01- . . . . . . . . . . . . . . . . Equality of the Sexes . . . . . . . . . . . . . . . .
02- . . . . . . . . . . . . . . . . (An Algebraic Proof) . . . . . . . . . . . . . . . .
03- . . . . . . . . . . . . . . . . by Donald R. Depew PE . . . . . . . . . . . . . . . .
04- Given that the science of Algebra is founded on the principle that an
05- equality remains an equality if both sides of the equation are manipulated
06- the same, we here prove that any female is the equal of any male.
07- Bring any female F together with any male M and equate their sum to the
08- couple C that results:
09- . . . . F+M = C . . . . . . . . . . . . . . . . . . . . .(1)
10- It is natural to presume, until proven otherwise, that the female F is not exactly
11- identical to the male M but that they are different by the degree (F-M). After all,
12- there are obvious differences. The French have even been known to exclaim,
13- "Vive la différence!”
14- Multiply each side of the equation (1) by this presumed difference (F-M):
15- . . . . (F+M)(F-M) = C(F-M). . . . . . . . . . . . .(2)
16- Expand the multiplication
17- . . . . F2-FM+MF-M2 = CF-CM . . . . . . . . . . .(2)
18- Simplify the terms
19- . . . . F2-M2 = CF-CM . . . . . . . . . . . . . . . . (2)
20- Segregate the sexes
21- . . . . F2-CF = M2-CM . . . . . . . . . . . . . . . . (2)
22- Now add the fourth part of a square couple C2/4 to each side of equation (2):
23- . . . . F2-CF+C2/4 = M2-CM+C2/4 . . . . . . . . (3)
24- These expressions, it is here shown, are perfect squares of the form (x-a/2)2:
25- . . . . (x-a/2)2 = (x-a/2)(x-a/2) = x2-ax/2-ax/2+a2/4 = x2-ax+a2/4
26- Therefore:
27- . . . . . F2-CF+C2/4 = (F-C/2)2 and M2-CM+C2/4 = (M-C/2)2
28- And it follows, therefore, that equation (3) can be rewritten:
29- . . . . (F-C/2)2 = (M-C/2)2 . . . . . . . . . . . . .(3)
30- Now, taking the square root of each side of the equation (3), the square-root
31- function [SQRT(~)] will cancel the included exponent-2, square function [(~)2]:
32- . . . . SQRT((F-C/2)2) = SQRT((M-C/2)2) . . . (4)
33- The square-root, canceling the square, leaves:
34- . . . . F-C/2 = M-C/2 . . . . . . . . . . . . . . . . .(5)
35- Then, adding half a couple C/2 to each side of the equation (5) finds:
36- . . . . F-C/2+C/2 = M-C/2+C/2 . . . . . . . . . . (6)
37- Which reduces to:
38- . . . . F = M . . . . . . . . . . . . . . . . . . . . . . (6)
39- This proves algebraically that any female F is the equal of any male M. This
40- unexpected result, which comes as a total surprise to the scientific community,
41- is said to have always been perfectly obvious to large numbers of lay women!
The math above is real. There's no sneaky semantics or tricks of any sort. It's all just elementary algebra. Whereas C is a dependent variable, that is, it's value depends on the values that were assigned independently to F and M, both F and M themselves are independent variables. That is, each of them can have any value one gives it, entirely independent of the other's value, and entirely independent of anything else. Yet the "proof" shows that F is always equal to M for every different value one might give to either.
(A) Of course this cannot be true; the proof cannot be correct. Indeed, there are mistakes in the math. The statements made on some of the numbered lines are untrue. In order to find the coordinates that will lead you to the cache, you must find all of the numbered lines (among all 41 of them) that are untrue. The line numbers are shown as 01-, 02-,...40-, 41-.
(B) Then also, you must find the first, numbered line that, though not untrue itself, is the imprudent line that introduces the likelihood of making those other mistakes, right down to the final untrue result. (So what's so tough about this. Come on now. How many of the lines actually introduce anything!)
(C) Having found these untrue line numbers, and the imprudent line number, first multiply all the untrue line numbers together, then divide that product by the imprudent line number.
(D) Ignoring the decimal point, round off the result to the nearest 6 significant digits. Now multiply that result by 81,869,081. Carry out this multiplication to 11 significant digits and round it off to the nearest tenth digit.
