## Monday, December 19, 2005

### Review of Matrices

In today's blog, I will review some very basic results in 2x2 and 1x2 matrices.

This represents a very basic introduction that is meant to provide background for my larger blog on Fermat's Last Theorem: n = 5 (see here).

Today's blog is based on the work by Harold M. Stark in his book An Introduction to Number Theory.

1. Matrix defined

A matrix is a grouping of numbers that allows working on all the numbers at the same time.

For example, let's consider a 2 x 2 matrix that can be based on a set of numbers: 1, 2, 3, 4.

The matrix itself looks like this:

2. Addition and subtraction of matrices

Addition and subtraction of matrices are exactly the same as if you added and subtracted the numbers independently:

3. Multiplication of Numbers with Matrices

Multiplication with an integer just applies the integer to all the values involved so that:

4. Product of Two Matrices

In addition to these properites, matrices have there own special operations. The product of 2 matrices is a bit confusing. We define a product of a 1 x 2 matrix with a 2 x 2 matrix as the following:

We define a product a 2 x 2 matrix with a 2 x 2 matrix as the following:

Now, here's where it gets a bit confusing. We normally refer to a matrix using a capital letter. So let's say we have two matrices A,B such that: A is a 2x2 matrix and B is a 2x2 matrix. We cannot assume that AB = BA. For example, if we reverse the matrices above, we get the following equation:

Another important point is that there is no product defined for a 2x1 matrix and a 2x2 matrix or a 2x2 matrix and 1x2 matrix (since order is important in matrix products) and for that matter, there is no product defined a 2x2 matrix with a 1x2 matrix. In the case of 2x2 matrices, you can only get a product for a 2x2 matrix with a 2x2 matrix or a 1x2 matrix with a 2x2 matrix.

5. Determinant

A determinant is a value that is derived from a 2x2 matrix. Here is the definition:

Lemma 1: det(AB) = (detA)(detB)

(1) Let A =

Let B =

(2) AB =

(3) det(AB) = (ae+bg)(cf+dh) - (af+bh)(ce+dg) = (acef + adeh + bcfg + bdgh) - (acef + adfg + bceh + bdgh) = adeh + bcfg - adfg - bceh.

(4) det(A) = ad - bc
(5) det(B) = eh - fg
(6) So det(A)det(B) = (ad - bc)(eh - fg) = adeh + bcfg - adfg - bceh

QED

6. Identity Matrix

The Identity Matrix is referred to as I and defined as:

Lemma 2: AI = IA = A

(1) Let A =

(2) AI =

(3) IA =

QED

7. Inverse

Let A =

We denote the inverse of A as A-1 and we define it as:
A-1 =

Lemma 2: AA-1 = A-1A = I

(1)

(2)

QED

Lemma 3: det A-1 = 1/(det A)

(1) (det A)(det A-1) = det(AA-1) [From Lemma 1]

(2) det(AA-1) = det(I) [From Lemma 2]

(3) det(I) = 1*1 - 0*0 = 1. [Definition of I, Definition of Determinant]

(4) So, (det A)(det A-1) = 1

(5) And dividing both sides by (det A) gives us:
det A-1 = 1/(det A)

QED

7. Final Points

The last point here is that while AA-1 = I, it is not necessarily true that ABA-1 = B. The reason is that AB does not necessarily equal BA and we are not allowed to change the order of the matrix elements.

## Monday, October 10, 2005

### Basic Lemmas Needed for FLT: n = 5

Here are some basic lemmas that are used by the proof for Fermat's Last Theorem: n=5.

Lemma 1: (p+q)5 + (p-q)5 = 2p(p4 + 10p2q2 + 5q4)

(1) Using the Binomial Theorem:
(p + q)5 = p5 + 5p4q + 10p3q2 + 10p2q3 + 5pq4 + q5
(p - q)5 = p5 - 5p4q + 10p3q2 - 10p2q3 + 5pq4 - q5

(2) Adding these two values together gives us:
2p5 + 20p3q2 + 10pq4 = 2p(p4 + 10p2q2 + 5q4)

QED

Lemma 2:

If:

t = q4 + 50q2r2 + 125r4
u = q2 + 25r2
v = 10r2

Then:

t = u2 - 5v2

(1) u2 = (q2 + 25r2)2 = q4 + 50q2 r2 + 625r4

(2) -5v2 = -5(10r2)2 = -500r4

(3) (q2 + 25r2)2 + -5(10r2)2 = q4 + 50q2 r2 + 625r4 + -500r4 = q4 + 50q2 r2 + 125r4

QED

## Sunday, October 09, 2005

### Some Simple Division Lemmas

In this blog, I want to outline some simple implications of the division which are very useful in basic mathematical reasoning:

Lemma 1:

If:

(a) c = a + b
(b) d divides a
(c) d divides b

Then:

d divides c

(1) By assumption (b) and (c), we know that there exists a',b' such that:

a = a'd
b = b'd

(2) So,

c = (a'd) + (b'd) = d(a' + b')

Lemma 2:

If

(a) c = a + b
(b) d divides c
(c) d divides a

Then

d divides b

(1) c = a + b → b = c -a = c + (-a)

(2) This results follows from Lemma 1 above.

## Monday, September 19, 2005

### Mathematical Induction

Mathematical induction is a method for proving a condition is true of an infinite set that obeys the well-ordering principle.

A set is considered well-ordered if for any subset of elements, one can always find one element which is the smallest. An example of a well-ordered set is the set of positive integers. One element is clearly the smallest. For example, if 1 is an element in the set, then it will be the smallest element.

Not all sets are well-ordered. For example, negative integers are not well-ordered. In such an infinite set, there is never one element which is the smallest. Likewise, real numbers are not a well-ordered set. For any number, it is possible to find another number which is smaller if only by a fraction. There is no smallest value.

It is useful to introduce some notation when talking about well-ordered sets. Sets are often refered to as capital letters such as S, T, R. Elements are often refered to as lowercase letters such as s,t,r.

We can imagine that s is an element of S, t is an element of T, and r is an element of R. Likewise, we can use a subscript to differentiate elements so that s1, s2, etc. are all elements of S. t1, t2, etc. are all elements of T.

