Hello, I am a Pomona College alumnus, and took an advanced calculus
class from Dr. Bentley. During class, he presented his proof, which
led to a lively discussion about its validity or lack thereof. I
was checking out the 47Society's webpage for the first time, and
read there that the Society has not yet been able to get a copy of
the proof. Here it is. Keep in mind, this is simply intended to be
a proof that any two numbers can be shown to be equal to each
other. He chose to prove the equality of all other numbers to 47
based on the already existing, and growing, Pomona College 47 lore.

Disclaimer: I am not a mathemetician, and any fault in the
explanation is based on either my inability to correctly undestand
it, or the fuzziness of my recollection (let's just say I was in Dr.
Bentley's class significantly more than a decade ago). Furthermore,
this proof is extremely simple to draw, but very hard to describe in
prose. Bear with me.

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 (at least to people like me)
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?" All I can
recall is that in the end, none of us students could present a valid
mathematical argument to convince Dr. Bentley that he was full of it.

To the best of my recollection, this is how he proved the "all
numbers are equal to 47" business. I remember thinking it was
pretty elegant. If anyone has a suggestion for how I could upload a
diagram of this, I'd be happy to. If this really is the first
posting of this proof, I'd just appreciate credit for bringing it to
light here, though obviously none of the credit for coming up with
it in the first place.

Cheers, and by the way, two of the last three street addresses I've
lived at, my driver's license, my medical license, my cell phone
number, and my son's social all contain....."47".

David Hart (Pomona '92).