Download Real Numbers - Will Rosenbaum

Real Numbers
Will Rosenbaum
Updated: October 7, 2014
Department of Mathematics
University of California, Los Angeles
In this essay, we characterize of the real numbers, R. We describe their algebraic properties—
that of a field—which determine arithmetic in R. In addition to the algebraic structure of
a field, R is ordered: given any two numbers x, y ∈ R, we can compare x < y, y < x, or
x = y in a way which is consistent with arithmetic. Further, the real numbers are complete.
Intuitively, completeness formalizes the idea that R contains no “holes” with respect to its
ordering.∗ We describe the completeness of R in terms of the least upper bound princi- ∗ To contrast, the rational
numbers Q are also and
ple. Finally, we describe an important relation between the natural numbers N and the real
ordered field, but Q is
numbers R: the Archimedean property.
not complete.
√ For
With our characterization of R complete, we turn our attention to sequences of real
/Q
example, 2 ∈
represents a “hole” in Q.
numbers. We define what it means for a sequence to converge to a limit, and describe some
basic properties of convergent sequences.
1 Properties of Real Numbers
1.1 Algebraic properties The real numbers R have the algebraic structure of a field. That is, R
is a collection of elements (denoted by x, y, z, etc.) along with two binary operations “+”
and “·” which satisfy
Closure Axioms :
Additive Closure (∀x, y ∈ R)[x + y ∈ R]
Multiplicative Closure (∀x, y ∈ R)[x · y ∈ R]
Additive axioms :
Commutativity (∀x, y)[x + y = y + x]
Associativity (∀x, y)[x + (y + z) = (x + y) + z]
Identity (∃0)(∀x)[x + 0 = x]
Inverses (∀x)(∃y)[x + y = 0] (We usually denote this y = −x.)
Multiplicative axioms :
Commutativity (∀x, y)[x · y = y · x]
Associativity (∀x, y)[x · (y · z) = (x · y) · z]
Identity (∃1)(∀x)[1 · x = x]
Inverses (∀x)(∃y)[x 6= 0 =⇒ x · y = 1] (We usually denote this y = x−1 or
y = x1 .)
Distributivity axiom :
(∀x, y, z)[x · (y + z) = x · y + x · z]
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The list of properties described above are called the field axioms, and any set with binary
operations + and · satisfying all of these properties is called a field. The rational numbers
Q and complex numbers C are also fields. The natural numbers Z = 0, ±1, ±2, . . . are not
a field because multiplicative inverses need not exist. For example, 2−1 = 21 ∈
/ Z.
Note that in the field axioms, we do not explicitly state basic facts like the uniqueness of
0 and 1, or that 0x = 0 for all x. Nonetheless, we show that these are consequences of the
field axioms.
Proposition 1. For every x ∈ R, 0x = 0.
Proof. The “trick” is to observe that by the additive identity property, 0 = 0 + 0. So we
compute
0x = (0 + 0)x
=⇒ 0x = 0x + 0x
(additive identity)
(distributive property)
=⇒ 0x + (−0x) = 0x + 0x + (−0x)
=⇒ 0 = 0x + 0
=⇒ 0 = 0x.
(additive inverses)
(additive identity)
This gives the desired result.
Notice that in the above proof, we explicitly state each axiom we apply in the algebraic
manipulations. This “two column” proof approach is a good habit to maintain, especially
when you are just beginning to write formal proofs. The proof above only references the
field axioms, hence the proof is valid for every field, not just the real numbers. Here is
another very useful fact.
Proposition 2. In R, 0 is the unique element satisfying 0 + x = x for all x ∈ F.
Proof. Suppose 0′ is an element which also satisfies 0′ + x = x for all x. Then we have
0′ = 0 + 0′
= 0′ + 0
(additive identity)
(commutativity of addition)
=0
(by hypothesis).
Putting everything together, we get 0 = 0′ . Thus the additive inverse in a field is unique.
Exercise 3. Prove that 1 is unique. Specifically, if 1′ is another element satisfying 1′ · x = x
for all x, then 1′ = 1.
Exercise 4. Prove that additive inverses and multiplicative inverses are unique.
Exercise 5. Prove that (−1) · (−1) = 1.
1.2 Ordering In addition to satisfying the field axioms, the real numbers are ordered—there
exists a relation, <, on R satisfying
Translation invariance (∀x, y, z)[x < y =⇒ x + z < y + z]
Transitivity (∀x, y, z)[(x < y) ∧ (y < z) =⇒ x < z]
Trichotomy (∀x, y)[(x < y) ∨ (y < x) ∨ (x = y)]
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Real Numbers
Scaling (∀x, yz)[(x < y) ∧ (z > 0) =⇒ zx < zy]
These properties are called the order axioms. Any field which also has an order relation
satisfying the order axioms is called an ordered field. The rational numbers Q are an ordered
field (as is R), but the complex numbers C are not an ordered field.
