65 new of 132 responses total.
Here we go again. if sign(I) then L(I) else T(I). if sign(II) then T(II) else L(II). "This room contains a tiger" cannot go on door I, because this would establish (sign(I) AND T(I)) OR ( (NOT sign(I)) AND L(I)) which evaluates as FALSE. So this sign goes on the other door. "Both rooms..." then goes on door I, and it can't be true, because it would put a kitty in room I as well as II, violating the "if sign(I) then L(I)" clause. So NOT sign(I), and the prisoner doesn't have to end up as cat food. However, since NOT sign(I), T(I). But fortunately, sign(II) isn't true either which leads to L(II).
Looks good, Drew, except that you need to invoke "process of elimination"
in the final sentence to draw the final conclusion. As in:
The fact that sign(I) is false implies both
1) There is a tiger in Room I, and
2) One of the rooms does not contain a tiger.
By process of elimination, Room II must contain a lady. (Sign(II) doesn't
help you at all.)
Ok, then, it's on to
THE THIRD DAY
"Confound it!" said the king. "Again all the prisoners won! I think
tomorrow I'll have *three* rooms instead of two; I'll put a lady in one
room and a tiger in each of the other two rooms. Then we'll see how the
prisoners fare!"
"Excellent idea!" replied the minister.
"Your conversation, though flattering, is just a bit on the repetitious
side!" exclaimed the king.
"Excellently put!" replied the minister.
The Ninth Trial
Well, on the third day, the king did as planned. He offered three rooms
to choose from, and he explained to the prisoner that one room contained a
lady and the other two contained tigers. Here are the three signs:
I II III
A TIGER A LADY A TIGER
IS IN IS IN IS IN
THIS ROOM THIS ROOM ROOM II
The king explained that at most one of the three signs was true. Which
room contains the lady?
Signs II and III are mutually exclusive. That is, if one is true, the other must be false. Therefore, one of them must be true, no matter what. Since at most one sign is true, then sign I must be false, and the Lady is in room I.
Hmmm... The fact that "if one is true, the other must be false" does not imply "one of them must be true".
Yes, but I think dang has only mis-articulated his analysis. II XOR III is true, so I must be false (to avoid having 2 true signs).
Sure it does. One sign must be true. Hense, two signs must be false. One of those two signs, then, must be false. If one of them is false, then the other one is true. Hense, one of those signs must be true, and sign I must be false.
It is in fact the case that if one of the two signs (II and III) is false, then the other must be true. *That* is what you need to conclude that one of them must be true. It's not enough to say that "if one is true, then the other is false", because that doesn't eliminate the possibility that they are both false. (And it's *not* the case that one sign must be true. All we know is that *at most one* of the signs is true.) With that correction, though, dang's (and i's) solution is a very nice one. I'll enter the next trial when i get home.
The Tenth Trial
Again there was only one lady and two tigers. The king explained to
the prisoner that the sign on the door of the room containing the lady was
true, and that at least one of the other two signs was false.
Here are the signs:
I II III
A TIGER IS A TIGER IS A TIGER IS
IN ROOM II IN THIS ROOM IN ROOM I
What should the prisoner do?
If the lady were in Room (II), then sign (II) would be false, contradicting the fact that the sign on the room containing the lady is true. Therefore, the lady is not in room (II). If the lady were in room (III), then tigers would be in the other two rooms, implying that signs (I) and (II) are both true, contradicting the fact that at least one of the signs on the tiger rooms is false. Therefore, the lady is not in room (III). Hence the lady is in room (I). (Note that this implies that sign (I) is true and sign (III) is false, consistent with the king's preconditions.)
Yes, John has put it very nicely. On to The Eleventh Trial:
First, Second, and Third Choice
In this more whimsical trial, the king explained to the prisoner that
one of the three rooms contained a lady, another a tiger, and the third
room was empty. The sign on the door of the room containing the lady was
true, the sign on the door of the room with the tiger was false, and the
sign on the door of the empty room could be either true or false. Here
are the signs:
I II III
ROOM III THE TIGER IS THIS ROOM
IS EMPTY IN ROOM I IS EMPTY
Now, the prisoner happened to know the lady in question and wished to
marry her. Therefore, although the empty room was preferable to the one
with the tiger, his first choice was the room with the lady.
