epdf.pub_the-nuts-and-bolts-of-proofs-third-edition-an-intr

Introduction and Basic Terminology 17
proof is constructed by considering two separate cases. Thus, one could
argue for its inclusion in the section on Multiple Hypotheses!
EXAMPLE 5. A five-digit number is divisible by 3 when the sum of its
digits is divisible by 3.
Discussion: This statement can be rewritten as: If the sum of the digits of a
five digit number is divisible by 3, then the number is divisible by 3.
A. Let n be an integer number with n = ±a4a3a2a\ao, 0 < a, < 9 for all
i = 0, 1, 2, 3, 4, and (247^0, such that ^4+ ^3+ ^2 + ^1 + ^0 = 3t,
where t is an integer number. (The fact that n is an integer number is
an impHcit hypothesis, as the concept of divisibihty is defined only for
integer numbers. The ± sign indicates that the number n can be either
positive or negative.)
B. The number n is divisible by 3; that is, n = 3s, where s is an integer
number.
Proof: As the hypothesis provides information about the digits of the
number, we will separate the digits using powers of 10. For the sake of
simplicity, let us assume that n is positive. Thus,
n = a4a3a2a\ao = 10"* a4 + 10^ ^3 + 10^a2 + lO^i + ao.
By hypothesis, (24 + ^3 + ^2 + ai + ^o = 3t, where t is an integer number.
Therefore,
^0 = 3t — (24 — ^3 — (^2 — «l
If we substitute this expression for ao into the expression for n and
perform some algebraic steps, we obtain:
Therefore,
n = 10'^fl4 + 10^fl3 + 10^ ^2 + 10^1 + ao
= 10^*^4 + 10^fl3 + 10^a2 + lOai + (3t -a4-a2-a2- ai)
= 9,999^4 + 999^3 + 99^2 + 9ai + 3t.
n = 9,999^4 + 999^3 + 99^2 + 9ai + 3t
= 3(3,333^4 + 333^3 + 33^2 + 3ai + t).
Because the number 3,333a4 + 333^3 + 33^2 + 3ai + f is an integer, we
proved that number n is divisible by 3.
If n is a negative number, just repHcate all the steps above, starting with
n — —a4fl3a2^i«o- •