Table A-1. Solution to Prob. 8 in Chap. 4. This S/R proof shows
how you can reverse the order in which four integers a,b,c, and d are
added, and get the same sum. As you read down the left-hand column,
each statement is equal to every statement that came above it.
Statements Reasons
a+b+c+d Begin here
a+ (b+c+d) Group the last three integers
a+ (d+c+b) Result of the “challenge” where it was proved
that you can reverse the order of a sum of three
integers
(d+c+b)+a Commutative law for the sum of a and
(d+c+b)
d+c+b+a Ungroup the first three integers
Q.E.D. Mission accomplished
Here are a couple of extra-credit exercises if you want a further challenge. Does the
original rule work when a= 0, but you let b and c be any integers? Does it work when
b= 0, but you let a and c be any integers? You’re on your own!
- This always works if c= 0. Then a and b can be any integers you want. Here’s the
original equation:
(a−b)−c=a− (b−c)
“Plug in” 0 for c. Then you get
(a−b)− 0 =a− (b− 0)
which simplifies to
a−b=a−b
Here are a couple more extra-credit exercises. Does the original rule work when a= 0,
but you let b and c be any integers? Does it work when b= 0, but you let a and c be any
integers? Have fun!
- See Table A-1. Follow each statement and reason closely so you’re sure how it follows
from the statements before. This proves that for any four integers a,b,c, and d,
a+b+c+d=d+c+b+a
- Let’s start with the original expression, and then morph it into the second one. We
begin with this:
(a+b+c)+d
Chapter 4 597
