The document discusses propositions, logical operations, predicates, quantification, and mathematical induction. It provides: 1) Definitions of predicates, propositions, universal and existential quantification, and the principle of mathematical induction. 2) Examples of applying predicates, quantification, and induction to prove statements about integers and sums. 3) The process of proving statements by mathematical induction, which involves showing the basis step and inductive step. It also introduces strong mathematical induction.

1. Propositions and Logical Operations
Definition: A predicate or a propositional function is a
noun/verb phrase template that describes a property of
objects, or a relationship among objects represented by
the variables:
Example: 𝑃𝑃 𝑥𝑥 : “𝑥𝑥 is integer less than 8.”
 𝑃𝑃 1 =
 𝑃𝑃 10 =
 𝑃𝑃 −11 =
1© S. Turaev, CSC 1700 Discrete Mathematics

2. Propositions and Logical Operations
Definition: The universal quantification of a predicate
𝑃𝑃 𝑥𝑥 is the statement “For all values of 𝑥𝑥 (for every 𝑥𝑥, for
each 𝑥𝑥, for any 𝑥𝑥), 𝑃𝑃 𝑥𝑥 is true” and is denoted by
∀𝑥𝑥𝑥𝑥 𝑥𝑥 .
Example: 𝑃𝑃 𝑥𝑥 : “− −𝑥𝑥 = 𝑥𝑥” is a predicate that is true
for all real numbers.
∀𝑥𝑥𝑥𝑥 𝑥𝑥 =
Example: 𝑄𝑄 𝑥𝑥 : “𝑥𝑥 + 1 < 4”.
∀𝑥𝑥𝑥𝑥 𝑥𝑥 =
2© S. Turaev, CSC 1700 Discrete Mathematics

3. Propositions and Logical Operations
A predicate may contain several variables.
Example: 𝑄𝑄 𝑥𝑥, 𝑦𝑦 : 𝑥𝑥 + 𝑦𝑦 = 𝑦𝑦 + 𝑥𝑥
∀𝑥𝑥∀𝑦𝑦𝑦𝑦 𝑥𝑥, 𝑦𝑦 =
Example: Write the following statement in the form of a
predicate and quantifier:
“The sum of any two integers is even number.”
3© S. Turaev, CSC 1700 Discrete Mathematics

4. Propositions and Logical Operations
Definition: The existential quantification of a predicate
𝑃𝑃 𝑥𝑥 is the statement “There exists a value of 𝑥𝑥, for
which 𝑃𝑃 𝑥𝑥 is true” and is denoted by ∃𝑥𝑥𝑥𝑥 𝑥𝑥 .
Example: 𝑃𝑃 𝑥𝑥 : “−𝑥𝑥 = 𝑥𝑥”.
∃𝑥𝑥𝑥𝑥 𝑥𝑥 =
Example: 𝑄𝑄 𝑥𝑥 : “𝑥𝑥 + 1 < 4”.
∃𝑥𝑥𝑥𝑥 𝑥𝑥 =
4© S. Turaev, CSC 1700 Discrete Mathematics

5. Mathematical Induction

6. Mathematical Induction
Suppose we have an infinite ladder:
1. We can reach the first rung of the
ladder.
2. If we can reach a particular rung of
the ladder, then we can reach the
next rung.
From (1), we can reach the first rung.
Then by applying (2), we can reach the
second rung. Applying (2) again, the
third rung, and so on. We can apply (2)
any # of times to reach any particular
rung, no matter how high up.
© S. Turaev, CSC 1700 Discrete Mathematics 6

7. Mathematical Induction
Principle of Mathematical Induction: To prove that 𝑃𝑃(𝑛𝑛)
is true for all positive integers 𝑛𝑛, we complete these
steps:
BASIS STEP: Show that 𝑃𝑃(𝑛𝑛0) is true.
INDUCTIVE STEP: Show that 𝑃𝑃 𝑘𝑘 → 𝑃𝑃(𝑘𝑘 + 1) is true
for all positive integers 𝑘𝑘 ≥ 𝑛𝑛0.
To complete the inductive step, assuming the inductive
hypothesis that 𝑃𝑃(𝑘𝑘) holds for an arbitrary integer 𝑘𝑘 ≥
𝑛𝑛0, show that must 𝑃𝑃(𝑘𝑘 + 1) be true.
© S. Turaev, CSC 1700 Discrete Mathematics 7

8. Mathematical Induction
Example: Show that
�
𝑖𝑖=1
𝑛𝑛
𝑖𝑖 =
𝑛𝑛(𝑛𝑛 + 1)
2
Solution:
BASIS STEP: 𝑃𝑃 1 is true since
1 1 + 1
2
= 1.
© S. Turaev, CSC 1700 Discrete Mathematics 8

