Euclid's Proof of the Infinitude of Primes

In today's blog, I will present a very well known proof. What makes this proof especially appealing is that it is not too complex. Even so, it is very powerful. This theorem was first presented in Euclid's Elements (Book IX, Proposition 20).

Theorem: There are an infinite number of primes.

Proof:

(1) Assume that there is only a finite number of primes.

(2) Then, there exists a prime p_n that is the largest prime.

(3) Let p₁, p₂, ..., p_n be the list of all primes that exist.

(4) Let x = p₁*p₂*...*p_n + 1.

(5) By the fundamental theorem of arithmetic (see Theorem 3, here), we know there is at least one prime that divides x. Let us call this prime p*.

(6) But none of the primes p₁ ... p_n divide x since x ≡ 1 (mod p_i) for any of the primes p₁ ... p_n

(7) Therefore, we have a contradiction. We have a prime p* that is not in the complete list of primes.

(8) So, we reject our assumption in step#1 and conclude that there are an infinite number of primes that exist.

QED

Modular Arithmetic: Additional Lemma

Lemma 1: if a ≡ b (mod p), then ap^n-1 ≡ bp^n-1 (mod pⁿ)

Proof:

(1) a ≡ b (mod p)

(2) So, there exists c such that pc = a - b

(3) So that a = pc + b

(4) So, ap^n-1 ≡ (pc+b)p^n-1 ≡ pⁿc + bp^n-1 ≡ bp^n-1 (mod pⁿ)

QED

Chinese Remainder Theorem

The Chinese Remainder Theorem gets its name from a 3rd century book written by the Chinese mathematician Sun Tzu. There are more details available on wikipedia.

In essence, it is a theorem about simultaneous congruences (see here if you need a review of modular arithmetic).

In today's blog, I will show the proof as it relates to standard integers.

Theorem: Chinese Remainder Theorem

Let a_i be a set of set of k integers consisting of a₁, a₂, ..., a_k.

Let n_i be a set of k coprime integers consists of n₁, n₂, ..., n_k where for any i,j if i ≠ j, then gcd(n_i,n_j)=1. [See here for review of greatest common divisor if needed]

Then, there exists an integer x such that for each a_i, n_i:

x ≡ a₁ (mod n₁)

x ≡ a₂ (mod n₂)

...

x ≡ a_k (mod n_k)

Proof:

(1) Let n = n₁ * n₂ * ... * n_k

(2) Let c_i = n / n_i for i = 1, ..., k

(3) We see that for all i, gcd(c_i,n_i) = 1 since:

(a) Assume gcd(c_i,n_i)=d where d is a number greater than 1.

(b) Then there exists a prime p that divides d (By the Fundamental Theorem of Arithmetic, see here)

(c) Since c_i = n₁ * ... * n_k, by Euclid's Generalized Lemma (see here), p must divides n_j where j ≠ i.

(d) But this impossible since by assumption gcd(n_i,n_j) = 1 when i ≠ j.

(e) So we reject that assumption in #2a.

(4) We further see that for all i,j when i ≠ j, n_j divides c_i.

This follows directly from c_i = n / n_i.

Which means that:

c_i ≡ 0 (mod n_i)

(5) Using Bezout's Identity (see here) and step #3, we know that for each c_i, n_i, there exists integers d_i,e_i such that:

c_id_i + e_in_i = 1.

(6) But this means that c_id_i ≡ 1 (mod n_i) since c_id_i - 1 = (-e_i)n_i

(7) Let x = a₁c₁d₁ + a₂c₂d₂ + ... + a_ic_id_i + ... + a_kc_kd_k

(8) We can show that x satisfies the assumption of the theorem since for any i:

x ≡ a₁(0)d₁ + a₂(0)d₂ + ... + a_i(1) + ... + a_k(0)d_k ≡ a_i (mod n_i)

QED

Group Theory: Quotient Groups

In today's blog, I will review some basic lemmas regarding normal subgroups and quotient groups. The concept of normal subgroups was first offered by Evariste Galois. The important idea behind normal subgroups is that when a subgroup is normal, then the set of left or right cosets is itself a group. This is what will be presented in today's blog.

I will use this result as part of Kummer's proof on Fermat's Last Theorem for regular primes.

Definition 1: Normal Subgroup

A subgroup H of a group G is called a normal subgroup of G if for all a ∈ G and all h ∈ H, there exists an element h' ∈ H such that ah = (h')a

NOTE: It is quite possible that h'=h but this is not necessary. From this perspective, normal subgroup is not necessarily an abelian group, that is, it may not be commutative (see here for definition of an abelian group).