When talking about induction, we are always trying to prove some proposition true about an entire set. When talking about a proposition or fact, I will use the following form p(). So, for example, p(s1) means that the proposition is true of the first element of the set S. Likewise p(S) means that the proposition is true of all elements of set S.

Finally, I need to use the concept of implication which is symbolized by . Implication doesn't say something is true or false. Rather, it describes a relationship. If a certain condition is true, then another condition follows. If I were president, I would be living in Washington, D.C. This doesn't mean that I am president and it doesn't mean that if I am living in Washington, D.C., then I am the president. It simply means that if the first condition were true (if I were the president), then the second condition would follow (then I would live in Washington, D.C.).

With this notation, I can now state the Principle of Mathematical Induction:

Theorem: Principle of Mathematical Induction: if p(s1) is true and if p(sn) → p(sn+1), then p(S).

(1) Let's start by assuming that the theorem is false.

(2) Let's let si be the first element in order where p(si) is not true.

(3) Well i cannot be the first element since we are assuming that p(s1) is true.

(4) So, this means that p(si-1) must be true since i is the first element where the proposition is not true.

(5) But if p(sn) is true, then p(sn+1) is true, so therefore p(si) must be true since p(si-1) is true.

(6) But this is a contradiction so we reject (1) and conclude that the theorem is true.

QED

## Tuesday, August 30, 2005

### Useful Equations

I thought it would be useful to review some basic equations. These are presented as a series of lemmas.

Lemma 1: (a + b)2 = a2 + 2ab + b2

(a + b)2 = (a + b)(a + b) = a(a + b) + b(a + b) = a2 + ab + ab + b2 = a2 + 2ab + b2.

QED

Lemma 2: (a + b)(a - b) = a2 - b2

(a + b)(a - b) = a(a - b) + b(a - b) = a2 - ab + ab - b2 = a2 - b2

Corollary 2.1: a4 - b4 = (a - b)(a + b)(a2 + b2)

By Lemma 2, a4 - b4 = (a2 - b2)(a2 + b2)

Applying Lemma 2 again, gives us:
a4 - b4 = (a - b)(a + b)(a2 + b2)

Lemma 3: a3 + b3 = (a + b)(a2 - ab + b2)

(a + b)(a2 - ab + b2 ) = a(a2 - ab + b2) + b(a2 - ab + b2) =
a3 - a2b + ab2 + a2b -ab2 + b3 = a3 + b3

Corollary 3.1: a3 - b3 = (a - b)(a2 + ab + b2)

Let b' = -b so that a3 - b3 = a3 + (b')3

By Lemma 3: a3 + (b')3 = (a + b')(a2 - ab' + (b')2) = (a - b)(a2 + ab + b2)

QED

Lemma 4: a5 + b5 = (a + b)(a4 - a3b +a2b2 - ab3 + b4)

(a + b)(a4 - a3b +a2b2 - ab3 + b4) =
=a(a4 - a3b +a2b2 - ab3 + b4) + b(a4 - a3b +a2b2 - ab3 + b4) =
=a5 - a4b + a3b2 - a2b3 + ab4 + a4b - a3b2 + a2b3 - ab4 + b5 =
=a5 + b5

QED

Lemma 4: n ≥ 5 → an + bn = (a + b)(a(n-1) - a(n-2)b + ... - ab(n-2) + b(n-1))

(a + b)(a(n-1) - a(n-2)b + ... -ab(n-2) + b(n-1)) =

a(a(n-1) - a(n-2)b + a(n-3)b2... -ab(n-2) + b(n-1)) +
b(b(n-1)+a(n-1) - a(n-2)b + ...+a2b(n-3) -ab(n-2)) =

an - a(n-1)b + a(n-2)b2 + ... -a2b(n-2) + ab(n-1) +
bn + a(n-1)b - a(n-2)b2 + ... + a2b(n-2) - ab(n-1) =
an + bn

QED

Lemma 5: a - b divides an - bn

Proof:

(1) Let n = be the highest number where this is true.

(2) We know that n is at least 2 since a2 - b2 = (a - b)(a + b)

(3) To complete this proof, we need to show that a - b divides an+1 - bn+1

(4) We know that (a-b)(an) = an+1 - anb and we know that (a-b)(bn) = abn - bn+1

(5) So that an+1 - bn+1 - (a - b)(an) - (a-b)(bn) = anb - abn = ab(an-1 - bn-1)

(6) So that:

an+1 - bb+1 = (a-b)(an + bn) - ab(an-1 - bn-1)

(7) Since n ≥ 2, we know that a-b divides an-1 - bn-1.

QED

### Factorials

A factorial is an equation that is abbreviated by a number followed by an ! such as:
2!

The idea behind a factorial is very simple. It is a shorthand for a series of multiplications where each multiple is one less than the previous value.

For example:

1! = 1
2! = 2 * 1 = 2
3! = 3 * 2 * 1 = 6
4! = 4 * 3 * 2 * 1 = 24

Factorials are often used in problems of permutations.

Let's start with a situation where we won't need factorials. Assuming that we have 5 roles and 5 people and each person can take any role and can take as many roles as they want. How many different permutations are there?

There are 5 ways to divide up the first role, 5 ways to divide up the second role, and so on. The answer turns out to be: 5 * 5 * 5 *5 *5 = 55 [If you need a review of exponents, review here]

What if we say that each person can only take one role. In other words, we are taking 5 people and dividing them up into 5 roles. In this case, the following happens:

There are 5 ways to divide up the first role, but only 4 ways to divide up the second role, and then 3 ways to divide up the third role, etc. The answer turns out to be 5 * 4 * 3 * 2 * 1 = 5!.

We can generalize this point and make it into an exact equation. For any set of n elements,
there are exactly n! ways of ordering it. This is easy to show.

Lemma 1: For a set of n elements, there are n! ways of ordering all n elements.

For the first element, we have n possible choices.
For the second element, we have n-1 possible choices.
And so on until we get the last element and we have only 1 choice.

