Table A-2, which breaks the process
    down into statements and reasons. The second expression can be simplified as shown in
    Table A-3. These are called S/R derivations. They’re not intended as proofs, but to show
    how an expression can be morphed into another expression. In this particular situation,
    the two original expressions turn out to be equivalent for all integers a,b, and c.

Table A-2. Solution to the first part of Prob. 10 in Chap. 4.

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

to every statement that came 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 outer sets of brackets; they 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”