Download Lecture 2 - Thursday June 30th

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Lecture 2 - Thursday June 30th
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Key words: Rational and irrational numbers, inductive sets, mathematical
induction, archimedean property, dense, triangle inequality, arithmetic mean
geometric mean inequality, geometric series, arithmetic series, difference of
powers formula, factorials, binomial coefficients, binomial series
Key concepts: Proofs from the axioms, proof by induction
2.1
Irrational numbers
A rational number is a real number x of the form m/n where m and n are integers. Recall
Q is the set of rational numbers. A real number is called irrational if it is not rational i.e.
it is not of the form m/n for any integers m, n. So the set of irrational numbers is exactly
R\Q. Do we know that any irrational numbers exist? The answer is yes, and this can be
proved as follows.
Theorem 1 The positive real number x such that x2 = 2 is irrational.
For if x = m/n for integers m, n, then we can assume m and n have no common factors.
Then x2 = m2 /n2 = 2 and so m2 = 2n2 . This means m is an even number, so m = 2k for
some k, but then 2n2 = (2k)2 and n2 = 2k 2 , so n is even too. That contradicts that m, n
have no common factor.
2.2
Inductive sets
A set S ⊂ R is inductive if 1 ∈ S and whenever s ∈ S we have s + 1 in S. The set
of natural numbers (positive integers) N = {1, 2, . . . } is clearly inductive. The rational
numbers Q, the integers Z = {. . . , −2, −1, 0, 1, 2, . . . } and the real numbers R are also
examples of inductive sets.
Theorem 2 The intersection of all inductive sets is N.
The proof involves showing that if x ∈ N, then x is contained in every inductive set. This
leads to the principle of mathematical induction.
Mathematical Induction.
For each k ∈ N, let P (k) be a logical statement. Suppose that P (1) is
true and for every natural number k, if P (k) is true then P (k + 1) is
true. Then P (k) is true for every natural number k.
This is true because A = {k ∈ N|P (k) is true} is an inductive set which is contained in
N, so by the preceding theorem, this means A = N.
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2.3
Examples.
As an example of induction, we can show 1 + 2 + · · · + k = k(k + 1)/2 for all k ∈ N.
Let P (k) be this statement. Then P (1) is the statement 1 = 1(1 + 1)/2 which is true.
If P (k) is true, can we show P (k + 1) is also true? Yes, since P (k + 1) is the statement
1 + 2 + · · · + (k + 1) = (k + 1)(k + 2)/2, and the first k terms on the left add up to
k(k + 1)/2 since P (k) is true. So we need to show
k(k + 1)/2 + (k + 1) = (k + 1)(k + 2)/2
and multiplying each side out confirms that this is true. Therefore P (k) being true implies
P (k + 1) is true. By the principle of induction, that means P (k) is true for all k ∈ N.
As a second example, we can show 1 + 14 + · · · + k12 ≤ 2 − k1 for k ∈ N. Let P (k) be this
statement. Then P (1) is the statement 1 ≤ 2 − 1/1 which is clearly true. Suppose P (k)
is true. Then we have to show P (k + 1) is true, which is
1+
1
1
1
+ ··· +
≤2−
.
2
4
(k + 1)
k+1
The first k terms are at most 2 − 1/k since P (k) is true. So it is enough to show
2−
This is the same as showing
1
1
1
+
≤2−
.
2
k (k + 1)
k+1
1
1
1
−
≥
.
k k+1
(k + 1)2
1
1
2
which is clearly larger than (k+1)
The left side is k(k+1)
2 because k(k + 1) < (k + 1) . So
we have shown P (k + 1) is true. By the principle of induction, P (k) is true for all k.
2.4
Distribution of rational numbers
The rational numbers are dense in R. This means that given any a, b ∈ R with a < b,
there is x ∈ Q such that a < x < b – so between any two real numbers lies a rational
number. We will prove this theorem in this section.
Theorem 3 (Archimedean Property) For any c ∈ R there is n ∈ N with n > c.
Proof. The proof is simple: if the statement is false, then c is an upper bound for N. By
the completeness axiom σ = supN exists. Since σ − 12 is not an upper bound for N, there
exists n ∈ N with n > σ − 21 . However n + 1 ∈ N and n + 1 > σ which contradicts that σ
is an upper bound for N. ¤
A maximum element of a set of integers S is an x ∈ S such that s ≤ x for all x ∈ S. In
other words, it is an upper bound for S which is contained in S.
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Proposition 4 The following are true:
(1) There is no integer in the interval (n, n + 1) when n ∈ Z.
(2) Any nonempty set of integers that has an upper bound has a maximum
element.
(3) There is exactly one integer in each interval [x, x + 1) for x ∈ R.
Sketch Proof. That (1) is true follows from the fact that (1) holds for n = 0, since
{k ∈ N|k ≥ 1} is an inductive set, so it equals N which shows k ≥ 1 for all k ∈ N.