So, the total number of thoices is n * (n-1) * (n-2) * ... * 1 which equals n!

QED

Lemma 2: For a set of n elements, there are n!/(n-m)! ways to to order m elements.

(1) There are two cases that need to be considered.

Case I: m = n
Case II: m is less than n

(2) For m = n, we know that there are n! ways to order the elements (Lemma 1)

n!/(n-n)! = n!/(0!) = n!/1 = n!. (Note: 0! is defined to equal 1)

This proves Case I.

(3) For m is less than n, we start by dividing all combinations by the first m elements.

(4) When we do this, we find that for each m sequence of elements, there are (n-m)! ways to select the remaining elements after we have selected the first m elements. (Lemma 1)

(5) Now, if each of these groups have (n-m)! elements, then we can get the total number of groups by diving n! by this value, that is: n!/(n-m)!.

QED

Lemma 3: For a set of n elements, there are n!/[m!(n-m)!] ways to select an unordered set.

(1) We can prove this using the same two cases:

Case I: m = n

Case II: m is less than n

(2) For Case I, there is only 1 way to select n items from a set of n elements.

n!/[n!(n-n)!] = n!/[n!0!] = n!/n! = 1.

(3) For case II, we take our n!/(n-m)! sets and organize them into groups where they have the same set of elements. For example, if we had 3 elements, would group {1,2,3} in the same group as {2,1,3} and {3,1,2} and {3,2,1}.

(4) For each group of m elements, there are m! ways to order them (Lemma 1)

(5) This means that the total number of groups in (3) is [n!/(n-m)!]/m! which is mathematically the same as: n!/[m!(n-m)!].

QED

Factorials are also used in the very important Binomial Theorem which I will talk about in a future blog.

## Saturday, August 06, 2005

### Modular Arithmetic

Modular arithmetic is a notation and set of mathematics that were first introduced by Carl Friedrich Gauss.

The major insight is that equations can fruitfully be analyzed from the perspective of remainders. Standard equations use the '=' sign. Modular arithmetic uses the '' sign. Two values that are '≡' to each other are said to be congruent relative the modulus. In the case below, the modulus is 3.

Here's an example of a modular equation:

7 ≡ 1 (mod 3).

By definition, this means that 3 divides 7 - 1.

Here's a more formal definition of a modular equation:

Definition 1: a ≡ b (mod c) if and only if c divides a - b.

This definition tells us the following is true:

7 ≡ 1 ≡ 10 ≡ -2 (mod 3).

Now, one of the most interesting things about '' is that it follows many of the same relations as '='.

Lemma 1: For any value a,b,c,d,n where a ≡ b (mod n) and c ≡ d (mod n):
(a) a + c ≡ b + d (mod n)
(b) a - c ≡ b - d (mod n)
(c) ac ≡ bd (mod n)

(1) n divides a - b, and n divides c - d. [definition of ≡ ]
(2) We know that n divides (a + c) - (b + d) since this is equal to: (a -b) + (c - d).
(3) We know that n divides (a - c) - (b - d) since this is equal to: (a - b) - (c - d).
(4) We know that n divides ac - bd since this is equal to : c(a - b) + b(c - d).

QED

Lemma 2: If a ≡ b (mod n) then:
(a) a + c ≡ b + c (mod n)
(b) a - c ≡ b - c (mod n)
(c) ac ≡ bc (mod n)

(1) So, we are given that n divides a - b.
(2) We know (a) since n divides a + c - (b + c) = a - b.
(3) We know (b) since n divides a - c - (b - c) = a - b.
(4) We know (c) since n divides ac - bc = c(a - b)

QED

Corrolary 2.1: a ≡ d (mod n), b ≡ e (mod n), c ≡ f (mod n), then:
a + b + c ≡ d + e + f.

(1) We know that a + b ≡ d + e from above.
(2) We therefore know that (a + b) + c ≡ (d + e) + f.

QED

Lemma 3: a + b + c ≡ 0, a ≡ 0 (mod p), then b + c ≡ 0 (mod p).(1) a + b + c ≡ 0 (mod p) [Definition of ]

(2) b ≡ c (mod p) → a + b ≡ a + c (mod p) [See above]

(3) So, 0 ≡ a + b + c ≡ 0 + b + c ≡ b + c (mod p).

QED

Lemma 4:  if gcd(k,p)=1 and kx ≡ ky (mod p), then x ≡ y (mod p)

Proof:

(1)  Since kx ≡ ky (mod p), there exists a such that ap = kx - ky = k(x-y)

(2)  Since gcd(k,p)=1, it follows that p divides (x-y)

(3)  So that x ≡ y (mod p)

QED

## Wednesday, July 27, 2005

### Euclidean Coprime Integers: xn + yn = zn

In a previous blog, I showed that given rational integers x,y,z such that:
xn + yn = zn

We can assume that x,y,z are relatively prime. See here for the details.

It turns out that we can make the same statement about all Euclidean Integers. For those not familiar with Euclidean Integers, please review here.

Lemma 1: For Euclidean Integers, relatively prime divisors of n-powers are themselves n-powers.

This theorem says that if gcd(v,w) = 1 and vw = zn
Then, there exists x,y such that v = xn, w = yn

(2) Assume that v is not equal to any number xn
(3) v ≠ 1 since 1 is an xn power
(4) Now, v is divisible by a prime number p. [Fundamental Theorem of Arithmetic for Euclidean Integers]
(5) So, there exists k such that v = pk
(6) p divides z since zn = vw = pkw [By applying Euclid's Lemma for Euclidean Integers]
(7) So, there exists m such that z=pm
(8) So, zn = vw = pkw = (pm)n = pnmn
(9) Dividing p from both sides gives us:
kw = p(n-1)mn
(10) From Euclid's Lemma, p divides k or w.
(11) It can't divide w since it already divides v and gcd(v,w)=1. Therefore, it divides k
(12) We can apply this same argument for each p in p(n-1)
(13) So, we can conclude that p(n-1) divides k.
(14) So, there exists V such that k = p(n-1)*V
(15) So, kw = p(n-1)mn = p(n-1)*V*w
(16) Dividing p(n-1) from both sides gives us:
vW = mn
(17) Now, gcd(V,w)=1 since V is a divisor of v and gcd(v,w) = 1
(18) Likewise, V cannot be an n-power. If it were, then v = pnV would make v an n-power which goes against our assumption.
(19) Finally, V is less than v since p(n-1) > 1.
(20) Thus, we have a contradiction by infinite descent.

