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Introduction to Real Analysis
Patrick Speissegger
McMaster University
Lesson 2
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Patrick Speissegger
Real Analysis: Lesson 2
using examples
Definition
A theorem (proposition, lemma) is a statement p that can
only take the truth value T . The argument that shows that p is a
theorem (proposition, lemma) is called a proof.
Example
Prove that ∃x 3 x 2 6= 9.
Technique 1
A statement of the form ∃xp(x) can be proved by finding a
number a such that p(a) is true.
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Patrick Speissegger
Real Analysis: Lesson 2
using examples
Example
Is ∀x, x 2 6= 9 a theorem?
Definition
To disprove a statement p is to show that p is not a theorem
(proposition, lemma).
Technique 2
To disprove a statement of the form ∀x, p(x), we find a number
a such that ∼ p(a) is true; such an a is called a
counterexample.
Note:
A statement of the form ∀x, p(x) cannot be proved by
establishing the truth of p(a) for some particular number a!
Patrick Speissegger
Real Analysis: Lesson 2
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deductive reasoning
Example
Let
h(n, m) := n2 − n − m.
Prove that ∀n, h(n, n − 1) = (n − 1)2 .
Technique 3
To prove a statement of the form ∀x, p(x), we prove p(x) in
a way that does not depend on x being any particular
number, that is, by only using properties known to be true
for every number x.
Applying a general property/principle to a particular case is
called deductive reasoning.
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Patrick Speissegger
Real Analysis: Lesson 2
hypothesis and conclusion
Definition
Most theorems (propositions, lemmas) are statements of the
form p ⇒ q. We usually call p the hypothesis and q the
conclusion.
Example
The previous example can be restated as a theorem of the form
p ⇒ q: take p to be the statement
∀n∀m, h(n, m) = n2 − n − m
and q to be the statement
∀n, h(n, n − 1) = (n − 1)2 .
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Patrick Speissegger
Real Analysis: Lesson 2
bridge of implications
Heuristic 1
To prove a theorem (proposition, lemma) of the form p ⇒ q, we
often attempt to build a bridge of implications
p ⇒ p1 ⇒ p2 ⇒ · · · ⇒ q2 ⇒ q1 ⇒ q.
To obtain such a bridge, we might try to
use deductive reasoning (”what can we conclude from p?”)
to get p1 from p, then again to get p2 from p1 , and so on;
use ”inverse” deductive reasoning (”what do we need to
know to conclude that q is true?”) to get q1 from q, then q2
from q1 , and so on.
Note:
Heuristics may or may not work...
Patrick Speissegger
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Real Analysis: Lesson 2
bridge of implications
Example
Prove the following theorem: for every > 0 there exists a
δ > 0 such that
1−δ <x <1+δ
implies that
6 − < 5x + 1 < 6 + .
Hint
Fix an arbitrary positive real number (that is, when
working with , we are only allowed to use properties that
are true of all positive real numbers).
Build a bridge of implications starting with the consequent,
by isolating the variable x.
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Patrick Speissegger
Real Analysis: Lesson 2
contrapositive
Example
Let m be an integer. Prove that if 11m is odd, then m is odd.
Technique 3
To prove a statement of the form p ⇒ q, we can prove the
statement ∼ q ⇒∼ p instead.
The statement ∼ q ⇒∼ p is calle the contrapositive of
p ⇒ q.
Technique 3 is justified by:
Proposition
Let p and q be statements. Then the statements p ⇒ q and
∼ q ⇒∼ p are logically equivalent.
Patrick Speissegger
Real Analysis: Lesson 2
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converse and inverse
Definition
The converse of p ⇒ q is the statement q ⇒ p.
The inverse of p ⇒ q is the statement ∼ p ⇒∼ q.
Questions
1
Are p ⇒ q and q ⇒ p logically equivalent?
2
Are p ⇒ q and ∼ p ⇒∼ q logically equivalent?
3
Are q ⇒ p and ∼ p ⇒∼ q logically equivalent?
Note:
Proving the converse or the inverse of p ⇒ q does not prove
p ⇒ q.
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Patrick Speissegger
Real Analysis: Lesson 2
contradiction
Example
Prove that if x is a negative real number, then 1/x < 0.
Technique 4
To prove a theorem of the form p ⇒ q by contradiction, we
assume both p and ∼ q and derive a statement that is known to
be always false (such as 1 = 0).
Proposition
Let c be any statement that is always false. Then for any two
statements p and q, the statement p ⇒ q is logically equivalent
to (p∧ ∼ q) ⇒ c.
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Patrick Speissegger
Real Analysis: Lesson 2
cases
Example
Prove that if x is a real number, then x ≤ |x|.
Technique 5
To prove a theorem of the form (p ∨ q) ⇒ r , we prove both
p ⇒ r and q ⇒ r .
This is called a proof by cases: p and q are each
considered a case of the statement p ∨ q.
Proposition
Let p, q and r be statements. Then (p ∨ q) ⇒ r is logically
equivalent to (p ⇒ r ) ∧ (q ⇒ r ).
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Patrick Speissegger
Real Analysis: Lesson 2