Which room contains the lady, and which room contains the tiger? If
you can answer these two questions, you should have little difficulty in
also determining which room is empty.
The lady can't be in room III, because this would make sign(III) false. If she were in room II, this would put the cat in room I with sign(I) being true. So she has to go in room I, making sign(I) true and room III empty, leaving the cat in room II.
Very nice, Drew. Ok, here's the grand finale:
THE FOURTH DAY
"Horrible!" said the king. "It seems I can't make my puzzles hard
enough to trap these fellows! Well, we have only one more trial to go,
but this time I'll *really* give the prisoner a run for his money!"
A Logical Labyrinth
Well, the king was as good as his word. Instead of having three rooms
for the prisoner to choose from, he gave him nine! As he explained, only
one room contained a lady; each of the other eight either contained a
tiger or was empty. And, the king added, the sign on the door of the room
containing the lady is true; the signs on doors of all rooms containing
tigers are false; and the signs on doors of empty rooms can be either true
or false.
Here are the signs:
I II III
THE LADY THIS ROOM EITHER SIGN V
IS IN AN IS EMPTY IS RIGHT
ODD-NUMBERED OR SIGN VII
ROOM IS WRONG
IV V VI
SIGN I EITHER SIGN II SIGN III
IS WRONG OR SIGN IV IS WRONG
IS RIGHT
VII VIII IX
THE LADY THIS ROOM THIS ROOM
IS NOT IN CONTAINS CONTAINS
ROOM I A TIGER A TIGER
AND ROOM IX AND SIGN VI
IS EMPTY IS WRONG
The prisoner studied the situation for a long while.
"The problem is unsolvable!" he exclaimed angrily. "That's not fair!"
"I know," laughed the king.
"Very funny!" replied the prisoner. "Come on, now, at least give me a
decent clue: is Room Eight empty or not?"
The king was decent enough to tell him whether Room VIII was empty or
not, and the prisoner was then able to deduce where the lady was.
Which room contained the lady?
Ugh. I got as far as figuring that if room VIII is not empty then the lady is in room one or seven, but then I decided to get a life.
(In order to "get a life", the prisoner's course of action would have to be rather different from yours... :-)
In order for sign(VIII) to yield any useful information, it must be ascertained to have some boolian value. Thus room VIII can't be empty. It can't have the lady either, since sign(VIII) would have to be true, and it would not be. So there's a kitty in room VIII. This is permitted by the AND function in sign(VIII) so long as there is something in room IX. Room IX also has to have a kitty, for similar reasons that room VIII does. This makes sign(VI) *right* (true). This of course makes sign(III) false. Sign(III) is an OR function (it was stated that either-or was *inclusive*-or in an earlier problem) which requires *both* clauses to be false. so, sign(VII), meaning that the lady is not in room I. Also, NOT sign(V), which is another OR function, both of whose switches must be turned off. Thus, NOT sign(IV); sign(I) is right, and the lady must be in either room I, III, V, VI, and IX. We've already eliminated rooms III, V, and IX, due to their signs being false, which leaves rooms I and VII; and we had already eliminated room I due to sign(VII) being true. This leaves the lady in room VII.
Sorry it has taken me so long to get back here; Drew is right, of course, and his solution is very succinct. I'm going away for a while - anyone else want to enter a puzzle?
Ok, so I've started reading Jack Chalker's Well World books. The Well world is supposed to be a planet which is divided into 1560 environments, each of which is shaped like a hexagon. "Each hex is identical in size - each one of the six sides is just a shade under three hundred fifty-five kilometers, and they're a shade under six hundred fifteen kilometers across." So the planet is tiled with hexagons. It's not clear whether 1560 is the total number of hexagons, though, because Chalker implies later that some hexes are cut in half by the equator, and I'm not sure whether those count in the 1560 number (I think not). So the question is: did Chalker think this out before he wrote it? Is it possible to tile a sphere with hexagons in a way that fits the description?
There are only five regular polyhedrons - all sides the same shape and a sphere touching each side at the center can be inscribed. I don't know, however, how close one might get to the Well World - say, a mismatch at just the poles.