QED

Lemma 2: Given xn + yn = zn and x,y,z are Euclidean Integers, we can assume that x,y,z are relatively prime.

To prove this, we will need to prove two things:

(1) If a factor divides any two values of this equation, then the n-power of it divides the n-power of the third value.

(2) If an n-power of a factor divides the n-power of a value, then the factor divides the value itself.

Step 1: For xn + yn = zn, the n-power of any common factor of two divides the n-power of the third.

Case I: Let's assume d divides x, d divides y

(1) There exists x', y' such that: x = d(x'), y = d(y')
(2) zn = xn + yn = (dx')n + (dy')n
= dn(x')n + dn(y')n
= dn[(x')n + (y')n]

Case II: Let's assume d divides z and d divides x or d divides y

(1) Let's assume d divides x (the same argument will work for y)
(2) There exists x', z' such that: x = d(x'), z = d(z')
(3) We now say yn = zn - xn
(4) We can now follow the same reasoning as above.

QED

Step 2: dn divides xn → d divides x

(1) Let c be the greatest common denominator (gcd) for d,x.
(2) Let D = d / c, X = x / c.
(3) Now the gcd of (X,D) = 1. [See here for the proof.]
(4) So, the gcd of (Xn,Dn) = 1.
(5) We know that there exists k such that xn = k * dn [Since dn divides xn ]
(6) Applying (2), we get (cX)n = k*(cD)n
(7) Which gives us: cnXn = k * cnDn
(8) Dividing cn from each side gives: Xn = Dn*k
(9) Now it follows that gcd(Dn,k) = 1.
(a) Assume gcd(Dn,k) = a, a > 1
(b) Then, a divides Dn and Xn [From 8]
(c) But gcd(Dn,Xn) ≠ 1.
(e) So, we reject our assumption.
(10) So, we can conclude that k is an n-power. [By the lemma above]
(11) Which means that there exists u such that un = k.
(12) And we get Dn * un = Zn
(13) And (Du)n = Zn
(14) Implying that Du = Z and multiplying by c that du=z.
(15) Which proves that d divides z.

QED

### Euclid's Method for the Greatest Common Denominator

The greatest common denominator (gcd) is the largest common factor shared by two or more positive integers. In the case of 4,2 the greatest common denominator is 2. Since all integers are divisible by 1, the greatest common denominator of any two integers is guaranteed to be at least 1. If the gcd of two integers is 1, those integers are said to be relatively prime or coprime. In other words, they do not have any common divisors.

It is perhaps obvious on the face there exists a gcd for any two integers and there exists a method for finding this value. Here is the proof. The method for finding the greatest common denominator is found in Euclid's Elements and is therefore known today as Euclid's algorithm. Here is a link to Euclid's proof in the Elements.

It turns out that if there is a division algorithm available, then for any two integers, there is necessarily a greatest common denominator. In other words, we can prove that Euclid's method for finding the gcd applies. Here is the proof (note: in the example referenced, it refers to Gaussian Integers, but the same proof applies to Eisenstein Integers and other types of quadratic integers that are characterized by a division algorithm).

For this reason, all quadratic integers that are characterized by a division algorithm are said to be Euclidean.

One result of this is that for any two Euclidean integers (quadratic integers that have a division algorithm) that are not relatively prime, it is possible to derive two smaller integers which are relatively prime.

Lemma: gcd(x,y)=d is greater than 1 → there exists X,Y such that x = Xd, y = Yd and gcd(X,Y)=1.

(1) Assume that gcd(X,Y) = D which is greater than 1.
(2) Then D divides X,Y such that there exists X = DX', Y=DY'
(3) And x = DX'd, y = DY'd
(4) So that Dd divides both x and y.
(5) But Dd > d which is impossible since d is the greatest common divisor.
(6) So we reject (1).

QED

Corollary: This result holds for all quadratic integers that are Euclidean.

This is true since the result only depends on the gcd which we showed above holds for all Euclidean Integers.

QED

Lemma: gcd(x,y)=1 → gcd(xn,yn)=1.

(1) Assume that gcd(xn,yn) = d which is greater than 1.
(2) Since d is greater than 1, there must exist a prime p that divides d. [See here for the proof of Fundamental Theorem of Arithmetic]
(3) if p divides xn, then p divides x. [See here for the proof of Euclid's Generalized Lemma]
(4) For the same reason, p would also divide y.
(5) But this is a contradiction since gcd(x,y)=1.
(6) So we reject (1).

QED

Corollary: This holds true for all Euclidean Integers

This proof depends on a proof for unique factorization and Euclid's Generalized Lemma.

(1) Euclid's Generalized Lemma holds true for all Euclidean Integers. [See here for proof].
(2) Unique Factorization holds true for all Euclidean Integers. [See here for proof].

QED

## Friday, July 22, 2005

### Division Algorithm

One of the most important and underappreciated theorems is the Division Algorithm. You can usually find it in any book on number theory as Theorem 1. The proof for the Division Algorithm for Integers can be found here.

It states that for any given integer and nonzero divisor, there exists two unique integers: a quotient and a remainder where the remainder is smaller than the divisor.

Perhaps this theorem gets underappreciated because it is so obvious. It does not seem on the surface that there is anything surprising about this statement. Likewise, the proof, when it is first seen, may appear to the amateur like an exercise in the obvious.

The standard division algorithm for rational integers (positive and negative whole numbers) is based on the Well Ordering Principle.

And yet, there are subtleties regarding this theorem that have important implications for the theory of numbers. Consider the situation where we add imaginary numbers (numbers in which -1 has a square root known as i, that is, i2 = -1.