NOTE #2: For all purposes definition 1 is the same as aH = Ha. Here's why:

For any x ∈ aH, there exists h ∈ H such that x = ah. From the definition, there exists h' such that h' ∈ H and ah=h'a which means that x=h'a which means that x ∈ Ha.

Lemma 1: A subgroup H is normal in G if and only if g^-1Hg ⊆ H for all g in G.

Proof:

(1) Assume H is a normal subgroup of G.

(2) Then, for any g ∈ G, h ∈ H, there exists h' ∈ H such that hg = gh' [See Definition of a Normal Subgroup above]

(3) Thus, g^-1hg = h' [Since g^-1gh' = eh' = h']

NOTE: We know that g^-1 ∈ G and that g^-1g = e (Inverse Property of Groups, see here). We know that eh' = h' (Identity Property of Groups, see here)

(4) Therefore for all g,h, we know that g^-1hg ∈ H [Since g^-1hg = h' and h' ∈ H]

(5) So that g^-1Hg ⊆ H [That is, for all x ∈ g^-1Hg → x ∈ H]

(6) Assume that g^-1Hg ⊆ H for all g ∈ G

(7) This means that Hg ⊆ gH [Since from #6, we have gg^-1Hg ⊆ gH and gg^-1Hg = eHg = Hg]

(8) Since g^-1 ∈ G, our assumption in #6 gives us:

(g^-1)^-1H(g^-1) = gHg^-1 ⊆ H

But this also means that:

gH ⊆ Hg since gHg^-1g = gH ⊆ Hg

(9) So combining step #7 and step #8 gives us: Hg = gH

NOTE: Hg = gH shows that for all h ∈ H, g ∈ G, there exists h' ∈ H such that: hg = gh'.

QED

Lemma 2: A subgroup of an abelian group is normal

Proof:

(1) Let G be an abelian group. [See here for definition of an abelian group]

(2) Let H be a subgroup of G. [See here for definition of a subgroup]

(3) g ∈ G, h ∈ H → gh = hg. [from the definition of abelian groups]

(4) Multiplying both sides by the inverse of g (see here) gives us:

g^-1gh = g^-1hg

This implies that:

h = g^-1hg

(5) Since h ∈ H, this gives us:
h ∈ H, g ∈ G → g^-1hg ∈ H.

(6) Using Lemma 1 above, we can now conclude that H is a normal subgroup of G.

QED

Definition 2: Quotient Group

Let G be a group and H a normal subgroup of G. The Quotient Group is defined as as the coset aH such that {aH : a ∈ G} and is represented as G/H

Example 1: Z/2Z under addition = { 0+2Z, 1+2Z}

Let Z be the set of integers.

We can see that the left coset 2z = the set of even integers = { ..., -2, 0, 2 ... }

Z/2Z = {a(2Z) : a ∈ Z} = two distinct cosets {0 + 2z, 1 + 2z}

Lemma 3: if aH = a'H, then there exists h ∈ H such that a' = ah

Proof:

(1) a' ∈ a'H [H is a group → e ∈ H → a'e ∈ a'H]

(2) a' ∈ aH [since aH = a'H]

(3) From #2, there exists h ∈ H such that a' = ah

QED

Lemma 4: aH = H if and only if a ∈ H

Proof:

(1) Assume aH = H

(2) a = ae ∈ aH

(3) So combining #1 and #2, a ∈ H

(4) Assume a ∈ H

(5) x ∈ aH → x=ah with h ∈ H

(6) But if a ∈ H and h ∈ H, then ah ∈ H (by property of closure, see here)

(7) From #5 and #6, we conclude that aH ⊆ H

(8) Let h ∈ H

(9) From #4 and #8, ah ∈ H and a^-1h ∈ H

(10) So, h = eh = (aa^-1)h = a(a^-1h)

(11) So, from #10, h ∈ aH and therefore H ⊆ aH

(12) Combining #7 (aH ⊆ H) and #11 (H ⊆ aH) gives us: aH = H

QED

Lemma 5: A Quotient Group is itself a group under the operation (aH)(bH) = abH.

Proof:

(1) Let H be a normal subgroup of G.

(2) Let G/H be the quotient group { aH : a ∈ G }

(3) First, we need to prove that (aH)(bH) = abH is a well-defined operation. In other words, we need to show that aH=a'H, bH = b'H → (a'b')H = (ab)H

NOTE: A function maps each of its input to exactly one output (see here for Wikipedia's review of a mathematical function) So, we need to prove if the inputs are equal, the outputs are equal.

