Definitions
Definition describe the objects and notions that is
used.
Precision is essential to any math-definition.
1
Mathematical statements
Mathematical statements expresses that some
object has certain property.
Statements may or may not be true, but it must be
precise.
2
Proof
Proof - is a convincing logical argument that a
statement is true.
3
Argument
Argument- a sequence of statements that end
with a conclusion.
Valid argument – if and only if it is impossible
for all the premises (preceding statements) to be
true and the conclusion (final statement of the
argument) to be false.
4
Rules of inference
Rules of inference - templates for constructing
valid arguments to deduce new statements from
statements already present.
They are basic tools for establishing the truth of
statements.
5
Rules of inference - Modus ponens
Modus ponens tells that if a conditional statement
and the hypothesis of this conditional statement
are both true, then the conclusion must also be
true.
6
Modus ponens – mathematic expression
p (hypothesis )
p→q(conditional statement)
:.q (conclusion)
or
(p /\ (p→q)) → q
Hypothesis
"You have a current password."
Conditional statement
"If you have a current password,
then you can log onto the
network."
Conclusion
Therefore,
"You can log onto the network."
7
Theorem
Theorem is a mathematical statement proved true.
8
Types of proof
1. Proof by construction,
2. Proof by contradiction,
3. Proof by induction.
9
Proof by construction
If a theorem states that a particular object exists,
this technique demonstrates how to construct the
object.
10
Example: Prove that for any integer a, b, if a
and b are odd, then ab is odd
Proof:
• An integer n is odd if there exists an integer x
such that n=2x+1
• Thus exists integers x and y for which a=2x+1
and b=2y+1. Show there is an integer z so that
ab=2z+1.
• Calculate ab=(2x+1)(2y+1)=2(2xy+x+y)+1. So
there is a z=2xy+x+y so that ab=2z+1
11
Proof by contradiction
It assumes that the theorem is false and then show
that this assumption leads to an obviously false
consequence called contradiction
12
Example
Jack sees Jill, who has just come in from outdoors. On
observing that she is completely dry, he knows that it is not
raining.
• His "proof" that it is not raining is that, if it were raining (the
assumption that the statement is false), Jill would be wet (the
obviously false consequence). Therefore it must not be
raining.
13
Prove that a statement p is true by
contradiction
Find a contradiction q such that ¬p→q is true.
• Since q is false, but ¬p→q is true, conclude that ¬p is
false, which means that p is true.
Because the statement r /\ ¬r is a contradiction
whenever r is a proposition, it can be proved that p is
true if it can be shown that ¬p→(r /\ ¬r) is true for
some proposition r.
14
Example
Prove: 2 is irrational
Proof:
Suppose 2 is rational - a contradiction. Then there are integers m’ and
n’ such that 2=m’/n’. By dividing both integers m’ and n’ by common
factor, obtain 2=m/n for some integers m and n having no common
factors. Squaring both side of equation we get m²=2n², so m² is even
and m is even, means for some k, m=2k. Therefore (2k) ²=2n². or
2k²=n². So n² or n is also even, so that n=2j for some j. We have shown
that both m and n are divisible by 2 (even numbers). This contradicts the
statement that m and n have no common factor. The assumption that 2
is rational therefore leads to contradiction and the conclusion that 2 is
irrational.
15
Proof by induction
Proof by induction is used to show that all
elements of an infinite set have a specified
property P.
16
Proof by induction - Idea
1. Proving the base of induction,
2. Forming the induction hypothesis, and
3. Proving that the induction hypothesis holds true
for all numbers in the domain.
• Induction hypothesis- the assumption in
induction step that P(i) is true
17
Proof by induction - Steps
1. The Basis step- it proves that P(1) is true.
2. The induction step- it proves that for each i ≥1,
assume that P(i) is true, and
3. Use this assumption to show that P(i+1) is true.
18
Example
Prove 1+2+...+n=n(n+1)/2
• n=1: 1=1(2)/2=1 checks.
• Assume n=k holds: 1+2+...+k=k(k+1)/2 (Induction Hypothesis)
• Show n=k+1 holds: 1+2+...+k+(k+1)=(k+1)((k+1)+1)/2
LHS: 1+2+...+(k+1)=1+2+...+k+(k+1)
=k(k+1)/2 + (k+1) by the Induction Hypothesis
=(k(k+1)+2(k+1))/2 by 2/2=1 and distribution of division over
addition
=(k+2)(k+1)/2 by distribution of multiplication over addition
=(k+1)((k+1)+1)/2 by commutativity of multiplication
19