When a mathematician looks at a problem like x2 + y2, it would be really convenient if there was an easy way to factor it. If we extend integers to include Z[i], then we can. In this case, we have the following factor (note: Z[i] is used to define an extended set set of integers. These are integers that have the form a + bi where a,b are rational integers):

(x - iy)(x + iy) = x2 + y2.

And, then the question arises, are integers of the type Z[i] characerized by a Division Algorithm? If they are, we need a new proof because the Well-Ordering Principle only applies to rational integers.

The path to a Division Algorithm requires the introduction of a special function known as a norm. It applies to any Z[α] value where α2 is a rational integer (such as i). It is based on the very simple identity:

(a - b)(a + b) = a2 - b2

Since we know that α2 is rational integer, we know that there exists a value c such that c = α2.

So, let us assume that we have an integer of the form a + bα. If we multiply it with its conjugate (an equivalent value where we change the sign), then we get:

(a + bα)(a - bα) = a2 - cb2.

If we had a value of a - bα, then its conjugate would be a + bα. In either case, we have found a function that maps any Z[α] integer to a rational integer.

With this function in place, we can prove for that given any Gaussian Integer and a nonzero divisor, there exists two unique values: a quotient and a remainder where the absolute norm of the remainder is smaller than the absolute norm of the divisor. Here is the proof.

We might ask if we will always find a division algorithm for all possible values of α.

Let us consider this question with respect to quadratic integers. These are integers of the form Z[α] where α is a irrational solution to an equation of the form x2 + bx + c = 0 where x = α and b,c are rational integers (an irrational number is any number that is not equal to a rational fraction. For example -3/4 is rational since it is a ratio of -3 to 4. i on the other hand is not rational).

In this situation, it turns out there are only 21 different nonsquare values where Z[α] has a division algorithm (by nonsquare, I am excluding values such as Z[√-4] because 4 is a square of 2 and -4 = 2i.

The 21 values are based on: -11, -7, -3, -2, -1, 2, 3, 5, 6, 7, 11, 13, 17, 19, 21, 29, 33, 37,41, 57, and 73.

Now, here's one more of the subtleties. It turns out that the form of this integer depends on the remainder when the number is divided by 4. If the remainder is any value but 1, then the form is simple: Z[√-2], Z[i], Z[√2],Z[√3], Z[√6],Z[√7],Z[√11], Z[√19].

If the remainder with 4 is 1, then the value takes a stranger form: Z[(-1 + √-11)/2], Z[(-1 + -7)/2], Z[(-1 + √-3)/2], Z[(-1 + √5)/2], Z[(-1 + √13)/2], Z[(-1 + √17)/2],
Z[(-1 + √
21)/2], Z[(-1 + √29)/2], Z[(-1 + √33)/2]. The details for why these values have this form can be found here.

The proof that many of these values are characterized by a Division Algorithm can be found here.

## Wednesday, June 15, 2005

### Fractions

A fraction is any ratio between whole numbers. In mathematical terms, it is called a rational number. An irrational number is any number such as π which cannot be represented as the ratio of two whole numbers.

In today's blog, I will go over a single lemma regarding fractions:

Lemma: for any given rational number let's say a/b, there exists an integer let's say c such that: absolute(a/b - c) ≤ (1/2).

(1) To prove this, we need only consider the case where abs(a) is greater abs(b). [If abs(a) ≤ abs(b), then the conclusion follows from Corollary 2.1, here]

(2) From the division algorithm (see Theorem 1, here), we know that a = bq + r where r ≥ 0 and less than abs(b).

[For example, if a=-3, b=-2, then q=2, r=1 where r is greater than b but less than abs(b).]

(3) Let a'=abs(a), b'=abs(b)

(4) We know that a' - b'q is less than b' (since a' - b'q = r and r is less than b')

(5) So, it follows that: a'/b' - q is less than 1.

(6) Now, if both a,b are positive or a,b are negative, it follows that a/b = a'/b' and abs(a/b - q) is less than 1.

(7) If a,b are of different sign, than -a/b = a'/b' and -a/b - q = -(a/b + q) so that abs(a/b + q) is less than 1.

(8) The conclusion follows from Lemma 2, here.

QED

Corollary: if a/b is a rational number, then there exists an integer c such that: absolute(a/b - c/2) ≤ (1/4).

(1) From the lemma above, we know that for 2*(a/b), there exists a number c such that:
abs(2*(a/b) - c) ≤ (1/2).

(2) Dividing both sides by 2, gives us:
abs(a/b - c/2) ≤ (1/4)

QED

Now, it turns out that any number with a repeating decimal can be represented as a rational number.

(1) Let's assume that we have a decimal such as 5.234523452345... We can represent this decimal as a repeating decimal such as 5.2345.

(2) Now, we know if we multiply the number by 104 we get:

52345.2345

(3) So, subtracting (2) by (1) gives us:

104 - 1 = 52345 - 5 = 52340.

(4) So, the rational form of this repeating decimal is:

52340/9999.

Now, let's look at the proof that demonstrates this:

Lemma: Any number with a repeating decimal is rational.

(1) Any number with a repeating decimal can be represented with the following form:

d1...dm.a1...an

NOTE: If a number has a nonrepeating portion, then we multiply this number by the number of nonrepeating digits, to get a number of the above form. Later, we divide our result by this same number.

(2) We can get an integer result by subtracting 10n*the number by the original number which after canceling for the repeating decimal gives us:

10n * (d1...dm.a1...an...) - (d1...dm.a1...an...) =
(d1..dma1..an) - (d1..dm).

(3) Now, our rational number is equal to the value in step #2 divided by 10n - 1.

QED

## Sunday, May 15, 2005

### Coprime Numbers: More Results

In this blog, I will review some basic results regarding coprime numbers.

Lemma 1: p,q coprime -> pq, p2 - q2 are coprime.

(1) Assume gcd(pq,p2 - q2) = d and d is greater than 1

(2) So, there is some prime P which divides d [Fundamental Theorem of Artihmetic]

(3) Since d divides pq. P either divides p or q. [Euclid's Lemma]

(4) Let's assume P divides p and the same argument will apply if it divides q. So that, there exists R such that p = PR.

