“Plug in” 1 for c. Then you get
(a/b)/1=a/(b/1)
which simplifies to
a/b=a/b
Here is an opportunity get some extra credit. Does the original rule work when a= 1, but
you let b and c be any integers except 0? Does it work when b= 1, but you let a be any
integer and c be any integer except 0?
- This, again, always works if c= 1. Then a and b can be any integers. Here’s the original
equation:
(ab)/c=a(b/c)
“Plug in” 1 for c. Then you get
(ab)/1=a(b/1)
which simplifies to
ab=ab
Now, can you guess what’s coming? Another extra-credit workout! Does the original rule
hold true when a= 1, but you let b be any integer and c be any integer except 0? Does it
work when b= 1, but you let a be any integer and c be any integer except 0?
- Let’s start with the distributive law of multiplication over addition in its left-hand form.
You’ll notice that we’re using different letters of the alphabet. That will help keep you
from sinking into an “abc rut” with the naming of variables. Otherwise, the law is stated
in exactly the same way. We can write the original form of the law like this:
p(m+n)=pm+pn
where n,m, and p are integers. Now, the solution is only a matter of applying the commutative
law for multiplication three times, once for each of the three products above:
(m+n)p=mp+np
Q.E.D. That’s all there is to it!
- The variables have unfamiliar names for the same reason as in Prob. 8. See Table A-6.
This proves that for any two integers d and g,
−(d+g)=−d−g
- Again, unfamiliar variable names can keep your attention on the way things work,
without getting stuck in the “abc routine” of rote memorization. See Table A-7. This
proves that for any two integers h and k,
−(h−k)=k−h
Chapter 5 601
