Ioan Despidespi@turing.une.edu.au

University of New England

July 5, 2013

Outline

Ioan Despi – AMTH140 2 of 1

Set Identities

Theorem

Let 𝐴, 𝐵 and 𝐶 be subsets of a universal set 𝑈 . Then

[Commutativity] 𝐴 ∪𝐵 = 𝐵 ∪𝐴 𝐴 ∩𝐵 = 𝐵 ∩𝐴[Associativity] (𝐴 ∪𝐵) ∪ 𝐶 = 𝐴 ∪ (𝐵 ∪ 𝐶)

(𝐴 ∩𝐵) ∩ 𝐶 = 𝐴 ∩ (𝐵 ∩ 𝐶)[Distributivity] 𝐴 ∪ (𝐵 ∩ 𝐶) = (𝐴 ∪𝐵) ∩ (𝐴 ∪ 𝐶)

𝐴 ∩ (𝐵 ∪ 𝐶) = (𝐴 ∩𝐵) ∪ (𝐴 ∩ 𝐶)[Identity] 𝐴 ∪∅ = 𝐴 𝐴 ∩ 𝑈 = 𝐴[Complementation] 𝐴 ∪𝐴′ = 𝑈 𝐴 ∩𝐴′ = ∅[De Morgan’s laws] (𝐴 ∪𝐵)′ = 𝐴′ ∩𝐵′ (𝐴 ∩𝐵)′ = 𝐴′ ∪𝐵′

[Double complement] (𝐴′)′ = 𝐴[Absorption] 𝐴 ∪ (𝐴 ∩𝐵) = 𝐴 𝐴 ∩ (𝐴 ∪𝐵) = 𝐴[Idempotent laws] 𝐴 ∩𝐴 = 𝐴 𝐴 ∪𝐴 = 𝐴[Difference repr] 𝐴−𝐵 = 𝐴 ∩𝐵′

Ioan Despi – AMTH140 3 of 1

Set Identities

Proof.

We are only going to exemplify the proofs of

𝐴 ∩𝐵 = 𝐵 ∩𝐴 (a)(𝐴 ∪𝐵)′ = 𝐴′ ∩𝐵′ (b)

(a) We need to show 𝐴 ∩𝐵 ⊆ 𝐵 ∩𝐴, and 𝐵 ∩𝐴 ⊆ 𝐴 ∩𝐵.

(i) ∀𝑥 ∈ 𝐴 ∩𝐵, we have 𝑥 ∈ 𝐴 and 𝑥 ∈ 𝐵, and hence 𝑥 ∈ 𝐵 ∩𝐴. Thus𝐴 ∩𝐵 ⊆ 𝐵 ∩𝐴.

(ii) ∀𝑥 ∈ 𝐵 ∩𝐴, then 𝑥 ∈ 𝐵 and 𝑥 ∈ 𝐴. Hence 𝑥 ∈ 𝐴 ∩𝐵 and thus𝐵 ∩𝐴 ⊆ 𝐴 ∩𝐵.

From (i) and (ii) we see (a) is valid.

(b) (i) We show first (𝐴 ∪𝐵)′ ⊆ 𝐴′ ∩𝐵′.∀𝑥 ∈ (𝐴 ∪𝐵)′, we have 𝑥 ∈ 𝑈 and 𝑥 ̸∈ 𝐴 ∪𝐵. Hence 𝑥 ̸∈ 𝐴 and 𝑥 ̸∈ 𝐵,which means by definition 𝑥 ∈ 𝐴′ and 𝑥 ∈ 𝐵′. Thus 𝑥 ∈ 𝐴′ ∩𝐵′ implying(𝐴 ∪𝐵)′ ⊆ 𝐴′ ∩𝐵′.

(ii) We need to show furthermore 𝐴′ ∩𝐵′ ⊆ (𝐴 ∪𝐵)′.∀𝑥 ∈ 𝐴′ ∩𝐵′, we have 𝑥 ∈ 𝐴′ and 𝑥 ∈ 𝐵′. Hence 𝑥 ∈𝑈−𝐴 and 𝑥 ∈𝑈−𝐵,implying 𝑥 ̸∈ 𝐴 and 𝑥 ̸∈ 𝐵. We thus have 𝑥 ̸∈ 𝐴 ∪𝐵. But since 𝑥 ∈ 𝑈 , weconclude 𝑥 ∈ 𝑈 − (𝐴 ∪𝐵) = (𝐴 ∪𝐵)′. Therefore 𝐴′ ∩𝐵′ ⊆ (𝐴 ∪𝐵)′.

