Special Assignment #1 — Solutions Math 315
1. Prove that for every natural number n > 3 that n!>2n. [You might want to use
the fact (that you need not prove!) that n! is defined recursively by: For each integer
n≥0
0! = 1 and (n+ 1)! = (n+ 1)n!]
Let Pnbe the statement n!>2n. We’ll prove that Pnis true for all n > 3.
(B) 4! = 24 >16 = 24, so P4is true.
(IS) Let n∈IN with n≥4 and assume that Pnis true. Then n+ 1 >4>2,
and
(n+ 1)! = (n+ 1)n!(IH)
>(n+ 1)2n>2·2n= 2n+1,
and Pn+1 is true.
So by induction Pnis true for all n > 3.
2. Prove the Generalized Triangle Inequality: For every finite sequence a1, a2, . . . , an∈
IR
|a1+a2+· · · +an| ≤ |a1|+|a2|+· · · +|an|.
[Of course, you may use the Triangle Inequality (see 3.7) itself to get started!]
Let Pnbe the statement
|a1+a2+· · · +an| ≤ |a1|+|a2|+· · · +|an|.
We’ll prove thatPnis true for all n∈IN.
(B) |a1| ≤ |a1|, so P1is true.
(IS) Let n∈IN and assume that Pnis true. Then for a1, a2, . . . , an+1 ∈IR
|a1+a2+. . . +an+an+1 |=|(a1+a2+. . . +an) + an+1 |
(3.7)
≤ |a1+a2+. . . +an|+|an+1 |
(IH)
≤(|a1|+|a2|+· · · +|an|) + |an+1 |
=|a1|+|a2|+· · · +|an|+|an+1 |
and Pn+1 is true.
So by induction Pnis true for all n∈IN.