Problem 6. Prove that C is not an ordered field. Specifically, there is no order relation < on
C which satisfies all of the order axioms.
The ordering of R is what allows us to visualize R as the “real number line.” Smaller
numbers are “to the left” while larger numbers are “to the right.”
1.3 Archimedean property The Archimedean property describes how the natural numbers N
fit into the real numbers, R. We identify the natural numbers
N = {0, 1, 2, . . .} ⊆ R
as being embedded in the real numbers. Informally, the Archimedean property says that
any positive real number can be scaled by a natural number to be arbitrarily large.
Archimedean property For every x, y ∈ R with x, y > 0, there exists a natural number
n ∈ N ⊆ R such that y < nx. Symbolically,
(∀x, y ∈ R)(∃n ∈ N)[y < nx].
For example, taking x = 1, we find that for any positive real number y, there is a natural
number n larger than y. We will use the Archimedean property of R frequently in proofs
of limits in the sequel.
1.4 Completeness From the perspective of calculus and analysis, the completeness property of
the real numbers is what makes everything work. Unfortunately, it is also the most technical to describe. There are several equivalent notions of completeness of R. We choose to
describe completeness in terms of the least upper bound principle.
We say that a subset A ⊆ R is bounded from above if there exists a number s ∈ R such
that for all a ∈ A, a ≤ s:
(∃s ∈ R)(∀a ∈ A)[a ≤ s].
In this case, we call s an upper bound for A. If A is bounded from above, we call s a least
upper bound for A, and write s = sup A (for supremum) if
1. s is and upper bound for A, and
2. for all t < s, t is not an upper bound for A.
Specifically, we have
s = sup A ⇐⇒ ((∀a ∈ A)[a ≤ s]) ∧ ((∀t ∈ R)[t < s =⇒ (∃a ∈ A)[t < a]]) .
We can now state the least upper bound principle for R:
Least upper bound principle (LUB) Every bounded subset A ⊂ R has a least upper bound.
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Real Numbers
o
n
1
1 2
Example 7. Let A = 1+x
2 x > 0 . Since x > 0 for all x, 1+x2 ≤ 1 for all x. Hence 1 is
an upper bound of A. We will show that, in fact, sup A = 1. To this end, suppose t < 1. We
must show that t is not an upper bound for A—that is, there exists some a ∈ A with t < a.
1
To find such an a, write a = 1+x
2 (we can do this because of how A is defined). Then
t<
Since t < 1,
q
1
t
1
1
⇐⇒ 1 + x2 <
2
1+x
t
r
1
⇐⇒ x <
− 1.
t
− 1 is well defined, so we may take, say, x =
a=
1
2
q
1
t
− 1. Then
1
1
=
> t.
1 + x2
1 + (1/t − 1)/4
Therefore, t is not an upper bound for A, hence 1 = sup A.
Example 8. Consider the set A = 1 − n1 n ∈ N, n > 0 . Again, 1 is an upper bound for
A, because n1 > 0 for all n > 0. To show that 1 is a least upper bound, we appeal to the
Archimedean property. Suppose t < 1. We will find an a ∈ A with t < a. Write a = 1 − n1 .
t≤1−
1
1
1
⇐⇒
< 1 − t ⇐⇒
< n.
n
n
1−t
1
> 0. By the Archimedean property, there exists n >
Since 1 − t > 0, 1−t
a = 1 − 1/n > t, so that t is not a lower bound.
1
1−t .
Thus
Remark 9. In general, the existence of a least upper bound for A depends not only on the
set A, but on the ambient set U . For example, we can consider the set
A = a ∈ Q a2 < 2 .
√
√
Viewing A as a subset of R, sup A = 2. However, as 2 ∈
/ Q, sup A does not exist if we
view A as a subset of Q.
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Sequences and Convergence
In this section, we describe convergent sequences in R. The notion of convergence is a
fundamental concept in mathematical analysis.
2.1 Convergent sequences A sequence (an ) in R is a set of numbers a0 , a1 , a2 , . . . ∈ R indexed by the natural numbers. Equivalently, we may view a sequence as a function from N
to R where n 7→ an . Informally, a sequence is convergent if it approaches some limit value
as n increases.
Definition 10. Let (an ) be a sequence in R. We say that (an ) converges to a limit a and
write
lim an = a
n→∞
if for every ε > 0, there exists N ∈ N such that for all n > N , |an − a| < ε. Symbolically,
lim an = a ⇐⇒ (∀ε > 0)(∃N ∈ N)(∀n > N ) [|an − a| < ε]
n→∞
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Real Numbers
One can think of this of the definition of limit as a game between a prover and an adversary. The prover wants to convince the adversary that (an ) is eventually—for sufficiently
large values of n—very close to the limit value of a. The parameter ε determines how close
an must be to a. The adversary makes the prover’s job difficult by choosing a small ε. In
order for the prover to convince the adversary that limn→∞ an = a, no matter what ε the
adversary chooses, the prover must produce a value of N ∈ N (depending on ε) such that
|an − a| < ε for all n > N .
