Download Lecture 6: real numbers One extremely useful property of R that

Lecture 6: real numbers
One extremely useful property of R that follows from the least upper bound property is
Theorem 0.1 (Archimedean property of R). Given x, y ∈ R with x 6= 0, there exists n ∈ Z
such that
nx > y .
Proof. First let x, y ∈ R such that x, y > 0 and assume that there is no such n. Then the
set
{nx : n ∈ N}
is bounded above by y. As it is clearly nonempty, it has a supremum s. Then s − x < s, so
s − x cannot be an upper bound, giving the existence of some m ∈ N such that
s − x < mx .
However this implies that s < (m+1)x, so s was actually not an upper bound, contradiction.
This proves the statement for the case x < y. The other cases can be obtained from this one
by instead considering −x and/or −y.
The Archimedean property implies
Corollary 0.2 (Density of Q in R). Let x, y ∈ R with x < y. There exists q ∈ Q such that
x < q < y.
Proof. Apply the Archimedean property to y − x and 1 to find n ∈ Z such that n(y − x) > 1.
We can also find m1 > nx and m2 > −nx, so
−m2 < nx < m1 .
It follows then that there is an m ∈ Z such that m − 1 ≤ nx < m. Finally,
nx < m ≤ 1 + nx < ny .
Dividing by n we get x < m/n < y.
Now we return to countability.
Theorem 0.3. The set Q is countable, whereas R is uncountable.
Proof. We already know that N × N is countable: this is from setting up the array
(1, 1) (2, 1) (3, 1) · · ·
(1, 2) (2, 2) (3, 2) · · ·
(1, 3) (2, 3) (3, 3) · · ·
···
1
and listing the elements along diagonals. On the other hand, there is an injection
f : Q+ → N × N ,
where Q+ is the set of positive rationals. One such f is given by f (m/n) = (m, n), where
m/n is the “reduced fraction” for the rational, expressed with m, n ∈ N. Therefore Q+ is
countable. Similarly, Q− , the set of negative rationals, is countable. Last, Q = Q+ ∪Q− ∪{0}
is a union of 3 countable sets and is thus countable.
To prove R is uncountable, we will use decimal expansions for real numbers. In other
words, we write
x = .a1 a2 a3 . . .
where ai ∈ {0, . . . , 9} for all i. Since we have not proved anything about decimal expansions,
we are certainly assuming a lot here, but this is how things go. Note that each real number
has at most 2 decimal expansions (for instance, 1/4 = .2500 . . . = .2499 . . .).
Assume that R is countable. Then as there are at most two decimal expansions for each
real number, the set of decimal expansions is countable (check this!) Now write the set of
all expansions in a list:
1 .a0 a1 a2 . . .
2 .b0 b1 b2 . . .
3 .c0 c1 c2 . . .
···
···
We will show that no matter what list we are given (as above), there must be a sequence
that is not in the list. This implies that there can be no such list, and thus R is uncountable.
Consider the diagonal element of the list. That is, we take a0 for the first digit, b1 for
the second, c2 for the third and so on:
.a0 b1 c2 d3 . . .
We now have a rule to transform this diagonal element into a new one. We can use many,
but here is one: change each digit to a 0 if it is not 0, and replace it with 9 if it is 0. For
example,
.0119020 . . . −→ .9000909 . . .
Note that this procedure changes the diagonal number into a new one that differs from the
diagonal element in every decimal place. Call this new expansion A = .â0 â1 . . .
Now our original list contains all expansions, so it must contain A at some point; let us
say that the n-th element of the list is A. Then consider the n-th digit ân of A. On the
one hand, by construction, ân is not equal to the n-th digit of the diagonal element. On the
other hand, by the position in the list, ân equals the n-th digit of the diagonal element. This
is a contradiction.
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Rn for n ≥ 2
A very important extension of R is given by n-dimensional Euclidean space.
Definition 0.4. For n ≥ 2, the set Rn is defined as
Rn = {~a = (a1 , . . . , an ) : ai ∈ R for all i} .
Addition of elements is defined as
~a + ~b = (a1 , . . . , an ) + (b1 , . . . , bn ) = (a1 + b1 , . . . , an + bn )
and multiplication of elements by numbers is
c~a = c(a1 , . . . , an ) = (ca1 , . . . , can ), c ∈ R .
Note that this definition gives us R for n = 1.
On Rn we place a distance, but to do that, we need the existence of square roots. We
will take this for granted now, since we will prove it later using continuity.
Lemma 0.5. For each √
x ∈ R with x ≥ 0 there exists a unique y ∈ R such that y 2 = x. This
element is written y = x.
Definition 0.6. On the set Rn we define the norm
q
|~a| = |(a1 , . . . , an )| = a21 + · · · + a2n
and inner product
~a · ~b = (a1 , . . . , an ) · (b1 , . . . , bn ) = a1 b1 + · · · + an bn .
Theorem 0.7. Suppose ~a, ~b, ~c ∈ Rn and c ∈ R. Then
1. |~a| ≥ 0 with |~a| = 0 if and only if ~a = 0.
2. |c~a| = |c||~a|.
3. (Cauchy-Schwarz inequality) |~a · ~b| ≤ |~a||~b|.
4. (Triangle inequality) |~a + ~b| ≤ |~a| + |~b|.
5. |~a − ~b| ≤ |~a − ~c| + |~c − ~b|.
Proof. Next time.
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