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