Discrete Structures

# Lecture Notes, 12 Nov 2010

 Last Update: 12 November 2010 Note: or material is highlighted

## §4.1: Mathematical Induction (cont'd)

General Notes:

• To make my typing easier, I will no longer italicize variables, so I'll write "∀x" instead of "∀x", etc.

• Also, please remember that when I write something like:

Σi ∈ {1,…,n}

that's just an easier way of typing:

i = n
Σ
i = 1

1. Mathematical Induction as a Proof Strategy (reminder):

To show:(∀nN)P(n):
show:P(0)
 "base case"
& show:(∀kN)[P(k) → P(k+1)]
 "inductive case"

• Notes:

1. To show P(0):

Identify P(n), & let n:=0; then use algebra

2. Sometimes you'll need to show (∀wW)P(w) instead,
so then just let the base case be: P(1), etc.

3. To show the inductive case,
just realize that it's a quantified conditional
and we already know how to prove those:

Choose an arbitrary k, and show P(k) → P(k+1)

And how do we prove a conditional? Using Direct Proof:

Suppose P(k), & try to show P(k+1).
Note: The antecedent "P(k)" is called the  "inductive hypothesis"

2. Another example of a proof by mathematical induction:

Thm (General Distributive Law)

Let a,b1,…,bnR = {x | x is a real number}.
Then: (∀n ∈ N)[a(b1+…+bn) = ab1+…+abn]

• Just to help you get used to the notation, here's the theorem re-stated using Σ notation:

(∀n ∈ N)[aΣi ∈ {1,…,n}bi = Σi ∈ {1,…,n}abi]

proof:

The "property" P that we're interested in proving that it holds for all natural numbers is:

P(n): "a(b1+…+bn) = ab1+…+abn"

Base Case (first (failed) attempt):

Show P(0):

Hmmm, what on earth could P(0) be?
I have no idea, but math inductions can begin with 1,
so let's re-state the theorem as:

(∀ n ∈ W)P(n)

Base Case (second (slightly more successful) attempt):

Show P(1):

i.e., show ab1 = ab1

That's trivially true! That was easy!

There's only one problem: It won't be of any help to us :-|

So, let's re-state the theorem one more time, as:

(∀ n ≥ 2)P(n)

(This is the only interesting case anyway.)

Base Case (third (finally successful) attempt):

Show P(2):

i.e., show a(b1+b2) = ab1+ab2:

Almost trivial(!), because…
…this is the distributive law of multiplication over addition, which is an axiom.

• Or it can be proved from the "recursive" definitions of + and *, which we'll get to later in the semester.

• If you want a geometric "proof",
it says that the area of a large rectangle consisting of two smaller rectangles piled one on top of the other,
one with dimensions a×b1 and one with dimensions a×b2,
equals the sum of the areas of the two smaller rectangles.

So now we're ready for the…

…Inductive Case:

Show ∀k[P(k) → P(k+1)]:

i.e., show ∀k[ (a(b1+…+bk) = ab1+…+abk) → (a(b1+…+bk+1) = ab1+…+abk+1) ]

To do this, suppose the antecedent: a(b1+…+bk) = ab1+…+abk

• (that's the "inductive hypothesis";
but really all it is is the assumption for a direct proof)

& show the consequent: a(b1+…+bk+1) = ab1+…+abk+1:

Up to here, it's all been fairly mechanical; now some "insight" is needed.

• (But it's only high-school-level, algebraic insight)

Let's compute a(b1+…+bk+1):

a(b1+…+bk+1) = a(b1+…+bk+bk+1)

• (just spelling out the "…")

= a( (b1+…+bk) + bk+1)

• (just putting some parentheses in, to clarify the structure)

= a(    B   + bk+1)

• (just letting B = (b1+…+bk) )

= aB + abk+1, by the base case!

= a(b1+…+bk) + abk+1

• (by the definition of B)

= (ab1+…+abk) + abk+1

• (by the inductive hypothesis)

QED

3. Another Example (which we did not have time for in class, so I'll record the entire proof here):

Thm (the cardinality of the power set of S = 2 raised to the cardinality of S)

• i.e., if S has n members, then it has 2n subsets.

(∀ n ∈ N)(∀ set S)[ |S|=n → |℘(S)| = 2n ]

proof:

Here, P(n) is: "(∀ set S)[ |S|=n → |℘(S)| = 2n ]"

Base Case:

Show P(0):

Show (∀ set S)[ |S|=0 → |℘(S)| = 20 ]:

Choose arb set S.
Suppose |S|=0, i.e., S = ∅
Show |℘(S)| = 20:

℘(S) = ℘(∅) = {∅}
∴ |℘(S)| = |{∅}| = 1 = 20.

[This is as far as we got in lecture; now for the rest of the proof:]

Inductive Case:

Show ∀k[P(k) → P(k+1)]:

i.e., show ∀k[ ∀S[ |S]=k → |℘(S)|=2k] → ∀S′[ |S′|=k+1 → |℘(S′)|=2k+1] ]:

• (I'm only using "S′" so that you can keep the S-in-the-antecedent separate from the S-in-the-consequent.)

Choose arb k
Suppose ∀S[ |S]=k → |℘(S)|=2k] (ind. hyp.)
& show: ∀S′[ |S′|=k+1 → |℘(S′)|=2k+1]:

Choose arb S′
Suppose |S′|=k+1
& show |℘(S′)|=2k+1:

We are assuming |S′|=k+1;
∴ we have the following picture:
S′, which has k+1 elements, consists of
a subset S′′ with only k elements,
unioned with one more element; call it "e".

• i.e., S′ = S′′ ∪ {e},
where |S′| = |S′′| + |{e}| = k+1

Describing this in FOL, we have:

∃S′′[S′=S′′∪{e} ∧ |S′′|=k]

∴ |℘(S′′)|=2k, by the inductive hypothesis

But to each subset of S′′, there correspond two subsets of S′:

i.e., in (sort of) FOL:

(∀ subset S* ⊆ S′′)(∃ 2 subsets of S′=S′′∪{e}), namely: S* itself & S*∪{e}.

∴ |℘(S′)| = |℘(S′&prime)|*2 (i.e., doubled)
= 2k*2 = 2k+1

QED