(5) Since d divides p2 - q2, there exists a value Q such that p2 - q2 = PQ

(6) This means that q2 = p2 - PQ = (PR)2 - PQ = P(PR2 - Q).

(7) But this means that P divides q by Euclid's Lemma which is a contradiction since P also divides p. [Because gcd(p,q)=1]

(8) Therefore, we reject our assumption.

QED

Lemma 2: p,q are relatively prime and are of different parity (one is odd, one is even), then p + q, p - q are relatively prime.

(1) Assume that gcd(p+q,p-q) = d and d is greater than 1.

(2) Then there exists a prime x which divides d. [Fundamental Theorem of Arithmetic]

(3) We know that this x is odd since p+q and p-q are odd. [Since they are of different parity]

(4) Now x divideds 2p since 2p = (p + q) + (p - q) which means x divides p. [By Euclid's Lemma Since x is odd]

(5) Likewise x divides 2q since 2q = (p + q) - (p - q) = p + q - p + q which means x divides q. [Same reason as above]

(6) But this is a contradiction since p,q are relatively prime.

QED

Lemma 3: if S2 = P2 + Q2 and P2 = T2 + Q2, then Q,S,T are relatively prime.

(1) We can assume that S,P,Q are relatively prime and P,T,Q are relatively prime. [See Lemma 1, here for details]

(2) We can also assume that S,P are odd which means that Q is even and that T is odd. [See details above]

(3) From the two equations, we can derive that:
S2 = T2 + 2Q2

(4) Now we need only prove that any factor that divides 2 values, will divide the third.

Case I: f divides S,Q

(a) There exists s,q such that S = fs, Q = fq
(b) T2 = S2 - 2Q2 = f2 [ s2 - 2q2 ]
(c) Which means that f divides T [See here.]

Case II: f divides T,Q

(a) We can use the same argument as Case I.

Case III: f divides S,T

(a) There exists s,t such that S = fs, T = ft
(b) 2Q2 = f2[ s2 - t2 ]
(c) Since S,T are odd, f must be odd and s,t must be odd.
(d) Then, s2 - t2 must be even.
(e) Then, there exists u such that 2u = s2 - t2
(f) So, we have:
2Q2 = (2)u(f2)
(g) And canceling out the 2s gives us:
Q2 = f2u
QED

Lemma 4: S,T relatively prime, both odd, 2u = S - T, 2v = S + T, then u,v are relatively prime.

(1) Assume f divides u,v such that u = fU, v = fV.
(2) The f divides S
S - T + S + T = 2S = 2u + 2v = 2f(U + V)
(3) And f divides T
S + T - (S - T) = 2T = 2v - 2u = 2f(V - U)l.
(4) This contradicts our S,T beings coprime so we reject our assumption.

QED

## Monday, May 09, 2005

### Fundamental Theorem of Arithmetic

Before, we can provide this theorem, we will need an initial theorem on division and one lemma which will we borrow from Euclid.

Theorem 1: Division Algorithm for Integers

This theorem shows that given any two integers, there exist a unique divisor and a unique remainder.

(1) Let S be a set such that { a - bc : c is an integer and a-bc ≥ 0}

(2) S is not empty.

Case I: b divides a

In this case, then there exists a c where a - bc = 0. Therefore, 0 is an element of this set.

Case II: b doesn't divide a

In this case, there exists an c where a - bc has a remainder is greater than 0. This remainder is then an element of this set.

(3) Then, then there exists d such that d is the smallest element of the set. [Well Ordering Principle]

(4) We know that d is less than b.

(a) Assume that d is greater than or equal to b.

(b) Then d - b is greater than or equal to 0.

(c) Since

d - b = (a - bc) - b = a - bc - b = a - b(c+1)

(d) We conclude that a - b(c+1) must be an element of S.

(e) But a - b(c+1) is less than d

(f) Which is a contradiction since d is the smallest element from step #3.

(5) Finally, for any a,b, the lowest value d and the remainder c are unique.

(a) Assume C,D are integers that can be derived from a,b.

a = bc + d, d is greater than 0 and less than b.
a = bC + D, D is greater than 0 and less than b.

(b) So

bc + d = bC + D --> b(c - C) = d - D

(c) Which means that b divides (d - D).

(d) Since d - D is less than b, we know d - D = 0 and that d = D.

(e) Which gives us b(c - C) = 0 or that c - C = 0.

(f) We can therefore conclude that c = C.

QED

Lemma 2: Euclid's Lemma

This lemma shows that given a prime that divides a product, either it divides one value or it divides the other.

(1) Let's assume a prime p divides a product ab.

(2) Let's assume that p does not divide a.

(3) Then gcd(p,a)=1. [See blog on Greatest Common Denominators for details]

(4) There exists s,t such that 1 = as + pt. [See blog on Greatest Common Denominators]

(5) Multiplying both sides with b gives us:

b(1) = b(as + pt) = b = abs + ptb

(6) Since p divides ab, there exists k such that ab = pk.

(7) Combining with (5), we get

b = pks + ptd = p(ks + td).

QED

We can also tag on a corollary.

Corollary 2.1: Generalized Euclid's Lemma

The idea here is that if a prime divides a product of n elements, then it is necessarily divides at least one of those elements.

(1) Let's assume that we have a product of n elements.

a = a1 * a2 * ... * an

(2) We can take any one of these elements and get:

a = a1 * (a2 * ... * an)

(3)So, by Euclid's Lemma (see Lemma 2 above), the prime either divides a1 or one of the rest of the elements.

(4) So, either it divides a1 or a2 etc.

(5) And we get to the last two elements, we are done by Euclid's Lemma.

QED

Theorem 3: Fundamental Theorem of Arithmetic

This theorem proves that each integer greater than 1 is the product of a unique set of primes.

(1) Let S be a set S of {primes or products of primes > 1}.

(2) 2 is an element of S since 2 is prime.

(3) From (1), (2) there exists a value n which is 3 or greater where all values less than n and greater or equal to 2 are in S.

(4) If n is a prime, then it is a member of S.

(5) If n is not a prime, then it is also a member of S.