(3) Assume that aH = a'H and bH = b'H with a,a',b,b' ∈ G

(4) Then a' = ah₁ and b' = bh₂ for some h₁,h₂ in H [By Lemma 3 above]

(5) Then a'b'H = (ah₁)(bh₂)H

(6) Using Lemma 4 above, since h₁, h₂ ∈ H we know that:

h₁H = H
h₂H = H

So that we have:

(ah₁)(bh₂)H = (ah₁)bH

(7) Since H is a normal subgroup of G, we know that b ∈ G → bH=Hb [See definition 1 above, note 2]

This gives us:

(ah₁)bH = (ah₁)Hb

Using Lemma 4 above gives us

(ah₁)Hb = aHb = abH

(8) Closure: x ∈ G/H, y ∈ G/H → xy ∈ G/H since:

x ∈ G/H → there exists a ∈ G such that x = aH.

y ∈ G/H → there exists b ∈ G such that y = bH

xy = (aH)(bH) = (ab)H [Definition of operation]

xy ∈ G/H since a ∈ G, b ∈ G → ab ∈ G so (ab)H ∈ G/H.

(9) Identity: H

e ∈ G → eH ∈ G/H

e ∈ H so eH = H [From Lemma 4 above]

Further, we see that:

(aH)(H) = (aH)(eH) = (ae)H = aH

(14) Inverse: a^-1H

x ∈ G/H → x = aH such that a ∈ G

a ∈ G → a^-1 ∈ G

So a^-1H ∈ G/H

Further,

(aH)(a^-1H) = (aa^-1)H = eH = H

(15) Associativity:

(aHbH)cH = (ab)HcH = (ab)cH [From the definition of the operation for this group]

= a(bc)H = [(ab)c = a(bc) since a,b,c ∈ G and G is a group so is characterized by associativity]

= aH(bc)H = [From the definition of operation for this group: aH(bc)H = a(bc)H]

= aH(bHcH) [From the defintion of operation for this group: bHcH = (bc)H]

QED

Group Theory: Lagrange's Theorem

In today's blog, I review the proof for Lagrange's Theorem. The theorem is named after Joseph-Louis Lagrange who first stated it. The first complete proof came 30 years later.

In today's blog, I also use the concept of order, subgroup, and coset. The concept of the coset was first proposed by Evariste Galois. The term "coset" was coined by G. A. Miller in 1910.

The content of today's blog is taken from Joseph A. Gallian's Contemporary Abstract Algebra.

Definition 1: Order of a Group

The number of elements of a group.

NOTE: If you need to review the definition of the group, see here.

Definition 2: Subgroup

If H is a subset of a group G and H is a group itself under the operation of G, then H is a subgroup of G.

Definition 3: Coset of H in G

Let G be a group and H a subgroup of G. For any a ∈ G, the set aH { ah : h ∈ H } is called the left coset of H in G containing a. The set Ha { ha : h ∈ H } is called the right coset of H in G containing a.

NOTE: An important idea behind the coset is that it is a distinct partition of elements (see Lemma 2 below). From the property of closure, we know that if two values are elements of a group, then their product (by this, I mean the result of the group operation) is also an element of the group.

From this, we know that the set of cosets can be divided up as follows: a₁H, a₂H, ..., a_rH where r is a positive integer and where for each coset i ≠ j → a_iH ∩ a_jH = ∅ [See Lagrange's Theorem below to see how this partitioning can be used]

Example 1 Coset: Z₉

Let G = Z₉ = { 0, 1, 2, 3, 4, 5, 6, 7, 8 } with operation '+'

NOTE: Z₉ is the set of integers modulo 9 [See here for a review of modular arithmetic]

Let H = { 0, 3, 6 } with operation '+'

We can see that H is a subgroup of G

The left coset in this case is a+H (note it would aH is the operation were '*')

Here are the cosets:

0 + H = { 0, 3, 6 } = 3 + H = 6 + H
1 + H = { 1, 4, 7 } = 4 + H = 7 + H
2 + H = {2, 5, 8} = 5 + H = 8 + H

We can see that all cosets are either equal or distinct. I present a proof of this in Lemma 2 below.

Example 2: Even integers

Let Z be the set of integers

Then 2Z is a left coset which includes { ..., 0, 2, 4 , 6, ... }

Lemma 1: aH = H if and only if a ∈ H

Proof:

(1) Assume aH = H

(2) Then, a = ae ∈ aH = H

(3) Assume a ∈ H

(4) aH ⊆ H since h ∈ H → ah ∈ H [By property of closure, see here if needed]

(5) H ⊆ aH since:

(a) Let h be any element of H.