From (i) and (ii) we see (b) is true.Ioan Despi – AMTH140 4 of 1

Example

Example

Let 𝐴, 𝐵 and 𝐶 be 3 sets. Prove the following set identities:

1 (𝐴−𝐵) ∩ (𝐶 −𝐵) = (𝐴 ∩ 𝐶) −𝐵.

2 (𝐴−𝐵) ∪ (𝐵 −𝐴) = (𝐴 ∪𝐵) − (𝐴 ∩𝐵).

Solution. The proofs are the easiest if we make use of the set identitiesS1–S10 given in the above theorem, although direct proofs are equallyacceptable.

(𝐴−𝐵) ∩ (𝐶 −𝐵) = (𝐴 ∩𝐵′) ∩ (𝐶 ∩𝐵′) S10= 𝐴 ∩ 𝐶 ∩𝐵′ ∩𝐵′ S1 & S2= (𝐴 ∩ 𝐶) ∩𝐵′ S9= (𝐴 ∩ 𝐶) −𝐵 . S10

(𝐴−𝐵) ∪ (𝐵 −𝐴) = (𝐴⋂︀𝐵′) ∪ (𝐵 ∩𝐴′) S10

= (𝐴 ∪ (𝐵 ∩𝐴′))⋂︀

(𝐵′ ∪ (𝐵 ∩𝐴′)) S3= ((𝐴 ∪𝐵) ∩ (𝐴 ∪𝐴′))

⋂︀((𝐵′ ∪𝐵) ∩ (𝐵′ ∪𝐴′))

= ((𝐴 ∪𝐵) ∩ 𝑈)⋂︀

(𝑈 ∩ (𝐵′ ∪𝐴′)) S5= (𝐴 ∪𝐵) ∩ (𝐵 ∩𝐴)′ S5 & S6= (𝐴 ∪𝐵) − (𝐴 ∩𝐵) .

Ioan Despi – AMTH140 5 of 1

Russell’s Paradox

The precise definition of a set has to be determined by a number ofaxioms.

One of them, the axiom of comprehension (or separability), states thatevery formula defines a set. That is

Axiom (comprehension)

If 𝑋 is a set, 𝑃 is a property, then there exists a set 𝑌 whose elements areprecisely those of 𝑋 having property 𝑃 .

A logical contradiction in set theory was discovered by Bertrand Russelland was called . . .

Russell’s Paradox: If 𝑅 is the set of all sets which don’t

contain themselves, does 𝑅 contain itself?If it does then it doesn’t and vice versa.

I For more details, see the book by Elizabeth J Billinton et al, DiscreteMathematics, Logic and Structures, Longman, 1990.

Ioan Despi – AMTH140 6 of 1

Russell’s Paradox

Let R be the set of all sets which are not members of themselves. Then R isneither a member of itself nor not a member of itself. Symbolically, let𝑅 = {𝑥 : 𝑥 ̸∈ 𝑥}. Then 𝑅 ∈ 𝑅 iff 𝑅 ̸∈ 𝑅.

Proof.

We now show that 𝑅 = {𝑥 |𝑥 /∈ 𝑥} is not a set. If otherwise, then𝑆 = {𝑥 ∈ 𝑅 |𝑥 /∈ 𝑥} must also be a set according to the above axiom ofcomprehension. This is not possible for the following reasons.

(i) If 𝑆 ∈ 𝑅 then 𝑆 /∈ 𝑆 by the definition of 𝑅. This then implies from thedefinition of 𝑆 that 𝑆 ∈ 𝑆 because 𝑆 ∈ 𝑅 and 𝑆 /∈ 𝑆, leading thus to acontradiction.

(ii) If 𝑆 /∈ 𝑅 then 𝑆 ∈ 𝑆 by the definition of 𝑅. This then implies 𝑆 ∈ 𝑅 fromthe definition of 𝑆, leading again to a contradiction.

Hence 𝑆 can’t be a set.To conclude, Russel’s paradox “𝑅 is neither an element of 𝑅 nor not anelement of 𝑅” would be true only if 𝑅 where considered a valid set. But wehave just shown that 𝑅 can not be a (valid) set, hence the paradox does notarise under the axiom of comprehension.

