This is Math Newsletter number 3; Wednesday, August 11, 2010.
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Algebraically derived numbers

Ye have heard prophets of old say unto you, "It's very
important to distinguish between numbers and names of
numbers," but I say unto you, "Names of numbers are also
numbers!"

This means that in algebra class when we write

(A + B) * (A + B)
= (A + B) * A + (A + B) * B
= A*A + B*A + A*B + B*B
= A**2 + 2*A*B + B**2

where the A**2 means A squared,

we are describing the properties of the numbers,
A and B.

We may imagine that A and B are numbers of a type,
not yet created. They are not necessarily integers,
or real rationals, or complex number rationals, or
square roots of complex number rationals.

We may imagine that A and B have no relationship to the
numbers that we have previously created. Yet, the property
shown for A and B, that

(A + B) * (A + B)
= A**2 + 2*A*B + B**2

is true for the integers, is true for the rational numbers,
real, imaginary, and complex.

It is because this property,

(A + B) * (A + B)
= A**2 + 2*A*B + B**2

depends only on the relationship between multiplication and
addition, expressed in the "distributive law of
multiplication over addition".

One favorite number within algebraic textbooks is the number
x.

x is sometimes a name of an integer, or rational number,
or the square root of a rational number.

Sometimes x has no given relationship to any number we have
prevously discussed.

Early in the first year algebra class, the student encounters
equations such as:


x + 5 = 9. Solve for x.



The student is expected to interpret this equation as:


What do I add to 5 to get 9?


To answer this question, the student subtracts 5 from 9 to
get:


x = 4


Near the end of first year algebra class the student is
introduced to quadratic equations.


Solve x**2 + 3 x - 10 = 0


The standard way to solve this problem is to ask, "What two
numbers have difference 3, and product of 10?"


We examine the factors of 10: 2 * 5 = 10. And we note that
2 and 5 have a difference of 3. Thus we know that

(x-2)*(x+5) = (x-2)*x + (x-2)*5
= x*x - 2*x + 5*x -2*5
= x**2 + 3 x - 10

Formally we solve this equation,

x**2 + 3 x - 10 = 0
by "factoring" x**2+3x-10.

There is also another way to look at this quadratic equation.

Put it in the form:

x**2 + 3 x = 10

x * (x+3) = 10

So now the equation is the question:

A number * that number plus 3, is equal to 10.
what is the number?

So we see immediately that one solution is

x = 2.

2 * (2 + 3) = 2 * 5 = 10

Another solution is

x = -5.

(-5) * (-5 + 3) = (-5) * (-2) = 10


Look again at the equation


x * (x+3) = 10.

It says that the product of two numbers is 10.

The product of two numbers is also the difference of two
squares.

The product of (A - B) and (A + B) is calculated as

(A - B) * (A + B)
= (A-B)*A + (A-B)*B
= A*A - B*A + A*B - B*B
= A**2 - B**2

We can convert the equation

x * (x+3) = 10

from a question about a product into a question about a
difference of squares.

Suppose x = A - B
and (x+3) = A + B

Then adding equations we get

2 A = x + x + 3 = 2 x + 3

Dividing by 2, we get

A = x + 3/2

Since x = A - B,

we also have

B = A - x = (x+3/2) - x = 3/2.

Putting this all together we get

x * (x+3)
= (A-B)*(A+B)
=A**2 - B**2
=(x+3/2)**2 - (3/2)**2

and the equation

x*(x+3) = 10

becomes

(x+3/2)**2 -(3/2)**2 = 10

Solving for (x+3/2)**2, we get

(x+3/2)**2 = 10 + (3/2)**2 = 10 + 9/4 = 49/4 = (7/2)**2

Leading to one solution,

x + 3/2 = 7/2 and a second solution

x + 3/2 = -7/2.

corresponding to

x = (-3/2 + 7/2) = 4/2 = 2
and
x = (-3/2 - 7/2) = -10/2 = -5.



Now I want to talk about x in polynomials.

X is a number, not related to any other number previously
defined.

x + 1 is therefore a new number.

x**2 is therefore a new number.

x**2 + x + 1 is therefore a new number.

etc

(x-1)*(x+1)=(x-1)*x+(x-1)*1 = x**2-x+x-1=x**2 - 1

(x-1)*(x**2 + x + 1)
=(x-1)*x**2 + (x-1)*x + (x-1)*1
= x**3 - x**2 + x**2 - x + x - 1
= x**3 - 1

(x-1)*(x**3 + x**2 + x + 1)
= (x-1)* x**3 + (x-1) * x**2 + (x-1)* x + (x-1)* 1
= x**4 - x**3 + x**3 - x**2 + x**2 - x + x - 1
= x**4 - 1

etc

This pattern continues, no matter how high an exponent we put
on x.

This relationship,

(x-1)*(1 + x + x**2 + x**3 + .... + x**n)
= x**(n+1) - 1, for each positive integer n,
is also show in the equation obtained by dividing through by
x-1.

1+x+x**2+x**3+...+x**n = (x**(n+1))/(x-1).

In second year algebra, this view of the relationship is
called the formula for a geometric series.

1 + 2 = (2**2 - 1)/(2-1) = (4-1)/1 = 3

1+2+2**2 = (2**3-1)/(2-1) = (8-1)/1 = 7

1+2+2**2 + 2**3 = (2**4 - 1) / (2-1) = (16 - 1)/(2-1) = 15


1 + 3 + 3**2 + 3**3 = (3**4 - 1)/(3 - 1)

1 + 3 + 9 + 27 = (81 - 1)/2

1 + 3 + 9 + 27 = 80/2 = 40.



Look again at the sequence of polynomials,

x + 1

x**2 + x + 1

x**3 + x**2 + x + 1

etc


If we set any of these polynomials to 0, then we are refining
the meaning of the number x.

x + 1 = 0 implies that
x = -1

x**2 + x + 1 = 0 implies that
x = (-1 + sqrt(-3))/2 or
x = (-1 - sqrt(-3))/2

x**3 + x**2 + x + 1 = 0 implies that
x = sqrt(-1) or
x = -1 or
x = -sqrt(-1).


We will revisit these polynomials in later issues.

Kermit Rose