(b) a^-1 ∈ H since a ∈ H [By the inverse property, see here if needed]

(c) a^-1H ∈ H [By property of closure, see here if needed]

(d) So h = eh = (aa^-1)h = a(a^-1h) [By identity property, inverse property, and associative property, see here if needed]

(e) But a(a^-1h) ∈ aH so that from (#5d), h ∈ aH.

(6) So H = aH since H ⊆ aH and aH ⊆ H.

QED

Lemma 2: if aH, bH are two cosets, then aH = bH or aH ∩ bH = ∅

Proof:

(1) Assume that aH ∩ bH ≠ ∅

(2) Let x ∈ aH ∩ bH since aH ∩ bH ≠ ∅

(3) Then there exists h₁, h₂ such that:

x = ah₁
x= bh₂

Since x is found in both aH and bH by assumption.

(4) Thus, a = xh₁^-1 = (bh₂)h₁^-1

(5) From this, aH = [(bh₂)h₁^-1]H = b(h₂h₁^-1H) = bH [By Lemma 1 above]

QED

Theorem: Lagrange's Theorem

If G is a finite group and H is a subgroup of G, then the order of H divides the order of G.

Proof:

(1) Let a₁H, a₂H, ..., a_rH denote the distinct left cosets of H in G. [See Definition 3 above for details on left cosets of H in G.]

NOTE: If two cosets are equal then we only count them once.

(2) For each a ∈ G, we have aH = a_iH for some i by our construction in step #1.

(3) a ∈ aH since:

a = ae and ae ∈ aH since H is a subgroup. [See Definition 2 above for details on subgroups]

(4) Thus, each member of G belongs to one of the cosets a_iH [from step #2 and step #3.]

(5) G = a₁H ∪ a₂H ∪ ... ∪ a_rH [This follows directly from step #4]

(6) From Lemma 2 above, we know a_iH = a_jH or a_iH ∩ a_jH = ∅ and since each a_i is distinct, we get:

order(G) = order(a₁H) + order(a₂H) + ... + order(a_rH)

(7) Since order(a_iH) = order(H) for each i (by the definition of cosets, see definition 3 above, we have:

order (G) = r*order(H)

QED

Modular Arithmetic and Unique Mappings

Today, I talk about a result using modular arithmetic that I use in the properties of cyclotomic integers.

Lemma 1: For a given odd prime λ, each value i, 2*i, 3*i, ... (λ-1)*i maps to a distinct value of 1,2,3,...,(λ-1) modulo λ

Proof:

(1) Let a,b be any integer between 1 and λ-1 where a ≠ b

(2) Let i be any integer between 1 and λ - 1.

(3) Assume that a*i ≡ b*i (mod λ) where a,b are distinct.

(4) So λ divides a*i - b*i and there exists c such that λc = ai-bi = i(a-b)

(5) By Euclid's Lemma, λ divides either i or λ divides a-b.

(6) We know that λ doesn't divide i since i is between 1 and λ -1.

(7) So λ divides a-b

(8) But this means that a-b=0 since a,b are between 1 and λ - 1 and if a,b are distinct then abs(a-b) is between 1 and λ-2 which is not divisible by λ

(9) So we have a contradiction and we reject our assumption in (3).

QED

modular arithmetic and functions

Today, I want to go over a single lemma which I use in my analysis of cyclotomic integers.

Lemma 1:

if α,β are integers such α ≡ β (mod γ) and f(x) = xⁿ + a₁x^n-1 + ... + a_n

then f(α) ≡ f(β) (mod γ)

Proof:

(1) f(α) - f(β) = [(α)ⁿ - (β)ⁿ] + [a₁(α)^n-1 - a₁(β)^n-1] + ... + [a_n-1α - a_n-1β] + [a_n - a_n]

(2) Now, since α ≡ β (mod γ), we also have:

(α)² ≡ (β)² (mod γ)

all the way up to:

(α)^n-1 ≡ (β)^n-1 (mod γ)

and:

(α)ⁿ ≡ (β)ⁿ (mod γ)

(3) So this means that γ divides all parts of
[(α)ⁿ - (β)ⁿ] + [a₁(α^n-1 - β^n-1)] + ... + [a_n-1(α - β)]

(4) Which means that γ divides f(α) - f(β)

QED

Fields and Rings

In today's blog, I will talk about the assumptions behind the idea of a field. I use the concept of a field in the proof for the Division Algorithm for Polynomials.

Definition 1: Ring

A ring R is a nonempty set with two binary operations: addition (a + b) and multiplication (ab).

