Recall that we just proved a theorem which says that for any interval of the

real numbers, (a, b) there exists a rational number q such that a < q < b: If

there is one, can we show there is 2? Sure, consider the new interval (q, b)

and use the theorem again. In fact, it seems like we could continue doing

this over and over again to get an infinite number of rationals in the interval.

Lets prove this is true but instead of inductively generating an infinite set

this way, lets prove it by using a contradction.

That is, we will assume there is a finite number of rationals in the
interval.

We will then create a new rational number in the interval by taking two

rationals in the interval and averaging them to get a new rational between

them.

**1. Prove there are an in nite number of rationals in any interval
(a,b)**

Suppose there are a nite number of rational numbers in the interval, call

this set A

Recall from the discussion above that we have at least two rationals in this

interval so this set has at least two members. Since this set is finite and has

at least two elements, we can choose two of its elements with out another

element in between them. For example, we could pick the two largest elements.

Lets call these m and n. But then, (m+n)/2 is also a rational number and

so,
and between m and n so we get a new rational not in our

original set A. This is a contradiction to our assumption that A had all of

the rationals between a and b in it. Hence, the set must be in finite.

**2. Prove using mathematical induction that
for all
**

We will prove this for an arbitrary a by using mathematical induction on n.

Base case We begin with the base case, n = 1

For this case we have a^{1}2^{1} 1 so the base case checks.

**Induction step: P(n) -> P(n+1)**

We assume P(n) is true and use this to prove P(n+1).

That is, assume a^{n}≥n for all a≥2:

Then, consider a^{n+1} = a^{n}a: We can now use P(n) to
substitute in for a^{n}. We

thus obtain,

We see that the last inequality is true because, since n≥1,

Hence, we have shown P(1) is true and P(n) -> P(n + 1) is
true. So, by the

Principle of Mathematical Induction, we know that P(n) is true

**3. Let S be a set of real numbers, and x be a real
number. Show
that x = sup(S) iff **

a .) x is an upperbound of S, and

Since this statement is of the form P <-> Q we will first
prove P -> Q

then Q -> P. First, we recall the definition of x = sup(S) requires that

1 .) x is an upperbound of S, and

2 .) for every upperbound x' of S, x≤x'

That is, I will assume that x = sup(S) and show a.) and b.).

a.) is immediate from the definition of sup. That is x =
sup(S) -> x is an

upperbound of of S by definition.

For part b.) we want to show that for any
we can find an element of S

greater than x-e: Since x is the least upperbound and
can

not be an upperbound of S. That is, there must be an element of S which is

greater than and so b.) must be true.

Hence, we have shown P -> Q now we must prove the other direction.

This direction is very similar to the other one. We are
going to assume a.)

and b.) and prove that x = sup(s).

First, requirement a.) is exactly the same as 1.) so we
are half way there

since we assumed a.

Next, we attempt to prove 2.) from b.) . That is, we want
to show that if

b.) is true we know that x must be the least upperbound.

Lets prove this by contraction, that is, lets assume b.)
is true and x is not

the least upperbound. So, there is another upperbound x' which is strictly

less than x. But then, set

Then, b.) tells me that there must be an element s ∈S for which

Recall, which was
assumed to be an upperbound so we have a

contradiction since there is an element of S which is greater than x' so it

can't be an upperbound.

Hence, we have proven a:)&b:) -> x = sup(S):

Finally, we have proven both a:)&b:) -> x = sup(S) and x =
sup(S) ->

a:)&b:) so we have a:)&b:) <-> x = sup(S):