(E) The first and second 5 digits of this product represent the 3-decimal minutes of latitude and of longitude to be added to 42° 00.000 latitude and to 84° 00.000 longitude.
(F) The resulting coordinates are those of a trailhead. Go there, enter the trail, and turning left at the tee, follow the trail clockwise until you find the Female boa constrictor, the one that has got hold of me in the photo. (Take that photo with you!).
(G) Finding her, then look for the Male boa constrictor, who has got a constricting grip of the geocache in his coils about 2 feet off the ground.
It is a microstash! That is, though it is just a 35-mm film can, it is a full-blown regular cache in that it contains a tiny logbook, a tiny pen for signing and dating the logbook, and tiny trinkets for you to trade. See the photo. Because the log book is so tiny, please just sign only your name and the date. Use just two lines, not a full page, and leave no blank lines between your's and the last finder's entry. By all means, do make a more extensive comment later, in your finder's report on this web page.
Also, because this microstash is kind of special, because it's so tiny, please do make a special effort ahead of time to think up and find a really nice and really tiny trinket to trade. Remember, there are a number of other trinkets in there already, so your's must be very tiny indeed to fit in there along with all the others. Don't leave money; that's so unimaginative, and way too big to fit. AND, do bring a towel or something to spread out the trinkets on so you don't lose some in the leaves. They are tiny.
PS: If you take that nice 14 karat gold ring with the big zircon solitaire stone (that I got out of one of Doc Ott's caches), go find and take the ring-box for it too that is hidden at the base of a tree 10-15 feet south of the cache. It ought to still be there somewhere, in a sandwidge bag, but I haven't been able to find it for years.
HINTS
Following are hints in the form of detailed discussions of each line of the "proof", discussions that I'd recommend you have with others to get the solution before you read these. do read them though and solve and find the cache. Good luck.
01- . . . . . . . . . . . . . . . . Equality of the Sexes . . . . . . . . . . . . . . . .
True: This line is just the title of a paper about equality of the sexes, making no claims.
02- . . . . . . . . . . . . . . . . (An Algebraic Proof) . . . . . . . . . . . . . . . .
False: This subtitle, referring to the title, claims that the paper is an algebraic proof. Though it is algebraic, the proof is in error, so it is not an algebraic proof as stated.
03- . . . . . . . . . . . . . . . . by Donald R. Depew PE . . . . . . . . . . . . . . . .
True: The author is Donald R. Depew and it can be shown that he was/is? a professional engineer, registered in Michigan. As you would have no simple way of checking these things, just assume they are truthful, or try it both ways.
04- Given that the science of Algebra is founded on the principle that an
True: This is a preamble accepting the premise that the following "algebra" is founded on some principle, notably the following one, which may or may not be a true principle.
05- equality remains an equality if both sides of the equation are manipulated
True: Hmm, though this line, standing alone, could be False, depending on how the sides of the equation are manipulated, whether the same or differently, the fact that this line is referring to what is being said in the following line, that the manipulation is "the same" for both sides of the equation, it is better construed as being a True line.
06- the same, we here prove that any female is the equal of any male.
False: This line's claim that a proof is accomplished is false because the acclaimed proof is erroneous.
07- Bring any female F together with any male M and equate their sum to the
True: This is merely a definition and naming of the variables.
08- couple C that results:
True: This too is merely a continuation of the definition and naming of the variables.
09- . . . . F+M = C . . . . . . . . . . . . . . . . . . . . .(1)
True: This is a proper statement of the addition of two independent variables to produce the resulting dependent-variable summation if them.
10- It is natural to presume, until proven otherwise, that the female F is not exactly
True: Since no two things are precisely identical, it is indeed natural to presume inequality until proven otherwise.
11- identical to the male M but that they are different by the degree (F-M). After all,
True: Indeed, not identical, so must be different by the difference, that is the net residual, plus or minus, of the constitution of the one taken away from the constitution of the other.
12- there are obvious differences. The French have even been known to exclaim,
True: Yes, obviously.
13- "Vive la différence!”
True: Yes indeed.
14- Multiply each side of the equation (1) by this presumed difference (F-M):
Imprudent: Not False, and perhaps good to advise it, with good results possibly forthcoming, but perhaps imprudent even to advise it, and much more so to actually do it, because by raising the "order" of the equations to now include second-power, squared and multiplied variables, it introduces spurious "roots" that must eventually be carefully sorted out to not cause spurious, erroneous results. OK, studying instruction (B), I think we should agree that this, line 14, is the "first" of the only two "Imprudent lines" that exist and that introduce something, in this case, line 14, the suggestion and direction to do it, and in the second case, line 15, to actually execute it and actually introduce the spurious roots.