It has the following properties:

1. Commutative Rule for Addition: a + b = b + a

2. Associative Rule for Addition: (a + b) + c = a + (b + c)

3. Additive Identity: there exists a value 0 ∈ R such that a + 0 = a for all a ∈ R

4. Additive Inverse: for all elements a ∈ R, there exists a value -a ∈ R such that a + -a = 0.

5. Associative Rule for Multiplication: a(bc) = (ab)c

6. Distributive Rule: a(b + c) = ab + ac and (b + c)a = ba + ca

Definition 2: Commutative Ring

A ring that has the following additional property:

7. Commutative Rule for Multiplication: ab = ba

Definition 3: Field

A field is a commutative ring R with unity in which every nonzero element is a unit.

It has all the properties of a commutative ring plus:

8. Multiplicative Identity (unity): there exists a value 1 ∈ R such that such that a * 1 = a for all a ∈ R

9. Multiplicative Inverse (every nonzero element is a unit): for all nonzero elements a ∈ R, there exists an inverse a^-1 ∈ R such that a*a^-1 = 1.

Examples:

1. The set of integers Z is a commutative ring with unity 1. [See here for more information on the integers]

2. The set Z_n is a commutative ring with unity 1. [See here for more information on modular arithmetic]

3. The set of 2x2 matrices with integer entries is a noncommutative ring with unity. [See here for more information on 2x2 matrices]

4. If p is prime, then the set Z_p is a field. [See here for more information on modular arithmetic]

5. The set of rational numbers Q is a field.

6. The set of real numbers R is a field.

7. The set of complex numbers is a field. [See here for more information on complex numbers]

8. The set of Gaussian Integers is a ring. [See here for more information on Gaussian Integers]

9. The set of Eisenstein Integers is a ring. [See here for more information on Eisenstein Integers]

References

• Joseph A. Gallian, Contemporary Abstract Algebra
• Rings, Wikipedia
  

Division Algorithm for Polynomials

In today's blog, I will go over a result that I use in the proof for the Fundamental Theorem of Algebra.

Today's proof is taken from Joseph A. Gallian's Contemporary Abstract Algebra.

Theorem: Division Algorithm for Polynomials

Let F be a field, f(x), g(x) ∈ F[x] with g(x) ≠ 0.

Then there exists unique polynomials q(x), r(x) in F[x] such that f(x) = g(x)q(x) + r(x) and r(x)=0 or deg r(x) is less than deg g(x).

Proof:

(1) Let f(x) = g(x)q(x) + r(x) with g(x) ≠ 0

(2) If f(x) = 0 or deg f(x) is less than g(x), then q(x)=0, r(x)=f(x)

(3) So, we can assume that f(x) ≠ 0 and deg f(x) ≥ deg g(x)

(4) Let f(x) = a_nxⁿ + ... + a₀

(5) Let g(x) = b_mx^m + ... + b₀

(6) Let f₁(x) = f(x) - a_nb_m^-1x^n-mg(x)

(7) We note that deg f₁(x) is less than deg f(x) since:

f(x) - a_nb_m^-1x^n-mg(x) =

= a_nxⁿ + ... + a₀ - a_nb_m^-1x^n-m(b_mx^m + ... + b₀) =

= a_nⁿ - a_nxⁿ + a_n-1x^n-1 + ... + a₀ - a_nb_m^-1b_m-1x^n-1 - ... - a_nb_m^-1b₀x^n-m =

= a_n-1x^n-1 + ... + a₀ - a_nb_m^-1b_m-1x^n-1 - ... - a_nb_m^-1b₀x^n-m

So that f₁(x) has a degree of n-1 while f(x) has a degree of n.

(8) Now, we are ready to prove this theorem by induction.

(9) The assumption is true for deg f(x) = 0

deg f(x) is 0 → f(x)=C where C is a constant.

If deg g(x) is 0, then g(x) = D where D is a nonzero constant and q(x) = C/D and r(x)=0.

If deg g(x) is greater than 0, then q(x)=0 and r(x)=C.

(10) We can now assume that the assumption holds for all polynomials up to degree n-1.

(9) We see that:

f(x) = a_nb_m^-1x^n-mg(x) + f₁(x)

where the degree of f₁(x) is n-1 [See step #7]

(10) But by the induction hypothesis (step #10), we can assume that there exists q₁(x) and r₁(x) where r₁(x) has a degree lower than g(x).

