FAILING
ALGEBRA RULES?
M. Mamikon, April 1,
2004
Recent cooperative research at
UCLA and CALTECH has lead to a shocking discovery: the
traditional multiplication rule "FOIL" (Firsts, Outers, Inners, Lasts):
(a + b)(c + d) = ac + ad +bc +
bd
is NOT 'really' true when one of
the numbers, say d, is negative.
In this case, a much simpler
SHORTER rule is found:
(a + b)(c  d) = ac  bd
called "FaiL" meaning
"Firsts
and, indeed, Lasts".
Here are some examples (you can
check them by hand or with calculator):
(3 + 2)(6  4) =
3^{.}6
 2^{.}4 (Check: 5^{.}2 = 18  8 = 10.
Yes! )
(5 + 2)(10  4) =
5^{.}10  2^{.}4 (Check:
7^{.}6
= 50  8 = 42. Yes!)
(6 + 5)(8  20/3) =
6^{.}8
 5^{.}20/3 ( 11^{.}4/3 = 48  100/3 =
44/3. Yes!)
Not only integers and rational
numbers but real and complex ones also "FaiL".

Now educators are NOT very sure
if FOIL always works. Are
you?
Such simplified rules are
possible for any algebraic
formula. Experiments support them
with infinitely many examples. Contact me if you want to know how
they are generated. Of course, one can find some exceptions
to these simplified rules, but according to two well known
proverbs 'there are no rules
without exceptions' and,
moreover, 'exceptions prove the
rule'.
More Failing
RULES

Students just love these new
algebraic rules, because they are easy, natural and have an
intuitive appeal to their early academic experiences.
Everyone is confused with
these new developments  parents, teachers, educators.
Scientists suspect that computers have gone wrong which will
cause the manned Mars Mission to fail. Politicians blame
mathematicians for misleading them to "War On Iraq", and
economists expect the gas prices go up high. Dollar will
fall below penny as the new
algebra rules confirm. Meanwhile,
school districts are seeking ways to keep millions of
excited students out of streets.
The Secret of
Failure
I have thoroughly
investigated what could cause the failure of
traditional rules in Theoretical and Practical
Arithmetic and came to the conclusion that the
problem is hidden in the roots of our numeration
system. We have 10 fingers, hence our counting is
in base10 numeration system. Thus, calculations
are NOT baseinvariant
!!! Therefore there
should be NO universal formula in algebra. All
depends on the base of the numeration system. So,
in our case the base number 10 must be explicitly
involved in the formulas.
Keeping this in mind, we
can easily fix all formulas. For instance the
formula for (a+b)^{2} should be
fixed for any Base of numeration like this:
(a+b)^{2} =
(Base)^{.}a +
b^{2}
Here are examples for
Base=10, Base=20 used
by Basques in Spain and former Soviet Georgians
(they do not wear shoes and count fingers and
toes), and Base=5
(when we use one hand if
the other one is busy):
Base 10:
(a+b)^{2} =
10^{.}a
+ b^{2}

Base 20:
(a+b)^{2} =
20^{.}a
+ b^{2}

Base 5:
(a+b)^{2} =
5^{.}a
+ b^{2}

(8+1)^{2} =
10^{.}8
+ 1^{2} =
81
(6+2)^{2} =
10^{.}6
+ 2^{2} =
64
(4+3)^{2} =
10^{.}4
+ 3^{2} =
49
(2+4)^{2} =
10^{.}2
+ 4^{2} =
36
(12+1)^{2} =
10^{.}12
+ 1^{2} =
121
(14+2)^{2} =
10^{.}14
+ 2^{2} =
144

(18+1)^{2} =
20^{.}18
+ 1^{2} =
361
(16+2)^{2} =
20^{.}16
+ 2^{2} =
324
(14+3)^{2} =
20^{.}14
+ 3^{2} =
289
(12+4)^{2} =
20^{.}12
+ 4^{2} =
256
(10+5)^{2} =
20^{.}10
+ 5^{2} =
225
You can easily continue
down, even for negative numbers

(3+1)^{2} =
5^{.}3
+ 1^{2} =
16
(1+2)^{2} =
5^{.}1
+ 2^{2} =
9
(7+1)^{2} =
5^{.}7
+ 1^{2} =
36
(9+2)^{2} =
5^{.}9
+ 2^{2} =
16
(11+3)^{2} =
5^{.}11
+ 3^{2} =
64
Warning: Chicken have 4
fingers, Goats have 2


I am working now on a universal
computer program "BaseToFace" that will fix Algebra
in all industrial computer systems before the year
2,999 DA (stands for new Dark Ages). Call me for
your promotions.