Example 11. Consider the sequence (an ) given by an =
1
n
for n ≥ 1. We will show that
lim an = 0.
n→∞
Given ε > 0, we must find N ∈ N such that n > N implies |an − 0| = n1 < ε. To this end,
we employ the Archimedean property. By the Archimedean property, for any ε > 0, there
exists N ∈ N with N > ε−1 . For this choice of N , we have for all n > N
1
1
<
< ε.
n
N
|an − 0| =
Therefore, limn→∞ an = 0, as desired.
Example 12. Consider the sequence given by
an =
3n2 + 1
.
4n2 + 1
We will show that limn→∞ an = 34 . First, we do a bit of scratch work, computing an − 34 :
2
1
an − 3 = 3n + 1 − 3 =
.
2
4
4n + 1 4 16n2 + 4
We must show that for any ε > 0, we can find N such that n > N implies (16n2 +4)−1 < ε.
Note that
1
< ε ⇐⇒ 16n2 + 4 > ε−1
16n2 + 4
1 p −1
⇐⇒ n >
ε − 4.
4
√
By the Archimedean property, we can find N ∈ N with N > 14 ε−1 − 1. Using this choice
of N , for n > N , we obtain
1
an − 3 =
< ε,
4 16n2 + 4
as desired.
2.2 Properties of convergent sequences We list some basic and useful properties of convergent
sequences.
Proposition 13. Suppose (an ) and (bn ) are convergent sequences with limn→∞ an = a and
limn→∞ bn = b, and suppose c ∈ R. Then
(a) limn→∞ can = ca
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(b) limn→∞ (an + bn ) = a + b
(c) limn→∞ an bn = ab
(d) limn→∞
1
an
=
1
a
so long as for all n, an 6= 0, and a 6= 0
Proof. We leave the proofs of (a) and (b) as exercises for the reader. For (c), we need to show
that |an bn − ab| is small for sufficiently large n. To this end, we employ the following trick:
note that
|an bn − ab| = |an − a| |bn − b| + |a| |bn − b| + |b| |an − a| .
((∗))
Since we know that limn→∞ an = a and limn→∞ bn = b, for large n, |an − a| and |bn − b|
are small. Therefore, we expect all three terms in on the right side of (∗) to be small.
Given ε > 0, since limn→∞ an = n, we can find N1 such that for all n > N1
|an − a| < min {1, ε/3, ε/3 |b|} .
Similarly, we can find N2 such that for all n > N2
|bn − b| < min {1, ε/3, ε/3 |a|} .
Choosing N = max N1 , N2 , we then have
|an bn − ab| = |an − a| |bn − b| + |a| |bn − b| + |b| |an − a|
ε
ε
+ |b|
< (1)(ε/3) + |a|
3 |a|
3 |b|
=ε
Therefore limn→∞ an bn = ab as desired.
For part (d), first note that
1
1 |a − an |
−
=
an
a
|an a|
Given ε > 0, find N such that for n > N ,
|an − a| < min
Then for n > N we have
1
1 |a − an |
an − a = |an a|
2 |a − an |
<
2
|a|
2
<
2 |a| ε
|a|2 2
= ε.
6
1 2 |a|
|a| ε,
2
2
.
(|an | > |a| /2)
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Example 14. Using these properties of convergent sequences, we can greatly simplify limit
proofs. Indeed, Proposition 13 justifies the following manipulations, giving an easier solution for Example 12
3+
3n2 + 1
= lim
n→∞ 4n2 + 1
n→∞ 4 +
lim
1
n2
1
n2
limn→∞ 3 +
limn→∞ 4 +
3+0
.
=
4+0
=
1
n2 1
n2
Definition 15. A sequence (an ) in R is called monotonically increasing if for all n ∈ N,
an ≤ an+1 . Similarly, (an ) is monotonically decreasing if an ≥ an+1 . The sequence (an )
is called bounded from above if there exists a number A ∈ R such that for all n, an ≤ A,
and bounded from below if there exists B ∈ R with B ≤ an for all n.
Proposition 16 (Bounded monotonic sequences are convergent). Let (an ) be a sequence in
R. Then
(a) If (an ) is monotonically increasing and bounded from above, then (an ) is convergent.
(b) If (an ) is monotonically decreasing and bounded from below, then (an ) is convergent.
We leave the proof of this important proposition as an exercise to the reader.
Problem 17. Define the sequence (an ) by a0 = 1 and
√
an+1 = 2 + an for n ≥ 1.
Use Proposition 16 to show that (an ) is convergent.
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