Ioan Despi – AMTH140 7 of 1

Sequences

A sequence 𝑎𝑚, 𝑎𝑚+1, 𝑎𝑚+2, · · · , is an ordered set of elements, denotedby {𝑎𝑖}𝑖∈𝐼 .𝐼 is the set of indices, that is a set whose members index (label) membersof another set.

I In other words, a sequence is a surjective function defined on an index set𝐼 and with values in any set 𝐴.

For example, sequence 𝑎0, 𝑎1, 𝑎2, · · · may be written as {𝑎𝑖}𝑖∈N or {𝑎𝑖}𝑖≥0

or {𝑎𝑖: 𝑖 ≥ 0}.Big-sigma and Big-pi notations are used a lot with sequences, e.g.,

4∑︁𝑗=2

(𝑗3 + 𝑗) = (23 + 2) + (33 + 3) + (43 + 4)

4∏︁𝑗=2

(𝑗3 + 𝑗) = (23 + 2) · (33 + 3) · (43 + 4)

𝑛∑︁𝑖=𝑚

𝑎𝑖 = 𝑎𝑚 + 𝑎𝑚+1 + · · · + 𝑎𝑛

𝑛∏︁𝑖=𝑚

𝑎𝑖 = 𝑎𝑚 · 𝑎𝑚+1 · · · 𝑎𝑛

Ioan Despi – AMTH140 8 of 1

Sequences

It is obvious that for 𝑚,𝑛 ∈ Z with 𝑛 ≥ 𝑚, one has

𝑛∑︁𝑖=𝑚

(𝑎𝑖 + 𝑏𝑖) =

𝑛∑︁𝑖=𝑚

𝑎𝑖 +

𝑛∑︁𝑖=𝑚

𝑏𝑖

𝑛∑︁𝑖=𝑚

𝑐 · 𝑎𝑖 = 𝑐

𝑛∑︁𝑖=𝑚

𝑎𝑖

𝑛∏︁𝑖=𝑚

(𝑎𝑖 · 𝑏𝑖) =(︁ 𝑛∏︁

𝑖=𝑚

𝑎𝑖

)︁·(︁ 𝑛∏︁

𝑖=𝑚

𝑏𝑖

)︁where 𝑐 is any constant, i.e., 𝑐 is independent of the dummy variable 𝑖.

Ioan Despi – AMTH140 9 of 1

Principle of Mathematical Induction

One of the most powerful technique for verifying assertions in all ofmathematics.

It is based on the fact that any non-empty subset of the natural numbershas a least element, generalised as follows:

If 𝑚 is an integer, and 𝑊 is the set 𝑊 = {𝑚,𝑚 + 1,𝑚 + 2, . . .},then every nonempty subset of 𝑊 has a least element.

The following theorem is the basis for the mathematical induction prooftechnique.

Ioan Despi – AMTH140 10 of 1

Principle of Mathematical Induction

Theorem

Let 𝑚 ∈ Z and 𝑊 = {𝑚,𝑚 + 1,𝑚 + 2, . . .}. Let 𝑆 be a nonempty subset of 𝑊such that the following two conditions hold:

1 𝑚 ∈ 𝑆

2 whenever 𝑘 ∈ 𝑆 then 𝑘 + 1 ∈ 𝑆

Then 𝑆 = 𝑊 .

Proof.

We will prove 𝑆 = 𝑊 by contradiction (reductio ad absurdum), that islet us suppose the contrary: 𝑆 ̸= 𝑊 .

Then 𝑊 − 𝑆 is not empty and contains a least element 𝑥 because everynonempty subset of 𝑊 has a least element (𝑥 ∈ 𝑊 and 𝑥 /∈ 𝑆).The first condition (1) tells us that 𝑚 ∈ 𝑆, so it must be the case that 𝑥 > 𝑚.Thus 𝑥− 1 ≥ 𝑚, and it follows that 𝑥− 1 ∈ 𝑆.Now we apply the second condition (2) to get that (𝑥− 1) + 1 ∈ 𝑆, that is𝑥 ∈ 𝑆.This is a contradiction, since we cannot have both 𝑥 ∈ 𝑆 and 𝑥 ̸∈ 𝑆 at thesame time. Therefore 𝑆 = 𝑊 .

Ioan Despi – AMTH140 11 of 1

Principle of Mathematical Induction

Two useful definitions from logic are:

A proposition is a statement that is unambigously true or false (even ifwe do not know which).A sequence of propositions is a rule (function) that associates witheach integer 𝑛 (index) a proposition 𝑆(𝑛), also denoted by 𝑆𝑛.