(a) If n is not a prime, then there exist a,b such that n=ab where a,b are greater than 1 and less than n. [Definition of a prime]

(b) But a,b must be elements of S since they are less than n. [From step #3 above]

(c) Then, n must also be a member of S since n is a product of two numbers which are either primes or products of primes.

(6) Now, all the primes that make up n are necessarily unique.

(a) Let n = p1 * p2 * ... pr = q1 * q2 * ... qs where p,q are all primes.

(c) Either p1 divides q1 or q2 or someother value. [see Euclid's Generalized Lemma above]

(d) Likewise, each value pr can be matched by some value qs.

(e) In other words, they must necessarily be the same combination.

QED

## Friday, May 06, 2005

### Greatest Common Divisor

The Greatest Common Divisor or gcd is the largest common factor between two numbers. In what follows, I will write this as function such as gcd(9,6) = 3. The gcd cannot be applied to 0.

For example, the gcd(2,4) = 2 and gcd(5,7)=1.

One assumption that I will use below is the Well Ordering Principle.

Postulate 1: Well Ordering Principle

For any set of positive integers, one value is the lowest.

For a small set of numbers, this principle is clearly true.

For the set of all negative numbers, it is not true. No matter which negative number we choose, we can always find another negative number which is lower.

For the set of all positive numbers, it is true. It is 1. Likewise, it is true for all subsets of this set.

This principle is normally called a postulate. A postulate is an assumption that mathematicians use without proof. It is the foundation of a proof.

Lemma 1: Bezout's Identity: If gcd(a,b) = c, then there exists s,t such that: c = as + bt.

(1) Let S be a set { am + bn : m,n are integers, am + bn > 0 }

In other words, let S be the set of all sums that can be created from am + bn. This includes (1)m + (1)n, (2)m + (1)n, etc. It includes any combination. So, we can also assume that (1000)m + (99)n are elements of this set if the sum is greater than 0.

(2) This set is not empty

In other words, for any a, b, we know that there is at least one element in this set.

If a + b > 0, then choose m=1,n=1.

If a + b = 0, then either a or b is negative. Let's assume a. Then choose m=-1,n=1.

If a + b is less than 0, then either a or b is the lowest. Let's assume a. Then choose m=-1,n=1.

(3) We know that one of these elements in the set is the smallest. [By the Well Ordering Principle]

(4) Let's call the smallest element d.

(5) d divides both a,b.

I will now show that d divides a. The same argument will also apply to b.

We know that there exists q,r such that a = dq + r where r is a positive integer less than d. [By the Division Algorithm]

In other words, q is the quotient and r is the remainder.

r = a - dq = a - (as + bt)q = a - asq - btq = a(1 - sq) - btq = a(1-sq) + b(-tq).

Assume that r > 0.

Then a(1 - sq) + b(-tq) is an element of the above set.

But this is a contradiction. d is the lowest element and yet here is an element smaller than d.

So, r = 0.

(6) d is the gcd for a,b.

Let d' = gcd(a,b)

Let's define h,k: a = d'h, b = d'k

So, d = as + bt = (d'h)s + (d'k)t = d'(hs + kt)

Which means d' divides d.

So, d' = d since d' is the gcd.

QED

Lemma 2: For all pairs of numbers, it is possible to reduce them to a form that is coprime.

(1) Assume that we have two numbers: x, y that are not coprime.
(2) Then there exists d such that d = gcd(x,y) and d > 1.
(3) Then there exists X,Y such that x = dX, y = dY.
(4) And finally, gcd(X,Y)=1
(a) Assume gcd(X,Y) = D and D > 1
(b) Then there exists a,b such that X = aD, Y = bD.
(c) And x = d(aD), y = d(bD)
(d) From this, we see that dD divides both x and y.
(e) We also know dD > d since d > 1 and D > 1.
(f) But this is a condradiction since d is the greatest common denominator.
(g) So we reject our assumption.
QED

## Wednesday, May 04, 2005

### CoPrime Numbers: xn + yn = zn

This blog is about the equation:
xn + yn = zn

I will show that given any three integers that satisfy this equation, either:

(a) all three of them are coprime with each other

or

(b) it is possible to cancel out common components and derive three numbers that are coprime.

Two numbers are coprime if they do not share any common divisors.

Definition 1: Coprime

Two numbers x,y are said to be coprime if and only if d divides x, d divides y -> d = 1

Two numbers x,y are said to not be coprime if and only if there exists a value d > 1 such that d divides x, d divides y

Now, here's the proof that was promised:

Lemma: We can reduce any solution to xn + yn = zn to a form where x,y,z are coprime.

To prove this, we will need to prove two things:

(1) If a factor divides any two values of this equation, then the n-power of it divides the n-power of the third value.

(2) If an n-power of a factor divides the n-power of a value, then the factor divides the value itself.

Step 1: For xn + yn = zn, the n-power of any common factor of two divides the n-power of the third.

Case I: Let's assume d divides x, d divides y

(1) There exists x', y' such that: x = d(x'), y = d(y')
(2) zn = xn + yn = (dx')n + (dy')n
= dn(x')n + dn(y')n
= dn[(x')n + (y')n]

Case II: Let's assume d divides z and d divides x or d divides y

(1) Let's assume d divides x (the same argument will work for y)
(2) There exists x', z' such that: x = d(x'), z = d(z')
(3) We now say yn = zn - xn
(4) We can now follow the same reasoning as above.

QED

Step 2: dn divides xn → d divides x

(1) Let c be the greatest common denominator (gcd) for d,x.
(2) Let D = d / c, X = x / c.
(3) Now the gcd of (X,D) = 1. [See here for the explanation]
(4) So, the gcd of (Xn,Dn) = 1.
(5) We know that there exists k such that xn = k * dn [Since dn divides xn ]
(6) Applying (2), we get (cX)n = k*(cD)n
(7) Which gives us: cnXn = k * cnDn
(8) Dividing cn from each side gives: Xn = Dn*k
(9) Now it follows that gcd(Dn,k) = 1.
(a) Assume gcd(Dn,k) = a, a > 1
(b) Then, a divides Dn and Xn [From 8]
(c) But gcd(Dn,Xn) ≠ 1.
(e) So, we reject our assumption.
(10) So, we can conclude that k is an n-power. [See blog on Infinite Descent for the proof]
(11) Which means that there exists u such that un = k.
(12) And we get Dn * un = Xn
(13) And (Du)n = Xn
(14) Implying that Du = X and multiplying by c that du=x.
(15) Which proves that d divides x.