15- . . . . (F+M)(F-M) = C(F-M). . . . . . . . . . . . .(2)
Imprudent: but not False to do this. But I suppose this would be the second imprudent line, not the first. On the other hand, until the act is actually done, which takes place here, no spurious roots were introduced yet, so perhaps this, and not the preceding line, is the first line where the imprudent action is really taken.
But, instruction (B) specifically instructs to find the "first numbered line…", not the only one, as if there were only one--so the first imprudent line must be line 14, not line 15, nor as will be seen to be third and fourth candidate imprudent lines, line 22 and line 23. So therefore, with the full emphasis of instruction (B) in mind, let the "first imprudent line" be deemed to be line 14!
16- Expand the multiplication
True: Simple direction to execute the simple operation.
17- . . . . F2-FM+MF-M2 = CF-CM . . . . . . . . . . .(2)
True: Correctly done.
18- Simplify the terms
True: Simple directive to execute the simple operation.
19- . . . . F2-M2 = CF-CM . . . . . . . . . . . . . . . . (2)
True: Correctly done.
20- Segregate the sexes
True: Simple directive to execute the simple operation.
21- . . . . F2-CF = M2-CM . . . . . . . . . . . . . . . . (2)
True: Correctly done.
22- Now add the fourth part of a square couple C2/4 to each side of equation (2):
True: Perhaps Imprudent: A legitimate directive to execute the simple operation, though introducing additional 2nd-power variables, which may be construed to be imprudent (though the 3rd such instance whereas only the 1st instance is significant to the solution).
23- . . . . F2-CF+C2/4 = M2-CM+C2/4 . . . . . . . . (3)
True: Perhaps Imprudent: Correctly done.
24- These expressions, it is here shown, are perfect squares of the form (x-a/2)2:
True: Correct observation and statement and introduction to the following demonstration of its claim.
25- . . . . (x-a/2)2 = (x-a/2)(x-a/2) = x2-ax/2-ax/2+a2/4 = x2-ax+a2/4
True: Using x and a in place of M and C and expanding the squaring multiplication to show that the expressions (x-a/2)2 and are x2 -ax+a2/4 are indeed synonymous.
26- Therefore:
True: Merely and correctly presenting that the expressions in line 23 are synonymous with the perfect-square forms claimed in line 24 to be so.
27- . . . . . F2-CF+C2/4 = (F-C/2)2 and M2-CM+C2/4 = (M-C/2)2
True: Correctly done.
28- And it follows, therefore, that equation (3) can be rewritten:
True: Truly predicted.
29- . . . . (F-C/2)2 = (M-C/2)2 . . . . . . . . . . . . .(3)
True: Correctly restated.
30- Now, taking the square root of each side of the equation (3), the square-root
False: It is OK to direct to and later to actually apply the square root function, but to go on with this direction, referencing the following line, to claim that the square-root function will totally cancel the included square function makes it FALSE because, even though the 1/2, SQRT exponent does indeed cancel the 2, square exponent, it does not also cancel the effects of the extraneous roots introduced by earlier multiplying by the variables.
31- function [SQRT(~)] will cancel the included exponent-2, square function [(~)2]:
False: It is OK to direct to and later to actually apply the square root function, but to go on with this direction, referencing the following line, to claim that the square-root function will totally cancel the included square function makes it FALSE because, even though the 1/2, SQRT exponent does indeed cancel the 2, square exponent, it does not also cancel the effects of the extraneous roots introduced by earlier multiplying by the variables.
32- . . . . SQRT((F-C/2)2) = SQRT((M-C/2)2) . . . (4)
True: It is OK to show that the SQRT is to be taken, but to actually later take it, correctly, the appropriate ones of the extraneous roots must be determined by some means and selected as the correct ones to extract.
33- The square-root, canceling the square, leaves:
False: The description in line 33 of what the so-called "cancelling action" of the SQRT function will show as having been left as will be shown in the following line 34, is false because what is shown there is not truly the total and correct dual effects of both extracting the roots and of selecting the correct ones.