Theorem (Principle of Mathematical Induction)

If 𝑚 ∈ Z and {𝑆(𝑛)}𝑛≥𝑚 is a sequence of propositions, then to prove that 𝑆(𝑛)is true for all integers 𝑛 ≥ 𝑚, perform the following two steps:

1 Prove that 𝑆(𝑚) is true.

2 Assume that 𝑆(𝑛) is true for arbitrary 𝑛 ≥ 𝑚. Prove that 𝑆(𝑛+ 1) is true.

Proof.

Let 𝑊 = {𝑛 |𝑛 ≥ 𝑚} and let 𝑆 = {𝑛 |𝑛 ≥ 𝑚 and 𝑆(𝑛) is true}. Assume wehave performed the two steps of Theorem ??. Then 𝑆 satisfies the hypothesisof Theorem ??, therefore 𝑆 = 𝑊 . So 𝑆(𝑛) is true for all 𝑛 ≥ 𝑚.

Ioan Despi – AMTH140 12 of 1

Principle of Mathematical Induction

Roughly speaking,

the first condition (1) in the Principle of Mathematical Induction (P.M.I.)ensures that 𝑆𝑛 is initially true, i.e., true when 𝑛 = the initial index 𝑚,while

the second condition (2) ensures that a true statement 𝑆𝑛 for any 𝑛 ≥ 𝑚will guarantee that at least the next statement 𝑆𝑛+1 is also true.

Ioan Despi – AMTH140 13 of 1

Examples

1. Suppose(a) Foxes can only give birth to foxes, and(b) Jim is a fox.Will Jim ever have a rabbit as one of his future offspring?

Let 𝑆𝑛 denote the statement that Jim’s offsprings of the 𝑛–th generation areall foxes. Then

(i) 𝑆0 is true, because Jim’s offsprings of 0–th generation consists of Jimhimself alone, who from (b) is a fox.

(ii) Assume 𝑆𝑘 is true with 𝑘 ≥ 0, i.e., the 𝑘–th generation of Jim’s offspringsare all foxes. Then, according to (a), all the children of this 𝑘–thgeneration can only be foxes. That is the (𝑘 + 1)–th generation of Jim’soffsprings will all be foxes. i.e., 𝑆𝑘+1 is true.

From the P.M.I., we conclude 𝑆𝑛 is true for all 𝑛 ≥ 0. Hence all Jim’soffsprings will be foxes.

Ioan Despi – AMTH140 14 of 1

Examples

2. Prove inductively for all integers 𝑛 ≥ 1,

1 + 2 + 3 + · · · + 𝑛 =𝑛(𝑛 + 1)

2. (*)

Solution. Let 𝑆𝑛 denote the statement (*). Since the initial index for 𝑛 ≥ 1is 𝑛 = 1, we first verify

(i) 𝑆1 is true because

l.h.s. of (*) = 1 , r.h.s. of (*) =1(1 + 1)

2= 1 = l.h.s. of (*) ,

Hence equality (*) holds for 𝑛 = 1, meaning 𝑆1 is true.

Note.l.h.s. (or LHS, L.H.S.) stands for left hand side andr.h.s. (or RHS, R.H.S.) stands for right hand side.

Ioan Despi – AMTH140 15 of 1

Examples

(ii) Assume 𝑆𝑘 is true for some 𝑘 ≥ 1, then by the definition of 𝑆𝑘 we have

1 + 2 + · · · + 𝑘 =𝑘(𝑘 + 1)

2. (**)

We now prove 𝑆𝑘+1 is also true. For 𝑛 = 𝑘 + 1,

l.h.s. of (*) =

𝑘(𝑘+1)2⏞ ⏟

1 + 2 + · · · + 𝑘 +(𝑘 + 1)use (**)

=𝑘(𝑘 + 1)

2+ (𝑘 + 1)

=(𝑘 + 1)(𝑘 + 2)

2= r.h.s. of (*) i.e., 𝑆𝑘+1 is true.

We thus conclude from the P.M.I. that 𝑆𝑛 is true for all 𝑛 ≥ 1. Thismeans (*) will hold for all integers 𝑛 ≥ 1.

Ioan Despi – AMTH140 16 of 1

Examples

3. Prove through the use of the P.M.I. that the following identity

12 + 22 + 32 + · · · + 𝑛2 =𝑛(𝑛 + 1)(2𝑛 + 1)

6(†)

holds for all positive integers 𝑛.

