MATHEMATICAL RECREATIONS 173

433. Now let us combine the theories of complements and
excesses, taking the base for excesses half the base for com-
plements, calling the latter the base for the combined theories.

434, Th. Then

n = m/2
c=m—a d=m—0>
e =a— mf2 f=b—m/2
c+ &= m2 d+f=ml2

c+d=2m— (¢ + D)
e+ f=@+0b) —m
c+d)+(e+f)=m
c+d=m— (e+f)
e+ f=m— (c+4d)

435. Th. Sincee + f = (@ + b) — m,
a+b=m+ (e +f)
In words: The sum of two numbers equals the base plus the
sum of their excesses over half the base.

436. Th. Since ab = [m — (¢ + d)Im + cd,
ab = (e + f)m + cd
That is, the product of two numbers equals the product of the
base and the sum of their excesses over half the base plus the product
of their complements.

437. Th. Since ab = [n + (e + f) In + ¢f, the product of
two numbers equals the product obtaind by multiplying the sum
of half the base and the excesses of the two given mumbers over
half the base by half the base plus the product of these excesses.

438. If a, b, and m are integers, since ¢ + f = (a + b) — m,
e + f is also an integer.

Hence, since (a + b) — (e + f) = m,
if m—=0, (@a+0b) —(e+f)>m

Hence a+ b=e+f (modm)

Th. That is, if the base is a non-zero integer, the sum of two
integers is congruent to the sum of their excesses over half the
base with respect to the base as a modulus.