2. 2 Steps:
• Step 1. Show it is true for the
first one
• Step 2. Show that if any one is
true then the next one is true
Then all are true
Mathematical Induction
Mathematical induction is a special way of proving
things.
3. Have you heard of
the “Domino Effect”?
• Step 1. The first domino
falls
• Step 2. When any
domino falls, the next
domino falls
So … all dominos will fall!
4. Induction Axiom
Let S be a subset of satisfying:
(i) 1 ∈ S, and
(ii) If k ∈ S, then also (k + 1) ∈ S.
Then S =.
This axioms leads
immediately to the
method of mathematical
induction.
5. Let P(n) be a statement concerning the natural integer n.
Suppose that:
1. P(1) is true, and
2. If P(k) is true, then P(k+1) is true.
Then, P(n) is true for all n ∈
Induction Axiom
Theorem: The Principle of Mathematical
Induction
Suppose the following three conditions are satisfied with
regard to a statement about natural numbers:
6. Theorem: The Principle of Mathematical Induction
CONDITION I: (Verification) The statement is true for the natural
number 1.
CONDITION II: (Induction Hypothesis) Assume the statement to be
true
for some natural number .
CONDITION III: (Proof of Induction) Show
that the statement is also true for the next
natural number + 1.
Then the statement is true for all natural
numbers.
8. CONDITION II: Induction
hypothesis. Assume true
n = k
Assume 1 + 2 + 3 + … +
is true.
CONDITION III: Proof of Induction: Show
true for n = k + 1.
Show 1+2+3+…+k+(k+1) =
Proof: 1+2+3+…+k+(k+1) = + (k+1)
𝑘 (𝑘 +1)
2 = +
=
=
Since all the conditions are true, then
1 + 2 + 3 + … + n is true for all natural
numbers n.
9. 2
Show that 1² + 2² + 3² + … + for n ∈
Example #2
Proof:
CONDITION I: Verification: Show true for
n = 1
1² = =
10. CONDITION II: Induction
hypothesis. Assume true for
n = k
Assume that 1² + 2² + 3² + … +
for k ∈ is TRUE
CONDITION III: Proof of Induction: Show true
for n = k + 1.
Show 1²+2²+3²+…+k²+(k+1)²= for k ∈
Proof: 1²+2²+3²+…+k²+(k+1)²= + (k+1)²
=
=
=
Since all the conditions are true, then
1² + 2² + 3² + … + n² is true for all natural
numbers n.
=
=
= ,
TRUE
12. Note the subtlety:
• Each and every element of A has
a single mapping.
• Each element of B may be
mapped to by several elements
in A or not at all.
Functions
• A function f from a set A to a set B is an assignment of exactly one element
of B to each element of A.
• We write f(a) = b if b is the unique element of B assigned by the function f to
the element a A
∈ .
• If f is a function from A to B, we write f:A. This can be read as ‘f maps A to
B’.
13. • b is the image of a
• a is the preimage
(antecedent) of b
• The range of f is the set
of all images of element
of A, denoted rng(f)
Terminology
Let f:A and f(a)=b. Then we use the
following terminology:
• A is the domain of f, denoted
dom(f)
• B is the co-domain of f
14. Injection
• A function f is said to be one-to-one or injective (or an injection) if x and y
in the domain of f, f(x) = f(y) x = y.
• Intuitively, an injection simply means that each element in the range has
at most one preimage (antecedent).
• It may be useful to think of the
contrapositive of this definition
x y f(x) f(y).
15. • A function f:A is called onto or surjective (or
surjection) if b B
∈ , a A
∈ with f(a)=b.
Surjection
• Intuitively, a surjection means that every element in the
codomain is mapped (i.e., it is an image, has an
antecedent).
• Thus, the range is the same as the codomain.
16. Bijection
• A function f is a one-to-one correspondence (or a bijection), if it is
both one-to-one (injective) and onto (surjective).
• One-to-one correspondence are important because they endow a
function with an inverse.
• They also allow us to have a concept
cardinality for infinite sets.
17. Example #1
Is this a function? Why?
- No, because each of and has
two images.
18. Is this a function
- one-to-one (injective)? Why?
- No, has two preimages.
- Onto (surjective)? Why?
- No, has no preimage.
Example #2
19. 2
Is this a function
- one-to-one (injective)? Why?
- Yes, all elements of A has images.
- onto (surjective)? Why?
- No, has no preimage.
Example #3
20. Is this a function
- one-to-one (injective)? Why?
- No, has two preimages.
- onto (surjective)? Why?
- Yes, every element of B has a preimage.
Example #4
21. Example #5
Is this a function? Why?
- Yes, it is a bijection or a one-to-
one correspondence.
22. Mathematical Induction
Evaluation
1. Prove that:
1³+2³+3³+…+n³=, for all n N
∈
Functions
Direction: Determine if the following function
is injective, surjective, and bijective.
1. Let A = {a,b,c,d} and B = {x,y,z}.
23. Evaluation
2. f:A B where A = {a,b,c,d} and B = {x,y,z}.
3. f:AB where A = {a,b,c,d} and B = {v,w,x,y,z}.
