Elementary Analysis. Homework 4

MATH5301 Elementary Analysis. Homework 4.

4.1
Using axioms of the ordered field, prove
(a) a > c ∧ b > d ⇒ a + b > c + d
(b) a > c > 0 ∧ b > d > 0 ⇒ ab > cd > 0
(c) a > b > 0 ⇒
1
a
<
1
b
(d) Denote |x| =
(
x, if x > 0
−x, else
. Then for any a, b it follows |a − b| > ||a| − |b||.
4.2
Determine which of the axioms satisfied by the set of real numbers are not satisfied by the following set:
(a) Set Q of all rational numbers.
(b) Set Q(
√
2) of all numbers, having the form a + b
√
2, where a, b ∈ Q
(c) Set C of all pairs of real numbers (a, b) with the addition defined as: (a, b) + (c, d) = (a + c, b + d), multiplication
defined as (a, b)·(c, d) = (ac−bd, ad+bc) and the order relation (a, b) < (c, d) if and only if (b 6 d)∨(b = d∧a < c).
4.3
Using the method of mathematical induction, prove the following statements:
(a) Bernoulli inequality: for all n ∈ N and for all x > −1 it follows that
(1 + x)
n > 1 + nx
(b) For all n ∈ N it follows
1
2
+
2
2
2
+ · · · +
n
2
n
= 2 −
n + 2
2
n
(c)
(1 + q)(1 + q
2
)(1 + q
4
)· · ·(1 + q
2
n
) = 1 − q
2
n+1
1 − q
(d)
1
3 + 33 + · · · + (2n + 1)3 = (n + 1)2
(2n
2 + 4n + 1)
(e)
Xn
k=0
(−1)k n!
k!(n − k)! = 0,
Xn
k=0
n!
k!(n − k)! = 2n
,
4.4
Show that for all n ∈ N, n > 2
(a)
1
√
1
+
1
√
2
+ · · · +
1
√
n
>
√
n
(b)
1
n + 1
+
1
n + 2
+ · · · +
1
3n + 1
> 1
(c)

n + 1
2
n
> n!
(d) The number 22
n
− 6 is divisible by 10.
4.5
(a) Show that √
2 ∈/ Q
(b) Show that for any a, b ∈ Q, a < b there exists x ∈ R \ Q : a < x < b
(c) Show that for any a, b ∈ R \ Q, a < b there exists x ∈ Q : a < x < b
4.6
Prove that for any n
(a) For the any configuration of n straight lines on a plane one could color the plane in two colors, so that every two
parts, having common boundary would have different colors.
(b) For any set of n squares, one can partition them on finite amount of pieces and which will constitute one square.
(c) What’s wrong with this ”theorem”?
Theorem 1. All the numbers are equal. In other words, the statement Pn is true for all n ∈ N, where Pn is ”if
{a1, a2, . . . , an} is a collection of n numbers, then a1 = a2 = · · · = an”.
Proof. P1 is definitely true: if {a1} is a collection of 1 number, then a1 = a1.
Assume that Pn is true. Let {a1, . . . , an+1} be a collection of n+ 1 number. Consider the collection {a1, . . . , an}.
Thanks to Pn it follows that a1 = · · · = an denote this number by b. Now, consider the collection {a2, . . . , an, an+1}.
It contains n numbers, and hence by Pn we get a2 = · · · = an = an+1. But a2 = · · · = an = b, therefore
an+1 = an = b. So a1 = · · · = an+1 = b and Pn+1 follows.