34- . . . . F-C/2 = M-C/2 . . . . . . . . . . . . . . . . .(5)
False: The result of a SQRT function is one of either the + root or the - root. The line should read +/-(F-C/2) = +/-(M - C/2). In this case, to avoid the absurd result of F = M, the signs of the roots should be one of each: either +(F-C/2) = -(M-C/2), or -(F-C/2) = +(M-C/2). After adding C/2 to each, of either of these, everything cancels out and one is left merely with what he started with: F + M = C, and not with F = M.
35- Then, adding half a couple C/2 to each side of the equation (5) finds:
True: This line merely directs what is to be done in the following line and introduces it.
36- . . . . F-C/2+C/2 = M-C/2+C/2 . . . . . . . . . . (6)
False: This line, though resulting correctly from the earlier referenced line, is incorrect both because it is the result of the previous incorrect line, and because one can foresee in it that the impossible equality of two independent variables will be shown.
37- Which reduces to:
True: It is true that what follows, though incorrect in itself, does correctly reduce from the preceding.
38- . . . . F = M . . . . . . . . . . . . . . . . . . . . . . (6)
False: This is false because it is clearly impossible for two stated independent variables to be equal for any and every value which could be assigned to F and to M. Of course there are an infinite number of instances when they do happen to have the same value and are equal. It can also be deemed to be false because it results from earlier equations that were erroneous in their manipulation.
39- This proves algebraically that any female F is the equal of any male M. This
False: It does not prove it. Errors have been shown.
40- unexpected result, which comes as a total surprise to the scientific community,
True: I should think a quick perusal by most anyone of the scientific community, though perhaps, suspicious of the SQRT claim had he paused there, would likely be surprised and amused by the sudden and certainly unexpected result.
41- is said to have always been perfectly obvious to large numbers of lay women!
True, maybe: This is a gag, which taken loosely, and along with the delimitating preface: "This is said to have been…", and with the further delimiting object: "…to large numbers of lay women…", not to all, and perhaps not to the most discriminating.
False, more likely: So even though this was a gag, and though many, women, and men, might agree that many women are equal or better than many men, nobody, women included, could truthfully hold that it is "perfectly obvious" that ANY female is the equal of ANY male, and that is to say, that EVERY female is EQUAL to EVERY male, neither better nor worse. Sorry ladies, but that you may not be equal is not to say that you are inferior to males, just different--and: "Vive la différence!”
Take your pick, whether line 41 is true of false, but try the other too to see which one gives you coordinates closer to the cache.
Solve For the Coordinates
You'll need a calculator that handles a lot of digits to solve this, unless you want to do it with old fashioned arithmetic. I wonder if anybody even knows how anymore. If you don't have a really high resolution, 15-digit calculator, and who does, here's one that has all the functions you will need (for this and other Don&Betty puzzles) and that will handle all the digits you have to punch in for this one, and all at one time. http://mdmetric.com/tech/scicalculator.htm
To use this calculator here, to get the coordinates you need:
(1) Left click the bottom register. With the blinking cursor appearing at the left side of the bottom register, in one unbroken line in the bottom register first left click "341332331087928-" on the digit pad then continue click-entering the string of digits of the several false-line numbers you found, each line number followed by a *, to multiply it times the next, then a * the imprudent line number you found, and finally click the =, like this:
341332331087928-f1*f2*f3*…*fn*P=
(2) A 14-digit number will appear in the upper register. Right in that register you can put a space after the 2nd and 7th and 9th digits and a decimal point after the 4th and 11th digits to display the latitude and longitude coordinates of the entrance to a woodsy trail. You shouldn't need a Geochecker verification because any wrong choices for the line numbers will not result in reasonable looking coordinates.
(3) Park on the side of the little-trafficked road just east of those coordinates.
(4) Enter the trailhead, at the coordinates, then turning left at the first T, continue following it clockwise until you come onto the giant mama boa, the one that is holding me in the picture.
(5) Then find the giant papa boa nearby which is gripping the tiny film-can geocache out of sight and low down near the ground.
(6) Open the can only over a jacket or something to not lose any of the very tiny trinkets in the leaves. And do bring a very very tiny trinket to trade. There's not much room in a film can for the actual log book and actual pen and all the trinkets. Sign only your name and date, and not on a new both sides of a page if there's room on the current one. It's a very small book, that's been logged into for a long time. Well. I guess it's been replaced a time or two.
You can check your answers for this puzzle on Geochecker.com.
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