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In 1964, Dr. Donald Bentley, Professor of Statistics, Pomona College, presented the following paradoxical proof: ”Why all numbers are equal to 47” (as recalled by David Hart, Pomona ’92)
Reference Sketch © Dr. Troglodyte 22AUG13
Dr. Troglodyte
ETA: Restored Image Link 13APR18/Dr. Troglodyte
“Quis custodiet ipsos custodes?“ - Decimus Junius Juvenalis
“Numero, Pondere et Mensura“
Edited 1 time(s). Last edit at 13-Apr-18 20:14 by Dr. Troglodyte.
Quote
”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.”
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”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).”
Dr. Troglodyte
ETA: Restored Image Link 13APR18/Dr. Troglodyte
“Quis custodiet ipsos custodes?“ - Decimus Junius Juvenalis
“Numero, Pondere et Mensura“
Edited 1 time(s). Last edit at 13-Apr-18 20:14 by Dr. Troglodyte.
Subject | Views | Written By | Posted |
---|---|---|---|
47 - Dr. Bentley’s Proof | 6408 | Dr. Troglodyte | 23-Aug-13 00:47 |
Re: 47 - Dr. Bentley’s Proof | 343 | spitfire888 | 23-Aug-13 00:58 |
Re: 47 - Dr. Bentley’s Proof | 296 | Dr. Troglodyte | 23-Aug-13 01:14 |
Re: 47 - Dr. Bentley’s Proof | 298 | spitfire888 | 23-Aug-13 01:16 |
Ahatmose get in here, my bro will get to the bottom of this | 222 | spitfire888 | 23-Aug-13 01:20 |
Re: 47 - Dr. Bentley’s Proof | 252 | spitfire888 | 23-Aug-13 01:30 |
Tom's proof Dr. Troglodyte mail me a check $74.00 I will mail you $47.00 | 264 | spitfire888 | 23-Aug-13 01:45 |
Re: 47 - Dr. Bentley’s Proof | 253 | Ahatmose | 23-Aug-13 03:48 |
Re: 47 - Dr. Bentley’s Proof | 291 | molder | 24-Aug-13 09:04 |
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