The set of real numbers

The set of real numbers
We finally get to the real number system. Instead of constructing the real number set from the
rational numbers, we simply state their existence as a theorem without proof. Notice that Q is an
ordered field.
Theorem 1.2.1. There exists a unique ordered field R with the least-upper-bound property such
that Q  R.
Note that also N  Q. As we have seen, 1 > 0. By induction (exercise) we can prove that n > 0
for all n 2 N. Similarly we can easily verify all the statements we know about rational numbers and
their natural ordering.
Let us prove one of the most basic but useful results about the real numbers. The following
proposition is essentially how an analyst proves that a number is zero.
Proposition 1.2.2. If x 2 R is such that x  0 and x < e for all e 2 R where e > 0, then x = 0.
Proof. If x > 0, then x=2 < x. Hence taking e = x=2 we get a contradiction. Therefore x = 0.
A more general and related simple fact is that any time you have two real numbers a < b, then
there is another real number c such that a < c < b. Just take for example c = a+b
2 (why?). In fact,
there are infinitely many real numbers between a and b.
The most useful property of R for analysts, however, is not just that it is an ordered field, but
that it has the least-upper-bound property. Essentially we want Q, but we also want to take suprema
(and infima) willy-nilly. So what we do is to throw in enough numbers to obtain R.
We have already seen that R must contain elements that are not in Q because of the least-upperbound
property. We have seen that there is no rational square root of two. The set fx 2 Q j x2 < 2g
implies the existence of the real number
p
2 that is not rational, although this fact requires a bit of
work.
Example 1.2.3: Claim: There exists a unique positive real number r such that r2 = 2. We denote r
by
p
2.
Proof. Take the set A := fx 2 R j x2 < 2g. First we must note that if x2 < 2, then x < 2. To see this
fact, note that x  2 implies x2  4 (use Proposition 1.1.8 we will not explicitly mention its use
from now on), hence any number such that x  2 is not in A. Thus A is bounded above. As 1 2 A,
then A is nonempty.
Uniqueness is up to isomorphism, but we wish to avoid excessive use of algebra. For us, it is simply enough to
assume that a set of real numbers exists. See Rudin [R2] for the construction and more details.
26 CHAPTER 1. REAL NUMBERS
Let us define r := sup A. We will show that r2 = 2 by showing that r2  2 and r2  2. This
is the way analysts show equality, by showing two inequalities. Note that we already know that
r  1 > 0.
Let us first show that r2  2. Take a number s  1 such that s2 < 2. Note that 2􀀀s2 > 0.
Therefore 2􀀀s2
2(s+1) > 0. We can choose an h 2 R such that 0 < h < 2􀀀s2
2(s+1) . Furthermore, we can
assume that h < 1.
Claim: 0 < a < b implies b2􀀀a2 < 2(b􀀀a)b. Proof: Write
b2􀀀a2 = (b􀀀a)(a+b) < (b􀀀a)2b:
Let us use the claim by plugging in a = s and b = s+h. We obtain
(s+h)2􀀀s2 < h2(s+h)
< 2h(s+1) (since h < 1)
< 2􀀀s2

since h <
2􀀀s2
2(s+1)

:
This implies that (s+h)2 <2. Hence s+h 2 A but as h>0 we have s+h>s. Hence, s<r =sup A.
As s  1 was an arbitrary number such that s2 < 2, it follows that r2  2.
Now take a number s such that s2 > 2. Hence s2􀀀2 > 0, and as before s2􀀀2
2s > 0. We can choose
an h 2 R such that 0 < h < s2􀀀2
2s and h < s.
Again we use the fact that 0 < a < b implies b2 􀀀a2 < 2(b􀀀a)b. We plug in a = s􀀀h and
b = s (note that s􀀀h > 0). We obtain
s2􀀀(s􀀀h)2 < 2hs
< s2􀀀2

since h <
s2􀀀2
2s

:
By subtracting s2 from both sides and multiplying by 􀀀1, we find (s􀀀h)2 > 2. Therefore s􀀀h =2 A.
Furthermore, if x  s􀀀h, then x2  (s􀀀h)2 > 2 (as x > 0 and s􀀀h > 0) and so x =2 A and so
s􀀀h is an upper bound for A. However, s􀀀h < s, or in other words s > r = sup A. Thus r2  2.
Together, r2  2 and r2  2 imply r2 = 2. The existence part is finished. We still need to handle
uniqueness. Suppose that s 2 R such that s2 = 2 and s > 0. Thus s2 = r2. However, if 0 < s < r,
then s2 < r2. Similarly if 0 < r < s implies r2 < s2. Hence s = r.
The number
p
2 =2 Q. The set RnQ is called the set of irrational numbers. We have seen that
RnQ is nonempty, later on we will see that is it actually very large.
Using the same technique as above, we can show that a positive real number x1=n exists for all
n 2 N and all x > 0. That is, for each x > 0, there exists a positive real number r such that rn = x.
The proof is left as an exercise.
1.2. THE SET OF REAL NUMBERS 27
1.2.2 Archimedean property
As we have seen, in any interval, there are plenty of real numbers. But there are also infinitely many
rational numbers in any interval. The following is one of the most fundamental facts about the real
numbers. The two parts of the next theorem are actually equivalent, even though it may not seem
like that at first sight.
Theorem 1.2.4.
(i) (Archimedean property) If x;y 2 R and x > 0, then there exists an n 2 N such that
nx > y:
(ii) (Q is dense in R) If x;y 2 R and x < y, then there exists an r 2 Q such that x < r < y.
Proof. Let us prove (i). We can divide through by x and then what (i) says is that for any real
number t := y=x, we can find natural number n such that n >t. In other words, (i) says that N  R is
unbounded. Suppose for contradiction that N is bounded. Let b := supN. The number b􀀀1 cannot
possibly be an upper bound for N as it is strictly less than b. Thus there exists an m 2 N such that
m > b􀀀1. We can add one to obtain m+1 > b, which contradicts b being an upper bound.
Now let us tackle (ii). First assume that x  0. Note that y􀀀x > 0. By (i), there exists an n 2 N
such that
n(y􀀀x) > 1:
Also by (i) the set A := fk 2 N j k > nxg is nonempty. By the well ordering property of N, A has a
least element m. As m 2 A, then m > nx. As m is the least element of A, m􀀀1 =2 A. If m > 1, then
m􀀀1 2 N, but m􀀀1 =2 A and so m􀀀1  nx. If m = 1, then m􀀀1 = 0, and m􀀀1  nx still holds
as x  0. In other words,
m􀀀1  nx < m:
We divide through by n to get x < m=n. On the other hand from n(y􀀀x) > 1 we obtain ny > 1+nx.
As nx  m􀀀1 we get that 1+nx  m and hence ny > m and therefore y > m=n.
Now assume that x<0. If y>0, then we can just take r =0. If y<0, then note that 0<􀀀y<􀀀x
and find a rational q such that 􀀀y < q < 􀀀x. Then take r = 􀀀q.
Let us state and prove a simple but useful corollary of the Archimedean property. Other
corollaries are easy consequences and we leave them as exercises.
Corollary 1.2.5. inff1=n j n 2 Ng = 0.
Proof. Let A := f1=n j n 2 Ng. Obviously A is not empty. Furthermore, 1=n > 0 and so 0 is a lower
bound, so b := inf A exists. As 0 is a lower bound, then b  0. If b > 0. By Archimedean property
there exists an n such that nb > 1, or in other words b > 1=n. However 1=n 2 A contradicting the fact
that b is a lower bound. Hence b = 0.