Math Newsletter number 12;Wednesday, October 13, 2010.

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Constructing Algebraic Identities

In first year algebra class two fundamental algebraic

identities are discovered.

For all integers, rational numbers, real numbers, complex

numbers, A, B,

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

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

In the first identity, replacing B by -B leaves the identity

essentially unchanged.

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

In the second identity, replacing B by -B makes it look

slightly different.

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

The first identity is spoken, in English, as

The product of the difference and sum is the difference of

the squares.

The second identity is spoken, in English, as

The square of the sum is the sum of the squares plus twice

the product.

or

The square of the difference is the sum of the squares minus

twice the product.

What makes these identities interesting.

These two identities are the foundation algebraic identities

for many algebras in which product A * B = product B * A.

In later issues, I will discuss algebras in which the product

depends on the order in which the multiplication is done.

For this issue, we focus only on algebras in which the

product does not depend on the order in which the

multiplication is done.

Example:

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

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

1 * 3 = 2**2 - 1**2

(7-1)*(7+1) = 7**2 - 1**2

6 * 8 = 7**2 - 1 = 48

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

(20 + 1)**2 = 20**2 + 2 * 20 * 1 + 1**2

21**2 = 400 + 40 + 1 = 441

21**2 - 10**2 = 441 - 100 = 341

21**2 - 10**2 = (21 - 10)*(21 + 10) = 11 * 31

11 * 31 = 341

From these two identities, we may also derive higher order

identities.

(A-B)(A+B)(A-B)(A+B) = (A-B)(A-B)(A+B)(A+B)

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

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

Note that one of the factors in

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

is the difference of A**2+B**2 and 2AB,

and the other factor is the sum of A**2+B**2 and 2AB.

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

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

From above, we repeat the identity,

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

Combining these two identities,

we get,

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

which can also be put in the form,

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

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

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

3**2 + 4**2 = 5**2

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

5**2 + 12**2 = 13**2

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

15**2 + 8**2 = 17**2

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

7**2 + 24**2 = 25**2

It is fairly easy to multiply two integers together.

11 * 31 = 341

293 * 3413 = (300 - 7) * (3400 + 13)

= 300 * 3400 - 7 * 3400 + 300 * 13 - 7 * 13

= 1020000 - 23800 + 3900 - 91

= 1020000 + 3900 - 23800 - 91

= 1023900 - 23800 - 91

= 1000100 - 91

= 1000009

More difficult is the problem of determining two integers

that multiply together to result in a given integer.

Given 1000009, find two integers x, y, such that x y =

1000009.

Of course, we just multiplied 293 and 3413 together to get

1000009, so we know the answer.

If we did not know that 293 divided 1000009, how could we

find out?

Factoring, the process of finding two integers that multiply

together is more difficult than multiplying together the two

integers found.

Factoring can be viewed as solving the equation

x y = z, for some know integer z, and unknown integers x and

y to be found.

Consider a more general looking equation,

M N + B m + C N = D

where B, C, and D are given, and M and N are to be found.

Example: M N + 13 M - 7 N = 1000100

Would you expect this equation to be more difficult than

factoring 1000009?

In fact, it is exactly as difficult as factoring 1000009.

Proof:

M N + 13 M - 7 N = 1000100

subtract 7 * 13 = 91.

M N + 13 M - 7 N - 7 * 13 = 1000100 - 91 = 1000009

M (N + 13) - 7 (N + 13) = 1000009

(M - 7) (N + 13) = 1000009

Since we know that 293 * 3413 = 1000009

we can set

M-7 to 293 or set M-7 to 3413.

Thus one solution is

M = 300, N = 3400.

Another solution is

M = 3420, N = 280.

In General,

M N + B m + C N = D

Add B C to both sides of the equation.

M N + B M + C N + B C = D + B C

(M + C) (N + B) = D + B C

Factor D + B C = x y

One solution is

M = x - C, N = Y - B

Another solution is

M = x - B, N = Y - C