QED

## Monday, May 02, 2005

### Exponents

1. Introduction to Exponents

An exponent is an elegant shorthand for multiplication.

Instead of 5 * 5 * 5, you can write 53

Instead of 3 * 3 * 3 * 3 * 3 * 3 * 3, you can write 37

The number that gets multiplied is called the base. The number of multiplications that occur is called the power. So, in the above example, 3 is the base and 7 is the power.

Of course, this method only applies when the power is a positive integer. Later on, I will discuss what it means when a power is 0, positive, or even a fraction.

So 42 = 4 * 4 = 16

And 43 = 4 * 4 * 4 = 64

And 41 = 4 = 4

2. x and y notation

In mathematics, when we want to talk about "any", we use a letter such as x or y or z. For example, if we wanted to say that 1 to any power equals 1, we could write this as follows:

1x = 1

Using x-and-y notation, we can create a definition for the positive exponents.

Definition 1: Positive Exponents
xy means x multiplied with itself y times.

x is called the base

y is called the power

3. Multiplication of Exponents

Multiplying exponents of the same base can be determined based on the above definition.

42 * 43 =
= (4 * 4) * (4 * 4 * 4)
= 4 * 4 * 4 * 4 * 4
= 45

So, when exponents get multiplied, if they have the same base, you can add the powers and create a new exponent.

Here are some more examples:

55 * 510 = 515

210 * 21000 = 21010

Of course, this does not work if two exponents have a different base.

In mathematics, a method such as this can be presented as a theorem. A theorem is any statement that can be derived from previous results.

In this case, we are able to prove a theorem regarding the method of adding the powers of the same base. Here's the theorem

Theorem 1: xy * xz = x(y+z)
(1) We know that xy = x multipled to itself y times and that xz = x multipled to itself z times. (Definition of Positive Exponents).
(2) Multiplying all those x's, we have (y + z) x's multiplied together.
(3) Now x multiplied to itself (x + z) times = x(y + z) by the Definition of Positive Exponents.

QED
QED is put at the end of a proof to show it is done. It is an abbreviation for a latin phrase that means basically that the proof is finished. It serves the same purpose in a proof as a period does in a sentence.

4. Division of Exponents

To talk about division, it is useful to introduce the following definition:

Definition 2: Division
a = b / c means a is equal to b divided by c.

a is refered to as the quotient.

b is refered to as the dividend.

c is refered to as the divisor.
Division with exponents of the same base can also be determined based on the definition for positive exponents:

42 / 41 =
= ( 4 * 4 ) / ( 4 ) =
= 16 / 4 = 4
= 41

To divide two exponents of the same base, you simply subtract the two powers.

Here are some examples:

53 / 51 = 52

410 / 45 = 45

Now, what happens if we are dividing by a number greater than the top (in other words, where the divisor is greater than the dividend)? In this case, we are left with a fraction.

51 / 53 = 1 / 52

45 / 410 = 1 / 55

This leads us to a third definition:

Definition 3: Negative Exponents
x(-y) means that we have a fraction of 1 over x multiplied by itself y times.
Here are some examples.

5-1 = 1 / 5

4-3 = 1 / 43

And what happens if the subtraction results in 0?

We can answer this with the following theorem:

Theorem 2: x0 = 1
(1) By basic arithemitic, we know that
x0 = x(1 - 1)
(2) Since 1 - 1 = 1 + (-1), we can rewrite this as:
x(1 + -1)
(3) Now x(1 + -1) = x1 * x(-1) by Theorem 1.
(4) Now, x(-1) = 1/x, by Definition 3.
(5) So, we are left with x * (1/x) = 1

QED
We can also introduce a corollary to this theorem. A corollary is a small proof that is derived directly from the logic of a theorem.

Corollary 2.1: x0 = 1 implies that x ≠ 0
(1) Now x0 = x(1 - 1)
(2) Which means that x0 = x / x
(3) But this implies that x ≠ 0 since division by 0 is not allowed.

QED
Another way of saying this result is that 00 just like 0/0 or even 1/0 is undefined.

We can summarize division of exponents with the following theorem.

Theorem 3: xy / xz = x(y - z)
Case I: y = z

In this case xy / xz = 1 = x0 = x(y - z).

Case II: y > z

In division, we are able to cancel out all the common factors. Since y > z, we cancel out z factors from both dividend and divisor and we are left with x(y-z).

Case III: y < z
Again, we cancel out common factors. Since z > y, we are left with a fraction of
1 / [x(z-y)] which, by definition 3, equals x(-(z-y)) = x(y-z)

QED
5. Fractional Exponents

There is more that we can talk about. What about fractional exponents such as x(1/2)?

It turns out that based on our definitions, corrolaries, and theorems, we are now ready to take on fractional exponent.

We know that x1/2 * x1/2 = x(1/2 + 1/2) by Theorem 1.
Now x(1/2 + 1/2) = x(1) = x.
So x1/2 is none other than the square root of x.

Let's start out by looking at a definition for what a root is.

Definition 4: an nth root of x is a number that multiplied n times equals x.

Sometimes, nth roots are whole numbers. The cube root of 27 is 3 since 3 * 3 * 3 = 27.

Likewise, the 4th root of 16 is 2.

1 is its own 5th root since 1 * 1 * 1 * 1 * 1 = 1.

This gives us our last theorem:

Theorem 4: x1/n = the nth root of x

(1) x1/n multiplied by itself n times equals x1/n + 1/n + 1/n + etc..
(2) Now 1/n + 1/n + etc. n times equals n/n which equals 1.
(3) Therefore x1/n multipled by itself n times equals x1
(4) And this is the very definition of an nth root.

QED