In Tuesday’s class we covered some challenge questions and some definitions and in this blog I will discuss about the following definitions that we addressed in class: Supremum and maximum, Theorem 1.4.2(Archimedean Property), Theorem 1.4.3(Density of Q in R) and Theorem 1.4.5 (There exists a real number α ∈ R satisfying α^2 = 2).

1.Supremum and maximum

Supremum: A real number s is the least upper bound for a set A ⊆ R if it meets the following two criteria:
(i) s is an upper bound for A;
(ii) if b is any upper bound for A, then s ≤ b.

Maximum: A real number x is a maximum of the set A if x is an element of A and x ≥ a for all a ∈ A.

Things to consider:

A maximum is a specific type of upper bound that is an element of the set while a supremum is not required to be an element of the set. The open interval (0, 2) does not possess 2 so the open interval (0,2) has a supremum of 2 but it doesn’t have a maximum. Thus, the supremum can exist and not be a maximum in this set, but when a maximum exists, then it is also the supremum.

2. Archimedean Property

(i) Given any number x ∈ R,
there exists an n ∈ N satisfying n > x.
(ii) Given any real number y > 0, there exists an n ∈ N satisfying 1/n < y.

Pf template:

Proving Archimedean Property relies on the Axiom of Completeness(AoC). For part (i) we can assume that N is bounded above and by AoC N should have a least upper bound. So we can set a = supN and a-1 will not be an upper bound since a is the least upper bound for N. So there exists an n∈ N satisfying a-1 < n. That means a < n + 1 but n + 1 ∈ N so we now have a contradiction that a is a least upper bound.

And for part (ii) we can set x=1/y. Since y is a real number and real number is closed under division we know that x is also a real number. Following part(i) we know that for any real number there exists a natural number that is larger than it, which means that 1/y < n for some n ∈ N. Then we can get 1/n < y by multiplying both sides by y and 1/n.

3. Density of Q in R

For every two real numbers a and b with a < b, there exists a rational number r satisfying a < r < b.

Pf template:

The formula can be changed to a < p/q < b for some p ∈ Z and q ∈ N. The basic steps of proving this theorem is that first we need to choose a denominator q large enough so that the consecutive increments of size 1/q will be close together enough to “step over” the interval of (a,b). Using the Archimedean Property to find a p large enough so that 1/p < b – a. The inequality a < p/q < b is equivalent to qa < p < qb. Then the idea is to find the smallest integer p so that it will be greater than qa, which means p-1< qa < p. And by transformation and combination of the two inequalities above (…) we should be able to come up with a < p/q and p/q < b. Combine these two we get a < p/q < b as desired.

Zehao Dong