A bit off-topic, but in a recreational-math kind of way. Plus,
I think we all need to step back a few paces from this USWest
thing, take a deep breath, regain our senses of humor (assuming
we had such to start with :) ...
My sister forwarded the following to me. The example problem
of the 1970s reminded me of something I experienced in elementary
school, so I couldn't resist some worked example in that vein -
perhaps you can come up with further "solutions" to the new math
problem I describe.
-----------------
Date: 13 Sep 98 07:43:42
From:Jooyun J Oh <[EMAIL PROTECTED]>
To:David Julie <[EMAIL PROTECTED]>
Subject:Math through the ages
For the US-educated geeks, with an (absurd) sense of humor ...
Teaching Math in 1950:
A logger sells a truckload of lumber for $100. His cost of
production is 4/5 of the price. What is his profit?
Teaching Math in 1960:
A logger sells a truckload of lumber for $100. His cost of
production is 4/5 of the price, or $80. What is his profit?
Teaching Math in 1970:
A logger exchanges a set "L" of lumber for a set "M" of money.
The cardinality of set "M" is 100. Each element is worth one
dollar. Make 100 dots representing the elements of the set "M."
The set "C," the cost of production, contains 20 fewer points
than set "M." Represent the set "C" as a subset of set "M" and
answer the following question: What is the cardinality of the set
"P" for profits?
Teaching Math in 1980:
A logger sells a truckload of lumber for $100. Her cost of
production is $80 and her profit is $20. Your assignment:
Underline the number 20.
Teaching Math in 1990:
By cutting down beautiful forest trees, the logger makes $20.
What do you think of this way of making a living? Topic for
class participation after answering the question: How did the
forest birds and squirrels feel as the logger cut down the trees?
There are no wrong answers.
Teaching Math in 1996:
By laying off 40% of its loggers, a company improves its stock
price from $80 to $100. How much capital gain per share does
the CEO make by exercising his stock options at $80? Assume
capital gains are no longer taxed, because this encourages
investment.
Teaching Math in 1997:
A company outsources all of its loggers. The firm saves on
benefits, and when demand for its product is down, the logging
work force can easily be cut back. The average logger employed
by the company earned $50,000, had three weeks vacation, a nice
retirement plan and medical insurance. The contracted logger
charges $50 an hour. Was outsourcing a good move?
Teaching Math in 1998:
A laid-off logger with four kids at home and a ridiculous alimony
from his first failed marriage comes into the logging-company
corporate offices and goes postal, mowing down 16 executives and
a couple of secretaries, and gets lucky when he nails a
politician on the
premises collecting his kickback. Was outsourcing the loggers a
good move for the company?
Teaching Math in 1999:
A laid-off logger, serving time in Folsom for blowing away
several people, is being rehabilitated as a COBOL programmer in
order to work on Y2K projects. What is the probability that the
automatic cell doors will open on their own as of 00:01,
01/01/2000?
Julie Oh
IBM Corporate Treasury
tel; 914 499 4011, t/l 641
fax: 914 499 4004
-----------------
Now, to my deeply traumatic grade school experience alluded to
in the header:
>Teaching Math in 1970:
>A logger exchanges a set "L" of lumber for a set "M" of money.
>The cardinality of set "M" is 100.
This is pretty close to capturing much of the "new math" craze
of the 70s. I remember in 5th grade at Arnold School once, they
let three or four of the brightest math students out of the math
class, gave us a new math textbook supposedly suitable for 5th
graders, put us in the currently vacant LD (learning disabled)
classroom, and told us to read through the book together for
two weeks and then tell the staff what we thought. (Apparently
they couldn't make heads nor tails of it, either - read on.)
Anyway, we opened the book, and on the very first page was the
following "problem" (no solution provided, probably because
to have a solution implies that one actually has a well-posed
problem, or worse, that one might actually come up with a wrong
answer, which is irreversibly damaging to one's self-esteem):
--------- /\
| | / \ (imagine a circle here,
| | / \ but that's tough to draw
| | / \ using ascii characters).
--------- ----------
Question: what equation does the above represent?
I'm not kidding - they gave us a picture of a square, a triangle
and a a circle, and the above question - no further instructions.
Anyway, we looked on in disbelief for about five minutes, then
had a good laugh and spent the remainder of the two weeks playing
checkers and with the A/V equipment in the room. I'm sure we
learned more than an actual class using the book would have.
In retrospect, there are many possible answers to the above question.
Here are a few samples:
1) 4 + 3 = 0, where we do the addition modulo the prime 7, said prime
modulus being equal to the number of words in the "Question" line.
2) 4 * 3 = 0, where we do the multiplication modulo 12, since most
people have 12 fingers (double-counting thumbs).
3) A square, an equilateral triangle, a circle - these are the three
simplest equal-sided polygons, where we treat the circle as a 2-sided
polygon, each side having been bent into a semicircle due to gravitational
lensing by a massive but invisible intervening object. Of course this
is not strictly speaking an equation, but hey, that's using constricted
old-math thinking.
4) Three cookie cutters
Laid side by side on a table
The still-pristine dough quivers.
Of course this is not strictly speaking a haiku, but hey, that's using
constricted old-poetry thinking.
5) Assuming all three objects have unit height, we observe:
- The square has unit side length, hence perimeter = 4 and area = 1.
- The triangle has unit height hence side length = x = sqrt(1+x^2/4).
The solution of this equation is x = sqrt(4/3), hence the triangle has
perimeter = 3*sqrt(4/3) = 3.464... and area = sqrt(4/3)/2 = .577... .
- The circle has unit diameter, hence perimeter = pi = 3.141... and
area = pi/4 = .785... .
Thus the perimeters of the objects, taken from left to right, form a
decreasing sequence, but their areas do not. A very deep result.
6) This is obviously the statement of Fermat's last theorem: the
two leftmost objects have 4 and 3 sides, respectively, and the
circle represents the empty set (we eschew the normal crossed-
circle notation as being too cluttered), so the statement should
be read (from right to left, of course) as
"An example of the empty set is the set of solutions in the positive
integers of the equation x^3 + y^3 = z^3 and x^4 + y^4 = z^4.
We leave the student to show that this is also the empty set
for all larger exponents - the margins of this page should be
wide enough for this purpose."
7) The circle can be considered as the limiting case of a regular
convex polygon as the number of sides tends to infinity. Thus, we
have three polygons, with respective sums of interior angles (reading
from left to right, of course) of a = 2*pi, b = pi, and c = infinity.
These three arguments satisfy the obviously intended equation
(a + b)/6 = arctan(c).
8) The circle can be considered as an abstract representation
of a 2-sided convex polygon (see solution #3 above).
The sum of interior angles of a regular convex polygon having N
sides is (N-2)*pi, so the respective sums of interior angles of
the three polygons (reading from left to right, of course) are
a = 2*pi, b = pi, and c = 0. The coeffcients of pi in a, b and
c, respectively, are 2, 1 and 0.
These three arguments satisfy the obviously intended relation
2 = 10, or, in words,
"the number two, written with respect to any base greater than
or equal to three (the number of objects depicted), equals the
binary number 10."
What could be more obvious?
Regards,
Ernst