(11) Therefore, we have:

a_nb_m^-1x^n-mg(x) + f₁(x) = a_nb_m^-1x^n-mg(x) + q₁(x)g(x) + r₁(x) =

= [a_nb_m^-1x^n-m + q₁(x)]g(x) + r₁(x)

(12) Which proves that degree r(x) is less than degree g(x) by principle of induction.

(13) Now, we still need to prove uniqueness of q(x),r(x)

(14) Suppose that:

f(x) = g(x)q(x) + r(x) = g(x)q'(x) + r'(x) where r(x),r'(x)=0 or deg r(x),r'(x) is less than deg g(x)

(15) Now, if we substract both equations, we get:

0 = g(x)[q(x) - q'(x)] + [r(x) - r'(x)]

which is the same as:

r'(x) - r(x) = g(x)[q(x) - q'(x)]

(16) Now since r'(x) and r(x) have degree less than g(x), the only way that this can be true is if r'(x) - r(x) = 0

(17) But then r'(x) = r(x) and q(x) = q'(x)

QED

References

• Joseph A. Gallian, Contemporary Abstract Algebra.
  

Properties of cos θ + i sin θ

In today's blog, I will go over some basic trigonometric properties that I use in my proof for the Fundamental Theorem of Algebra.

Lemma 1: [(cos α + i sin α) ]^a = cos (a*α) + i sin (a*α)

Proof:

(1) From Euler's Formula, we know that:

e^iα = cos α + i sin (α)

(2) So, we see that:

(e^iα)^a = e^i*a*α

(3) Finally,

e^i(a*α) = cos (a*α) + i sin (a*α)

QED

Lemma 2: (cos α + i sin α)(cos β + i sin β) = cos(α + β) + i sin(α + β)

Proof:

(1) Again, using Euler's Formula, we have:

e^iα = cos α + i sin α

e^iβ = cos β + i sin β

(2) So multiplying these two values together gives us [see here if a review of exponents are needed]:

(e^iα)(e^iβ) = e^{iα + iβ} = e^{i(α + β)}

(3) Finally,

e^{i(α + β)} = cos(α + β) + i sin(α + β)

QED

Argand Diagram

An Argand Diagram is a way of visualizing complex numbers. It involves graphing a complex number to the Cartesian coordinates.

There are two ways to represent a complex number:

x + iy

r (cos φ + i sin φ)

Both of these ways are equivalent since:



Here is the Argand Diagram:

One important point to remember is that in the r (cos φ + i sin φ) form, all values can be positive since r is an absolute magnitude and φ is a value between 0 and 2π using radians (or in degrees, between 0 and 360).

References

• Complex Number, Wikipedia
  

Using Maclaurin Series to define exponents

It is easy to define exponents in terms of positive integers.

xⁿ = x₁ * x₂ * ... * x_n

It is also straight forward to define exponents for 0, negative integers, and rational numbers. I wrote more about this in a previous blog.

But what about the complex numbers? What does it mean to have a value such as 5ⁱ. In fact, a very well known equation is Euler's Identity (see here):

e^iπ + 1 = 0

To handle this, we need a new definition for exponents that is consistent with their use for rational numbers but which can also handle the complex domain.

One way is to use the Maclaurin Series. In a previous blog, I showed that the Maclaurin series is:

f(x) = f(0) + (x/1!)f'(0) + (x²/2!)f''(0) + (x³/3!)f'''(0) + ... + (xⁿ/n!)fⁿ(0)


Now, if we define an exponent as a function so that e^x becomes:

f(x) = e^x

Applying derivatives, we get:

f(0) = e⁰ = 1

And likewise, all fⁿ(0) = 1

Substituting these results into the above equation gives us:

e^x = 1 + (x/1!) + (x²/2!) + ... + xⁿ/n!

Now a^y = e^y*ln(a) since:

(a) e^ln(a) = a [See here for details on ln if needed]

(b) ln(a^y) = y*ln(a) [See here for detalis on ln if needed]

So, we can use the equation for e^x to define a^y

a^y = e^y*ln(a) = 1 + (y*ln(a))/1! + (y*ln(a))²/2! + ... + (y*ln(a))ⁿ/n!

So for example,

1³ = 1 + (3*ln(1))/1! + (3*ln(2))²/2! + ... + =

= 1 + (0)/1! + (0)/2! + ...

2¹ = 1 + (1*ln(2))/1! + (1*ln(2))²/2! + ... + =

1 + ln(2) + ln(2)*ln(2)/2 + ln(2)*ln(2)*ln(3)/6 + ...

