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

156 The Nuts and Bolts of Proof, Third Edition
not to be distinct. Similarly, the numbers qj need not be distinct, but
no Pi can equal any qj. Using the factorization of p above and the
properties of multiplication, we have p" = (PiTiPiT- • • (Pr-iTiPrT-
Because the prime factors of p" are the same as the prime factors of p,
p" and q have no common factors. Therefore, q cannot divide p".
10. Step 1: Is the formula true for n= 1, the smallest number we can use?
When n = 1, we obtain
11 1
12 1 + 1
Therefore, the formula is true in this case.
Step 2. Assume that the formula is true for an arbitrary number n;
that is,
11 11 11 11 n
12 23 34 nn-\-l n+T
Step 3. Show that the formula is true when we use it for the next
integer number (namely, n +1). So, we need to prove that:
11 11 11 11 1 1 n + 1
T7^ + ;;T + X-: + ---+ —^ + -
12 23 34 nw+1 (n+l)(n + 2) (n+l)+l'
or, equivalently,
11 11 11 1 1 1 1 _n+l
J2'^23'^34^"'^nn+l^(n-\-l)(n + 2)~n-j-2'
We will manipulate the expression on the left-hand-side of the
equation using first the associative property of addition, then the
inductive hypothesis, and then some algebraic steps:
11 11 11 11 1 1
12 23 34 nn-\-l (n+l)(n + 2)
11 11 11 111 1 1
12 23 34 nn+lj (n+l)(n + 2)
1 n(n + 2) + 1
• + -
n+1 (n+l)(n + 2) (n+l)(n + 2)
_ n-\-1
~n + 2*
Because this is exactly the equahty we were trying to prove, the
formula is indeed true for all positive integer numbers by the principle
of mathematical induction.