Therefore since every integer x > 0 is a natural number and we just showed x ≥ 1,
(0, 1) contains no integer. Now we consider (n, n + 1), and apply the above proof to
{k ∈ N|k − n ≥ 1}. If k ∈ (n, n + 1) is an integer, then n < k < n + 1 and so k − n is
an integer and 0 < k − n < 1 and so k − n ∈ (0, 1), which is a contradiction to what we
first proved. To prove (2), let σ = sup(S), which exists by the completeness axiom. Then
σ − 1 is not an upper bound for S, so there is an integer n > σ − 1 in S. In other words,
σ < n + 1 and so every element of S is less than n + 1. By (1) there is no element of S in
(n, n + 1), and so n is the maximum element of S and σ = n. Finally, (3) will be obtained
from the archimedean property. If S = {n ∈ Z|n < x + 1}, then S is nonempty: if x ≥ 0
then 1 ∈ S, if x < 0 then by the archimedean property there is n with n > −x and
therefore −n < x < x + 1, so −n ∈ S. Also x + 1 is an upper bound for S, so (2) applies
and there exists a maximum element m ∈ S. If m < x, then m + 1 ≥ x + 1 since m is
the largest element of S. This is impossible, so we conclude m ≥ x and so m ∈ [x, x + 1).
Finally there cannot be another integer n ∈ [x, x + 1), otherwise n < m and it is not hard
to see m − n ∈ (0, 1), contradicting (1). ¤
The main proof of this section is that the rationals are dense in R.
Theorem 5 Q is dense in R.
Sketch Proof. One has to show for any interval (a, b) there is x ∈ (a, b) such that
1
. This means n1 < b − a.
x ∈ Q. By the archimedean property choose integer n > b−a
Then in [nb − 1, nb) there is an integer m. So in other words nb − 1 ≤ m < nb. Now
1
b − 1/n ≤ m
< b. Since n > b−a
, b − 1/n > a and so a < m
< b as required. From this
n
n
one shows easily that the irrationals are dense
an irrational number in (a, b),
√ too.
√ To find √
note that if p/q is a rational number in (a/ 2, b/ 2) then 2p/q is an irrational in (a, b)
(check that the product of a rational and an irrational number is irrational). ¤
2.5
Inequalities
We define the absolute value of a real number x to be −x if x < 0 and x otherwise. From
the positivity axioms, |x| ≤ c is equivalent to −c ≤ x ≤ x and −|x| ≤ x ≤ |x| for any
x ∈ R.
Theorem 6 (Triangle inequality) For x, y ∈ R, |x + y| ≤ |x| + |y|.
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Proof. To see this, start with −|x| ≤ x ≤ |x|. Therefore −|x| − |y| ≤ x + y ≤ |x| + |y|.
This is exactly |x + y| ≤ |x| + |y|. It could also be proved by noting that the positive
square root of x2 is |x|, so since
(x + y)2 ≤ x2 + y 2 + 2|x||y| = (|x| + |y|)2
we get |x + y| ≤ ||x| + |y|| = |x| + |y| taking positive square roots. This relies on x2 ≤ y 2
implying |x| ≤ |y|. ¤
Another useful inequality is
Theorem 7 For x, y ∈ R, |x − y| ≥ ||x| − |y||.
Proof. Since (x − y)2 = x2 − 2xy + y 2 ≥ x2 − 2|x||y| + y 2 = (|x| − |y|)2 we are done by
taking positive square roots. ¤
Another useful inequality is
Theorem 8 For positive real x and k ∈ N, (1 + x)k ≥ 1 + k|x|.
Proof. This is easily proved by induction on k. Clearly P (1) is true. If P (k) is true, then
P (k+1) is (1+|x|)k+1 ≥ 1+(k+1)|x|. It is enough to show (1+|x|)(1+k|x|) ≥ 1+(k+1)|x|,
which is easy. ¤
A final inequality with absolute values is the arithmetic-geometric mean inequality:
Theorem 9 (Arithmetic-Geometric Mean Inequality). For any positive real numbers x1 , x2 , . . . , xk ,
x1 + x2 + · · · + xk
≥ (x1 x2 . . . xk )1/k .
k
Proof. We prove this by induction on k. Let ak denote the arithmetic mean and gk the
geometric mean of x1 , x2 , . . . , xk , for convenience. Clearly P (1) is true, since a1 = x1 and
g1 = x1 and so a1 ≥ g1 . Suppose P (k − 1) is true, i.e. ak−1 ≥ gk−1 . We prove P (k). This
is the same as showing
(x1 + · · · + xk )k ≥ k k x1 x2 . . . xk .
We may assume xk ≥ xk−1 ≥ · · · ≥ x1 . Now since P (k − 1) is true
µ
¶
xk − gk−1
x1 + x2 + · · · + xk ≥ (k − 1)gk−1 + xk = kgk−1 1 +
= kgk−1 (1 + y).
kgk−1
Since xk ≥ ak−1 , because xk is the largest, P (k − 1) shows xk ≥ gk−1 and so y ≥ 0. By
the last theorem, and since gk−1 (1 + ky) = xk ,
k−1
k−1
xk = k k x1 x2 . . . x k .
gk−1 (1 + ky) = k k gk−1
(x1 + · · · + xk )k ≥ (kgk−1 )k (1 + y)k ≥ k k gk−1
This completes the proof. ¤
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