9. Mathematical Induction
INDUCTIVE STEP: Assume true for 𝑃𝑃(𝑘𝑘).
The inductive hypothesis is
�
𝑖𝑖=1
𝑘𝑘
𝑖𝑖 =
𝑘𝑘 𝑘𝑘 + 1
2
1 + 2 + ⋯ + 𝑘𝑘 + 𝑘𝑘 + 1
=
𝑘𝑘 𝑘𝑘 + 1
2
+ 𝑘𝑘 + 1 =
𝑘𝑘 𝑘𝑘 + 1 + 2(𝑘𝑘 + 1)
2
=
(𝑘𝑘 + 1)(𝑘𝑘 + 2)
2
© S. Turaev, CSC 1700 Discrete Mathematics 9

10. Exercises
Prove that
• 12 + 22 + 32 + ⋯ + 𝑛𝑛2 =
𝑛𝑛 𝑛𝑛+1 2𝑛𝑛+1
6
• 12 + 32 + 52 + ⋯ + 2𝑛𝑛 − 1 2 =
𝑛𝑛 2𝑛𝑛+1 2𝑛𝑛−1
3
• 1 + 21 + 22 + ⋯ + 2𝑛𝑛 = 2𝑛𝑛+1 − 1
© S. Turaev, CSC 1700 Discrete Mathematics 10

11. Mathematical Induction
Example: Prove that 𝑛𝑛 < 2𝑛𝑛 for all positive integers 𝑛𝑛.
Example: Prove that 𝑛𝑛! ≥ 2𝑛𝑛−1
for all positive integers 𝑛𝑛.
Example: Prove that 𝑛𝑛3 − 𝑛𝑛 is divisible by 3, for every
positive integer 𝑛𝑛.
Example: Let 𝐴𝐴1, 𝐴𝐴2, 𝐴𝐴3, … , 𝐴𝐴𝑛𝑛 be any sets. Show that
�
𝑖𝑖=1
𝑛𝑛
𝐴𝐴𝑖𝑖 = �
𝑖𝑖=1
𝑛𝑛
𝐴𝐴𝑖𝑖
© S. Turaev, CSC 1700 Discrete Mathematics 11

12. Strong Mathematical Induction
Strong induction tells us that we can
reach all rungs if:
1. We can reach the first rung of
the ladder.
2. For every integer 𝑘𝑘, if we can
reach the first 𝑘𝑘 rungs, then we
can reach the (𝑘𝑘 + 1)st rung.
© S. Turaev, CSC 1700 Discrete Mathematics 12

13. Strong Mathematical Induction
To conclude that we can reach every rung by strong
induction:
BASIS STEP: 𝑃𝑃 1 holds.
INDUCTIVE STEP: Assume 𝑃𝑃 1 ∧ 𝑃𝑃 2 ∧ ⋯ ∧ 𝑃𝑃 𝑘𝑘
holds for an arbitrary integer 𝑘𝑘, and show that
𝑃𝑃 𝑘𝑘 + 1 must also hold.
We will have then shown by strong induction that for
every positive integer 𝑛𝑛, 𝑃𝑃 𝑛𝑛 holds, i.e., we can reach
the 𝑛𝑛th rung of the ladder.
© S. Turaev, CSC 1700 Discrete Mathematics 13

14. Strong Mathematical Induction
Example: Show that if 𝑛𝑛 is an integer greater than 1, then
𝑛𝑛 can be written as the product of primes.
Solution: Let 𝑃𝑃 𝑛𝑛 be the proposition that 𝑛𝑛 can be
written as a product of primes.
BASIS STEP: 𝑃𝑃 2 is true since 2 itself is prime.
INDUCTIVE STEP: The inductive hypothesis is 𝑃𝑃 𝑖𝑖 is
true for all integers 𝑖𝑖 with 2 ≤ 𝑖𝑖 ≤ 𝑘𝑘. To show that
𝑃𝑃 𝑘𝑘 + 1 must be true under this assumption, two
cases need to be considered:
© S. Turaev, CSC 1700 Discrete Mathematics 14

15. Strong Mathematical Induction
INDUCTIVE STEP: The inductive hypothesis is 𝑃𝑃 𝑖𝑖 is
true for all integers 𝑖𝑖 with 2 ≤ 𝑖𝑖 ≤ 𝑘𝑘. To show that
𝑃𝑃 𝑘𝑘 + 1 must be true under this assumption, two
cases need to be considered:
1. If 𝑘𝑘 + 1 is prime, then 𝑃𝑃 𝑘𝑘 + 1 is true.
2. Otherwise, 𝑘𝑘 + 1 is composite and can be written
as the product of two positive integers 𝑎𝑎 and 𝑏𝑏
with 2 ≤ 𝑎𝑎 ≤ 𝑏𝑏 < 𝑘𝑘 + 1. By the inductive
hypothesis 𝑎𝑎 and 𝑏𝑏 can be written as the product
of primes and therefore 𝑘𝑘 + 1 can also be written
as the product of those primes.
© S. Turaev, CSC 1700 Discrete Mathematics 15