Solution. Let 𝑆𝑛 denote the statement (†). Then

(i) 𝑆1 is true because

l.h.s. of (†) = 12 = 1, r.h.s. of (†) =1 × (1 + 1)(2 × 1 + 1)

6= 1

implies l.h.s.=r.h.s., i.e., (†) is valid for 𝑛 = 1.

Ioan Despi – AMTH140 17 of 1

Examples

(ii) Assume 𝑆𝑘 is true for some 𝑘 ≥ 1, then

12 + 22 + 32 + · · · + 𝑘2 =𝑘(𝑘 + 1)(2𝑘 + 1)

6. (‡)

Hence12 + 22 + · · · + 𝑘2 + (𝑘 + 1)2 =

= (12 + 22 + · · · + 𝑘2) + (𝑘 + 1)2

from (‡) =𝑘(𝑘 + 1)(2𝑘 + 1)

6+ (𝑘 + 1)2

=(𝑘 + 1)

6

[︁(2𝑘2 + 𝑘) + (6𝑘 + 6)

]︁=

(𝑘 + 1)(2𝑘2 + 7𝑘 + 6)

6=

(𝑘 + 1)(𝑘 + 2)(2𝑘 + 3)

6

=(𝑘 + 1)((𝑘 + 1) + 1)(2(𝑘 + 1) + 1)

6,

i.e. ,𝑆𝑘+1 is true. Hence the P.M.I. implies 𝑆𝑛 is true for all integers 𝑛 ≥ 1.

Ioan Despi – AMTH140 18 of 1

Strong Principle of Mathematical Induction

The Principle of Mathematical Induction can also be equivalentlyrepresented in other forms.One of them is the following Strong Principle of M. I. (S.P.M.I.).

Theorem (Strong Principle of Mathematical Induction)

Suppose 𝑚,𝑚′ ∈ Z and for any integer 𝑛 ≥ 𝑚, 𝑆𝑛 is a statement.

(i) Prove that 𝑆𝑚, · · · , 𝑆𝑚′ with 𝑚′ ≥ 𝑚 are all true.

(ii) Assume that 𝑆𝑚, · · · , 𝑆𝑛 are all true and assume 𝑛 ≥ 𝑚′ and prove that𝑆𝑛+1 is true.

Then 𝑆𝑛 is true for all 𝑛 ≥ 𝑚.

Proof.

S.P.M.I. implies P.M.I. is obvious. As for the converse one only needs to define𝑇𝑛 to be the statement that 𝑆𝑘 for 𝑚 ≤ 𝑘 ≤ 𝑛 are all true. Then the P.M.I. on𝑇𝑛 will imply the S.P.M.I. on 𝑆𝑛. We shall however skip the simple detailshere.

Ioan Despi – AMTH140 19 of 1

Examples

Sometimes knowing the previous step just isn’t enough.

Strong induction is induction where you assume that all previous casessatisfy your induction hypothesis, not just the most recent case.

4. Divisibility by a Prime:Every integer greater than 1 is divisible by a prime number.

To see, via strong induction, that every integer greater than 1 is divisible by aprime number, we note that

the basis value of 2 is trivially divisible by a prime (itself).

if we assume the strong inductive hypothesis that every integer up to 𝑘 isdivisible by a prime, when we look at 𝑘 itself, either

I it is prime (and hence divisible by itself), orI can be factored as lesser integers, each of which having a prime factor.

Thus 𝑘 is divisible by a prime.

Ioan Despi – AMTH140 20 of 1

Examples

Let 𝑛, 𝑟 ∈ N with 𝑟 ≤ 𝑛. We define (︂𝑛

𝑟

)︂,

called “𝑛 choose 𝑟” (binomial coefficients), to be the total number of𝑟–element subsets that can be chosen from a set of 𝑛 elements.Or, more simply, the number of ways of choosing 𝑟 items from a list of 𝑛items. We thus have

(a)

(︂𝑛

0

)︂= 1, ∃ only 1 empty set

(b)

(︂𝑛

𝑛

)︂= 1, the only subset with all elements in it is itself

(c)

(︂𝑛

𝑟

)︂=

(︂𝑛− 1

𝑟

)︂+

(︂𝑛− 1

𝑟 − 1

)︂. if 𝑟 ≥ 1, 𝑛 > 1

Recall the convention 0! = 1.