= 1 + 0.693... + 0.240... + 0.0555... + ... = 2

Finally, let's ask the question about aⁱ, which equals:

aⁱ = 1 + (i*ln(a)) + (i*ln(a))²/2! + ... + (i*ln(a))ⁿ/n!

= 1 + i*ln(a) - [ln(a)]²/2 -i*[ln(a)]³/3! + [ln(a)]⁴/4! + ...

= (1 - [ln(a)]²/2 + [ln(a)⁴/4! + ...) + i(ln(a) - [ln(a)³/3! + ln(a)⁵/5! + ...)

radians

The idea of radians is to define degrees in terms of pi. In a previous blog, I showed how Archimedes' proof for the area of a circle can be used to establish pi.

Definition Radian: There are 2π radians in a circle.

In other words 2π radians = 360 degrees.

A straight line has π radians and a right angle has π/2 radians.

This is useful because it enables trigonometric functions to be expressed in terms of taking π as a value.

Here are some examples:

sin (π/2) = 1
sin (π) = 0
sin(2π) = 0

cos (π/2) = 0
cos(π) = -1
cos(2π) = 1

Lemma 1: sin(nπ) = 0 where n is any integer

Proof:

(1) sin(0) = 0 and sin(π)=0 [See Property 1, here]

(2) sin(x + &2pi;) = x [See Property 5, here]

(3) So, we can see that:

if n is even, then n = 2x and sin (nπ) = sin(x*2π) = sin(0 + x*2π) = sin(0) = 0

if n is odd, then n = 2x+1 and sin(nπ) = sin(x*2π + π) = sin(π) = 0

QED

Lemma 2: The area of a sector = (1/2)θr²

Proof:

(1) We know that the area of a circle = πr² (see Corollary 2, here)

(2) An angle of a circle = (θ/2π) of a circle.

(3) So therefore, the area of a sector = (θ/2π)(πr²) = (1/2)θr²

QED

Euclid's Proof for the Pythagorean Theorem

In today's blog, I will go over Euclid's proof for the Pythagorean Theorem as presented in Euclid's elements. There is a very large number of different proofs for this result. The simplest involve tiles that show how squares of the two sides can be refitted to form a square on the hypotenuse. One proof was discovered by American president, James Garfield.

Theorem: In a right triangle, the hypotenuse squared is equal to the sum of the other sides squared.





























Proof:

(1) Since BAC is a right angle, we can extend AC to make a square GFBA and we can extend AB to make a square ACKH. (See here for details on the construction)

(2) We can also build a square based on BC. (See here for details on the construction)

(3) Let AL be a line that is parallel to BD and CE. (See here for details on the construction)

(4) ∠ DBA ≅ ∠ FBC since ∠ FBA and ∠ DBC are right angles and we get this result if we add ∠ ABC to both.

(5) triangle DBA ≅ triangle FBC by Side-Angle-Side (see here) since:

(a) ∠ DBA ≅ ∠ FBC (#4)

(b) FB ≅ AB since they are sides of the same square.

(c) BC ≅ BD since they are sides of the same square.

(6) The parallelogram BL is double the triangle ABD since they have the same base (BD) and since they are in the same parallel. (see Corollary 3.1 here for details)

(7) Likewise, the parallelogram GFBA is double the triangle FBC for the same reason as step #6.

(8) From step #5, we can conclude that the parallelogram BL is congruent to the parallelogram GFBA.

(9) We can follow the same steps to show that parallelogram CL is congruent to parallelogram ACKH since:

(a) ∠ KCB ≅ ∠ ACE since ∠ BCE, ∠ ACK are right angles and we can add ∠ ACB to each.

(b) AC ≅ CK and BC ≅ CE since the sides of the same square are congruent.

(c) So that triangle KCB ≅ triangle ACE by Side-Angle-Side.

(d) parallelogram CL is double the area of triangle ACE

(e) parallelogram CA is double the aera of triangle KCB

(f) So therefore parallelogram CL ≅ parallelogram CA

(10) So that we have the square of BC is congruent to the square of AB added to the square of AC since:

(a) the square BDEC = parallelogram BL + parallelogram CL

(b) parallelogram BL ≅ square of AB

(c) parallelogram CL ≅ square of AC

QED

Corollary 1: In a right triangle, the hypotenuse is longest of the three sides of a triangle.

Proof:

(1) Assume that the hypotenuse h is equal or less than one side s₁

(2) Then, h² is less than s₁² + s₂²

(3) But this contradicts the Pythagorean Theorem so we can reject our assumption.

QED

Corollary 2: sin²(θ) + cos²(θ) = 1.

