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