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

Special Kinds of Theorems 55
Using the factorization technique for differences of two powers (this is
possible because n + 1 > 2), we can write:
^n+l _ 4^+1 ^ (7 _ ^^^yn ^ jn-1 >< 4 + . . . + 7 x 4"-^ + 4")
^ 3(7" + 7«-i X 4 + • • • + 7 X r-^ + 4").
As the number in parentheses is an integer (it is a combination of
integers), we proved that 7""^^ — 4"+^ = 3s for an integer number s.
Clearly, in the proof of the deductive step we have not used the inductive
hypothesis to reach the conclusion. This might imply either that the proof
of the original statement can be constructed without using the principle of
mathematical induction (in which case we need to redo the entire proof
to make sure that the logic of it is correct) or that we made a mistake in
the third step (in which case we can work on it and still use the previous
two steps).
Correct Proof by Induction
1. Base case. The statement is true for the smallest number we can
consider, which is 1, because 7^ — 4^ = 3.
2. Inductive hypothesis. Assume the statement is true for an arbitrary
number n; that is 7" - 4" = 3t for an integer number t.
3. Deductive proof. Is the statement true for n+ 1? Is 7""^^ — 4^'^^ = 3s
for an integer number s?
We will use properties of exponents and other rules of algebra to
obtain the expression 7" — 4" so we can use the inductive hypothesis.
Thus, we have:
yn+l _ 4n+l ^ 7 x 7" - 4 X 4"
= (3 + 4) X 7" - 4 X 4" = 3 X 7" + 4 X 7" - 4 X 4"
-= 3 X 7" + 4 X (7" - 4").
At this point we can use the inductive hypothesis to write:
yn+l _ 4^+1 :::, 3 X 7" + 4 X (7" - 4")
= 3x7" + 4x3t = 3(7" + 40-
Because the number T + 4t is an integer number, we proved that
7«+i _ 4^+1 _ 35 fQj. 2in integer number s.
Therefore, by the principle of mathematical induction the original
statement is true. •