Proof:

(1) From the main theorem, for any right triangle with base sides x and y and with hypotenuse z, we have:

x² + y² = z²

(2) If θ is the angle between z and x, then we have (see here):

sin θ = y/z

cos θ = x/z

So that:

sin²(θ) = y²/z²

cos²(θ) = x²/z²

(3) Now if we divide the equation in step #1, by z² on both sides we get:

x²/z² + y²/z² = 1.

(4) Now inserting the values in step #2 gives us:

sin²(θ) + cos²(θ) = 1.

QED

Archimedes and the area of a circle

In today's blog, I will go over Archimede's proof that the area of a circle is (1/2)rc. When this proof is combined with Euclid's proof (most likely from Eudoxus), it is possible to show that for all circles, the ratio of C/D is constant.

Postulate 1: For a given chord of a circle, the segment of the circumference is longer than the chord.













In the example above, BDC is longer than BC.

Postulate 2: For a given segment of the circumference, lines connected above it combined are longer.





























In the example above, AG + GE is greater than the segment of the circumference AE.

Lemma 1: The area of a regular polygon is (1/2)h*Q where Q is the sum of the perimeter.





















Proof:

(1) A regular polygon can be divided up into a sum of triangles.

(2) The area of each triangle is equal to (1/2)*(GH)*(DE) [See here for details if needed]

(3) So, the area of the polygon = (# of sides)*(1/2)*(GH)*(DE)

(4) Let Q = the sum of the bases (for example, for the hexagon above, Q = DE + EF + FA + AB + BC + CD)

(5) Then, the area of the polygon = (1/2)*(GH)*Q

QED


Theorem: The area of a circle is (1/2)circumference * radius

Proof:

(1) Let K = (1/2)*C*R where C = circumference and R = radius.

(2) Let A = the area of the circle.

(3) Assume that A is greater than K

(4) We can inscribe a polygon inside A that is greater than K and less than A. [See Lemma 2 and the Method of Exhaustion here for details]

(5) So, the area of the polygon = (1/2)*Q*h where h is the distance from the center to the base and where Q is the perimeter of the polygon. [See Lemma 1 above]

(6) But Q is less than C (see Postulate 1 above) and h is less than R.

(7) So we have Area Polygon = (1/2)Q*h which is less than (1/2)*C*R

(8) But this contradicts step #4 so we reject step #3.

(9) Now, let's assume that K is greater than A.

(10) We can circumscribe a polygon P around A such that P is greater than A but less than K. [See Lemma 3 and Method of Exhaustion here for details.]

(11) From Lemma 1 again, we know that the area of this polygon is (1/2)*Q*h where Q is the perimeter of the polygon and h is the height.

(12) In the case of the circumscribed polygon (see diagram for Postulate 2), h = R.

(13) Using Postulate 2 above, we see that Q is greater than C.

(14) But then the area of the polygon is greater than K since (1/2)*Q*R is greater than (1/2)*C*R

(15) But this contradicts step #10 so we reject our assumption at step #9.

(16) Now, we apply the Law of Trichomoty (see here) and we are done.

QED

Corollary 1: For all circles, the ratio of Circumference to Diameter is constant

Proof:

(1) We know from Euclid (see here) that for any two circles C₁ and C₂ that:

Area₁/Area₂ = (Diameter₁)²/(Diameter₂)²

(2) We know from Theorem 1 above that:

Area of circle = (1/2)*radius*circumference.

(3) Combining step #1 with step #2 and using R=(1/2)D gives us:

[(1/2)*(1/2)D₁*C₁]/[(1/2)*(1/2)D₂*C₂] = [D₁]²/[D₂]²

(4) Canceling out (1/4)/(1/4) gives us:

[D₁*C₁]/[D₂*C₂] = [D₁]²/[D₂]²

(5) Multiplying both sides by (D₂/D₁) gives us:

C₁/C₂ = D₁/D₂

which means that:

C₁*D₂ = C₂*D₁

and finally that:

C₁/D₁ = C₂/D₂

QED

Definition 1: π

From the Corollary, we know that the ratio of the circumference to the diameter is constant. π is this ratio from circumference to diameter. In other words, π = C/D.

Corollary 2: The area of a circle is πr².

Proof:

(1) From the theorem, the area of a circle is (1/2)(circumference)(radius)

(2) From the definition above, π = C/D

This means that C = D*π = 2*r*π

(3) Putting this all together gives us:

area = (1/2)(circumference)(radius) = (1/2)(2*r*π)(r) = πr²

QED

References

• Archimedes, The Measure of a Circle
• William Dunham, Journey Through Genius
• David Joyce, Euclid's Elements