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The Irony of Mitt Romney's 47% Tape-- Star Trek Style! (Original Post)
Segami
Nov 2012
OP
wandy
(3,539 posts)1. Funny, I always thought the answer was 42. See! Romney WAS wrong!
Jack Rabbit
(45,984 posts)2. Everyone knows the answer is 42 . . . but what is the question?
Bill Drat
(11 posts)3. Hilarious
I liked it
MrNice100
(28 posts)4. Draw an isosceles triangle, with points ABC...
A mathematical proof, written in 1964 by Professor Donald Bentley, supposedly demonstrates that all numbers are equal to 47. However, Bentley offered it as a "joke proof" to further a popular student research project that listed real and imaginative "47 sightings". Bentley used the invalid proof to introduce his students to the concept of mathematical proofs. Joe Menosky graduated from Pomona College in 1979 and went on to become one of the story writers of Star Trek TNG.
en.wikipedia.org/wiki/47_(number)
So here's the proof:
Draw an isosceles triangle, with points ABC, with line AC
representing the "base" of the triangle. By definition, side AB and
side BC are equal length, and angle BAC and angle BCA are equal to
each other.
Assign the smaller of whatever two numbers you intend to prove to
the base, line AC. Let's choose...oh, I don't know, maybe 47.
Assign the larger number (let's say 74) to the pathway A-B-C. In
other words, divide the larger number by 2 and assign that (37 in
our example) to line AB; by definition, line BC is then also 37.
So, the distance along line AC is 47. The distance along the path
>from A to B to C equals 74.
Next, bisect each of the three sides. Assign points D,E,F to the
midpoints of lines AB, BC, and AC respectively. Draw lines DF and
EF. Now, by definition, line AF and line FC are equal (both equal
to 23.5, half of 47). Also by definition, line AD and line EC are
equal (both half of 37, therefore 18.5). We already established
that angles BAC and BCA are equal to each other, therefore by the
geometry rule "side-angle-side", triangles ADF and FEC are congruent
triangles. Therefore, line EF and line DF are also congruent, and
both are equal to 18.5.
Now, step back and answer the question: What is the total length of
the path along the line ADFEC? Well, it's four congruent lines,
each equal to 18.5, for a total of 74. This is the same length as
the original path along A to B to C.
Now, bisect lines AD, DF, FE, EC, AF, and FC. Label the midpoints
as follows: midpoint of AD = G. AF = H. DF = I. FE = J. FC =
K. EC = L. Now draw the lines GH, GI, JK, and KL. By the same
geometric rules and problem solving, triangles AGH, HIF, FJK, and
KLC are all congruent, and every line AG, GH, HI, IF, FJ, JK, KL,
and LC is equal to 1/2 of 18.5 = 9.25. Now the total path along the
line AGHIFJKLC is still 74.
If you continue this process, making infinitely smaller isosceles
triangles, and still calculating the path along the line of these
triangles from A to C, you still get 74. Now, here's the crux of
the argument: In the *limit*, you arrive at two paths along the
route from point A to point C. The original line AC is still 47.
But, the path from point A to point C along the other route is
always equal to 74. Therefore, 47 and 74 are equal to each other.
By the same logic, any number can be shown to be equal to 47 (or any
other number).
This all gets down to very obscure issues about limit theory and
calculus, and the old problem of "if you start out two feet from
the wall, and every time you take a step you get halfway to the
wall, do you ever reach the wall?"
hrmjustin
(71,265 posts)5. Welcome to DU!