Hi, Math always the worst experience in college . I get E on both Calculus I & II. The only way to study it is to do 10,000 exercises. (take a long time). The 2nd year, I got A and B. It is a nightmare! Remembering all those Integral formula etc. They just prove that Newton is wrong in Gravitation theory. There is a WEAK force that cause in-accurate result. Fortunately physic is an interesting subject (nuclear physic) Regards AW
Hi, about the book upgrade: what about buying CD, instead of books? IMO it will be more exciting, easy to understand and save tons of wood from Tropical rainforest. Perhaps they can use rewritable CD to keep the cost down? What about bookstore with CD-Writer? IMO it will be cheaper than printed books. Regards (AW)
I don't think you can tile a sphere with any number of regular hexagons. You can tile a plane with them, though. You need a few pentagons in order to get it to curl up into a sphere. If you look at a geodesic dome, a la Buckminster Fuller, you will see that it is made up of triangles, with 5 or 6 coming together at each point. The "dual" of this structure is made by using the center of each triangle as a node and connecting each new node to the adjacent ones. The dual of a geodesic dome is made up of hexagons and pentagons that tile a sphere. You can have many more hexagons if you like, but you need to have exactly 12 pentagons to complete the sphere. If you leave the hexagons out altogether, you have a dodecahedron. The hexagons don't contribute any curling, which is why you can tile a plane with them. if they are regular, each one forms an inside angle of 120 degrees at each vertex. Since there must be no more nor less than 3 coming together at each point, they always sum to 360 degrees. Hence, no curling. Curiously, although you can't tile a sphere with hexagons, you can tile a torus with them. To see this, section the torus and unwind it to form a cylinder, then cut the cylinder the long way and unwind it to form a rectangle. Tile the rectange as you would a plane, then wind it and stitch it back up the way you cut it.
That implies that I could roll a cylinder out of paper, and then wind it into a torus. I can't. Or, please explain further.
You're stretching & compressing the surface of the torus in your unwrapping & wrapping operations - yes, they're all hexagons, but goodbye regularity. I'm not familiar with all the official terminology used, but if you're dividing up the surface of a sphere (curved pieces) instead of building polyhedral solids (flat pieces), you CAN do it with hexagons. Only 2 are needed, and those unfamiliar with non-Euclidean geometry may have problems with calling one of them a hexagon.... But they're regular!
The original proposition (#84) was that the hexagons were "all identical in size" - that would imply all identical in shape, too, but they were not stated to be regular (spherical) hexagons, thought the sides are nearly all the same size, and all the hexagons are about the same 'diameter'.... So..what are the shapes of the two irregular hexagons (ala #90) with which to (spherically) tile a sphere?
Re #91: I think what Walter is thinking of in #91 is: take a sphere, draw a small hexagon on it. Then the one you drew and its complement are two hexagons which tile the sphere. Are you sure, Rane, that "same size" (which I take to mean same area) implies "same shape"? Can you prove that? Steve W., can you prove that you can't tile a sphere with hexagons? I am assuming here that the sides of each hexagon should be "lines", which on a sphere means segments of a great circle which are less than half a circumference in length.
No, I can't. The logic I ran off in resp:88 did assume that exactly 3 and no more or fewer hexagons came together at any one vertex. In order to drop that restruction, you would have to distort the shape of the hexagons. Since your hinting above allows that the hexes may be distorted, it seems that my assumption may have been too great. Perhaps these must be the rules. (1) Every shape is bounded by 6 straight sides (straight is "geodesic" or great circle on a sphere). All sides are the same length. (2) Every shape has the same size. That is, the same total area. (3) The entire sphere is covered, so that every point on the sphere is either in exactly one of these shapes, on the boundary of exactly two, or at an intersection of three or more. Under those conditions, I believe I can prove that it cannot be done, but I am not posting such a proof here yet. The proof will assume that at least three hexagons come together at each vertex. Perhaps this is not a requirement. Let me rewrite rule (3) above as follows... (3) The entire sphere is covered, so that every point on the sphere is either in exactly one of these shapes, on the boundary of exactly two, or at an intersection of two or more. This allows for two hexagons to share two adjacent pairs of edges and the vertex between them. Such a vertex I might choose to label "bogus", because only two edges meet. Topologically, these hexes are now a lot like pentagons, as they may have five (or fewer) neighboring hexagons. If "bogus" vertices are allowed, then it very likely can be tiled. In fact, I can modify the example Mark presumed Walter was proposing to the construct the following tiling of a sphere with 2 hexagons: Take a sphere of radius r and draw 6 straight line segments of radius pi*r/3 along the equator so that they are laid out end-to-end and touch at 6 vertices, equally spaced along the equator. This tiles the sphere into two equally sized hexagons of area 2*pi*r^2 (OK, it is a bit degenerate, because the line segments are colinear, but it makes the point.)
