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

162 The Nuts and Bolts of Proof, Third Edition
20. Let £ > 0 be given. Is it possible to find a (5 > 0 such that \f{x) — 10| < £
for all X with |x — 1| <5? Observe that
\f(x) - 10| = |3x2 + lx- 10| = \x- l||3x + 10|.
How large can the quantity |3x + 10| be, if x is sufficiently close to 1?
Start by using values of x closer than 2 units to 1 (this is a completely
arbitrary choice); that is, -1 < x < 3. Then -3 < 3x < 9, and
7 < 3x + 10 < 19. Therefore, for these values of x we have:
[/(x)-10| = |x-l||3x+10| < 19|jc-l|.
This quantity is smaller than s when \x — 1|<£/19. Thus, choose
5 = min{2, 8/\9]. When \x - 1| <5, it will follow that \f{x) - 10| <8.
Note that different choices of the interval around 1 will produce
different choices for 5.
21. This statement will be proved using mathematical induction.
a. We will show that the formula holds true when /c=: 1, the smallest
number we are allowed to use.
1^(1 + \f
V = l =
Therefore, the equality is true for /c= 1.
b. The inductive hypothesis states that the formula holds true for an
arbitrary number n; that is.
3 , o3 , i3 , ^ ^3 _ n\n-^X)
r + 2^ + 3'+ • • • + n'-
c. We want to prove that:
l. + 2= + 3= + ^^^ + „> + ,„+,)3 = <^l±«!Kl±i)±i£,
Using the associative property of addition and the inductive
hypothesis we obtain:
1^ + 2^ + 3^ + • • • + n^ + (n + 1)^
= [1^ + 2^ + 3^ + • • • + n^] + (n + 1)^
= !'!fiL+i)! + ,„ + i)3.