Ioan Despi – AMTH140 21 of 1

Examples

5. Prove that (︂𝑛

𝑟

)︂=

𝑛!

𝑟!(𝑛− 𝑟)!, 𝑛 ≥ 𝑟 ≥ 0 .

Solution. Let 𝑆𝑛 denote the following statement(︂𝑛

𝑟

)︂=

𝑛!

(𝑛− 𝑟)!𝑟!is true for all 0 ≤ 𝑟 ≤ 𝑛 .

(i) 𝑆0 is true from either (a) or (b), because

(︂0

0

)︂=

0!

0! · 0!= 1.

Ioan Despi – AMTH140 22 of 1

Examples

(ii) Assume for 𝑛 ≥ 0, 𝑆𝑘 is true for all 0 ≤ 𝑘 ≤ 𝑛 and we want to prove that𝑆𝑛+1 is true. Then from (c) for 1 ≤ 𝑟 ≤ 𝑛 we have

𝑆𝑛+1 :

(︂𝑛 + 1

𝑟

)︂=

(︂𝑛

𝑟

)︂+

(︂𝑛

𝑟 − 1

)︂=

𝑛!

𝑟!(𝑛− 𝑟)!+

𝑛!

(𝑟 − 1)!(𝑛− 𝑟 + 1)!(from induction assumption)

=𝑛!

(𝑟 − 1)!(𝑛− 𝑟)!

[︂1

𝑟+

1

𝑛− 𝑟 + 1

]︂=

(𝑛 + 1)!

𝑟!(𝑛 + 1 − 𝑟)!

As for 𝑟 = 0 and 𝑟 = 𝑛 + 1, we see from (a) and (b)(︂𝑛 + 1

0

)︂=

(︂𝑛 + 1

𝑛 + 1

)︂= 1 =

(𝑛 + 1)!

(𝑛 + 1)!0!.

Hence 𝑆𝑛+1 is true. From (i) and (ii) and the S.P.M.I. (with 𝑚 = 𝑚′ = 0), weconclude 𝑆𝑛 is true for all 𝑛 ≥ 0.

Ioan Despi – AMTH140 23 of 1

Examples

6. For integer 𝑛 ≥ 1 show the following statement (often called binomialtheorem or binomial expansion) is true𝑆𝑛:

(𝑥+𝑦)𝑛 =

(︂𝑛

0

)︂𝑥𝑛+

(︂𝑛

1

)︂𝑥𝑛−1𝑦+

(︂𝑛

2

)︂𝑥𝑛−2𝑦2+· · ·+

(︂𝑛

𝑛− 1

)︂𝑥𝑦𝑛−1+

(︂𝑛

𝑛

)︂𝑦𝑛

Solution.

(i) 𝑆1 is true because

(𝑥 + 𝑦)1 = 𝑥 + 𝑦 =

(︂1

0

)︂𝑥 +

(︂1

1

)︂𝑦 .

(ii) Assume 𝑆𝑘 is true for some 𝑘 ≥ 1, then

(𝑥 + 𝑦)𝑘+1 = (𝑥 + 𝑦)[︀(𝑥 + 𝑦)𝑘

]︀= (𝑥 + 𝑦)

[︂(︂𝑘

0

)︂𝑥𝑘 +

(︂𝑘

1

)︂𝑥𝑘−1𝑦 + · · · +

(︂𝑘

𝑘

)︂𝑦𝑘

]︂

Ioan Despi – AMTH140 24 of 1

Examples

=

(︂𝑘

0

)︂𝑥𝑘+1 +

[︂(︂𝑘

0

)︂+

(︂𝑘

1

)︂]︂𝑥𝑘𝑦 +

[︂(︂𝑘

1

)︂+

(︂𝑘

2

)︂]︂𝑥𝑘−1𝑦2

+ · · · +

[︂(︂𝑘

𝑘 − 1

)︂+

(︂𝑘

𝑘

)︂]︂𝑥𝑦𝑘 +

(︂𝑘

𝑘

)︂𝑦𝑘+1

=

(︂𝑘 + 1

0

)︂𝑥𝑘+1 +

(︂𝑘 + 1

1

)︂𝑥𝑘𝑦 + · · · +

(︂𝑘 + 1

𝑘 + 1

)︂𝑦𝑘+1

i.e., 𝑆𝑘+1 is true. Hence from the P.M.I. we conclude 𝑆𝑛 is true for all 𝑛 ≥ 1.

Ioan Despi – AMTH140 25 of 1