Download Real Analysis Lecture 2

Real Analysis Lecture 2
Maksim Maydanskiy
Fall 2012
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Rationals numbers.
The more you want to do, the bigger your number system has to be. Adding and
multiplying can be done with natural numbers. If you want to be able to subtract
any two natural numbers you need to introduce zero and negative integers. If you
also want to be able to divide by any non-zero number, you expand the number
system to rational numbers. If you want to do calculus you need real numbers.
We will go over these number systems in turn. One does not have to stop there if you want to be able to solve any equation you may continue to expand to get
complex numbers, and even go beyond that. We shall be content with the reals,
however.
1.1
Integers.
I assume you are familiar with those. The notation is Z = {0, 1, −1, 2, −2, . . .}.
1.2
Rational Numbers.
with n 6= 0. This inThe rational numbers Q are all numbers of the form m
n
cludes all decimals. The ancient Greeks were shocked to discovered that not all
numbers are rational. For them numbers were associated to lengths. After they
co-discovered the Pythagorean theorem∗ and found that the length of the diagonal
of the unit square can not be expressed as m
, which we will show, they were quite
n
disturbed. The modern framework for dealing with irrational numbers had to wait
untill 19th century.
∗
The Pythagoren theorem has been used by Babylonians and Indians before it was found by
the greeks.
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√
Exercise 1.1.
Prove
2 is irrational the classic (and I do mean classic) way:
m 2
Suppose n = 2, where m
is the fraction in lowest terms (that is, m and n have
n
no common factors, n > 0). Conclude that m is even, m = 2k, then that n is
even, and arrive at a contradiction.
√
√
√
This works, but is not easy to generalize. Is 3 18 irrational? Is 2 + 3? A
powerful tool to answer these sorts of questions is the Rational Zeros Theorem.
Theorem 1.2. Suppose a rational number r =
a polynomial equation
p
q
(in lowest terms) is a solution of
an xn + an−1 xn−1 + . . . + a1 x + a0 = 0
where ai are integers with an 6= 0, a0 6= 0 and n ≥ 1. Then q divides an and p
divides a0 .
This means that to see whether a given polynomial equation does not have a
rational root we only have to check a finite set of candidates.
√
Example 1.3. 2 represents a root of x2 − 2 = 0. But by the the Rational Zeros
Theorem only possible candidate rational roots are
√±1, ±2, neither of which is
actually a root. So there are no rational roots, and 2 is irrational.
Not only is this arguably easier, it is also more algorithmic, meaning we can
repeat this argument for other numbers.
√
Exercise 1.4. Prove 3 18 is irrational. Hint: If you do not want to check every
divisor, you can use 23 < 18 < 33 .
Or even abstract away:
√
Exercise 1.5. Prove that for any natural number n the number n is either an
integer or is irrational.
√
√
Exercise 1.6. Prove x = 3 + 2 is irrational. Hint: Find an equation with
integer coefficients that x satisfies.
The numbers in the exercises of this section, while irrational, all satisfy polynomial equations with integer coefficients. Such numbers are called algebraic.
Numbers that are not algebraic are called transcendental. In a precise sense, most
real numbers are transcendental† , but it is very hard to prove that any given number is transcendental - the first example was only constructed by Joseph Liouville
in 1844. Math is full of such strange things. In addition, while the numbers π and
e are known to be transcendental, whether π + e is transcendental is unknown!
†
There are uncountably many transcendental numbers and only countably many algebraic ones.
2
Exercise 1.7. Suppose r 6= 0 is algebraic. Show r−1 is also algebraic.
Remark. In fact, if s and t are algebraic, then so are s + t, and st, but his is not
so easy to prove! You can look up ”resultant” on wikipedia for a glimpse of cool
math that can be used to prove this.
1.3
Properties of rational numbers.
As with N, a fully rigorous development of Q would require the following:
(i) Come up with a set of axioms for Q.
(ii) Construct candidate Q from some previously defined thing. (Note: Q is
usually constructed from the integers, while N is constructed directly from
set theory; set theory is a basic axiomatic foundation of modern mathematics and is not constructed out of anything. )
(iii) Check that the constructed candidate object satisfies the axioms from 1.
(iv) Deduce other properties of Q from the axioms.
(v) (optional) Show that the axioms define Q uniquely in the appropriate sense.
When talking about N we chose to focus on the well-ordering principle and
forego detailed discussion of everything else. For Q we will again suppress most
of the steps, but shall go into more details about 1 and 4.
The axioms are:
A1. Associativity of addition: a + (b + c) = (a + b) + c.
M1. Associativity of multiplicatin: a(bc) = (ab)c.
A2. Commutativity of addition: a + b = b + a.
M2. Commutativity od multiplication: ab = ba.
A3. Additive unit: a + 0 = a for all a.
M3. Multiplicative unit a · 1 = a for all a.
A4. Additive inverse: For each a there is an element −a such that a + (−a) =
0.
M4. Multiplicative inverse: For each a 6= 0 there is an element a−1 such that
aa = 1.
−1
DL Distributive law: a(b + c) = ab + ac.
In a course on algebra one learns that properties A1 through A4 make Q into
a commutative group with respect to addition, and M1-M4 make the non-zero
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rationals into a commutative group with respect to multiplication. The distributive
law binds the two operations together. All together these properties make Q into
a field. Algebraic properties of Q (i.e. those that involve only basic things about
addition multiplication subtraction and division) can be proved directly from these
nine fundamental properties. They are then valid in any field. This is the magic of
abstract algebra.
Example 1.8.
(i) Prove a + c = b + c implies a = b.
(ii) Prove ac = bc and c 6= 0 implies a = b.
(iii) Prove a · 0 = 0 for all a.
(iv) Prove (−a)b = a(−b). Deduce (−1)a = −a.
(v) Prove (−a)(−b) = ab.
(vi) Prove ab = 0 implies a = 0 or b = 0.
Example 1.9. Prove a2 = b2 implies a = b or a = −b.
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