Yeah, I think it's OK to assume that every vertex is at the corner of at least 3 edges. And each edge is a segment of a great circle. I don't know if you can call what I was doing "hinting", because I don't know the answer to this one.
Re: #92, 91 - aruba has the idea. Regularity presents no challenge whatever. You can even make the two hexagons the same size if you don't mind degenerate n-gons on a sphere...as steve notes. If what steve calls bogus vertices are allowed, then 3, 4, and 5-gons can be called hexagons, and the regular polyhedra provide plenty of regular & symmetrical tilings. Disallowing 180-degree vertices (let's call the results "true" n-gons) messes up the regularity and symmetry, but the tilings persist.
I don't think it has yet been pointed out that all the interior angles of a regular spherical hexagon smaller than half the sphere, are greater than 120 degrees. Hence no three regular spherical hexagons, regardless of their size, can have a common vertex. My mind boggles, however, at irregular hexagons of less than 6 edges....
Those would be about as irregular as it's possible to get.
I think Chalker's plan is pretty hopeless. Tiling a sphere is hexagons is really pretty much the same problem as putting a grid of squares on it (you can convert from a grid of squares to a grid of hexagons by shifting the odd numbered rows half a square left and turning the horizonatal boundaries into zig-zags). But we know how successfull geographers have been at putting a grid on the earths surface - The squares near the poles are awfully long and skinny for squares. I think the same problem arrises with hexagons. The polar hexes get very long and skinny - not even vaguely regular. I seem to recall some of Chalker's later books admitted that the polar hexes aren't as regular, and I think there are large polar regions that aren't tiled.
The interior angles of a regular spherical hexagon *larger* than half the sphere are still greater than 120 - the size clause in #96 is unneeded. Convexity (all interior angles <180) seems to be a popular assumption here. If you don't get hung up with that, then degeneracy (an 180 interior angle, thus 2 co-linear edges) is a very un-boggling point discontinuity in an angle vs. # of sides graph. For greater irregularity, just allow 2-dimensional projections of 3-d n-gons. You get X-intersections of lines at non-vertex points, double vertices, 0 length sides, and all sorts of fun. (But this is boring from the tiling point of view, because ANY tiling of the sphere can be represented as such a projection of a 3-d n-gon!)
Someone recently came out with a 100-sided die. I'm not sure whether hexagons could be circumscribed about the faces, but if a 100 sided die is possible, then maybe something like hexagon-world could be done if you don't mind some irregularity and mismatch.
If you're just interested in generating random numbers from 1 to n, then a fair n-sided die can easily be made. Besides the common 6-siders, 4-, 8- 12-, and 20-sided dice (based on the regular polyhedra) are quite common. (Dungeons & Dragons uses all these, as do a number of other games.) The results are often (mathematically and visually) uglier for other values of n. My guess is that commercial 100-sided dice would either split the faces of a 20-sided die into 5 pieces (each) or follow an apple-peeler pattern. But I wouldn't be surprised by something really odd like a little 10-sided die inside a large, clear 10-sided die (read as 1's and 10's).
Okay, change-of-pace time! Here's a reasonably straightforward
geometry problem:
Three billiard balls of diameter 1 are resting on a flat
tabletop. Each ball is touching the other two. A fourth ball of
diameter 1 rests on top of the first three. How high above the
surface of the table is the apex of the fourth ball?
(Supply reasoning to justify your answer.)
It is the diameter of a ball plus the height of a tetrahedron with sides equal to the diameter of a ball. (Reduced to previously solved problem....)
re 100: 100 sided die have been out for a while...I picked one up in like 5th grade to use with D&D.. I recently finished "Flatland" by Edwin Abbott Abbott, is the sequel "spehere Land" and good?
I have a "die" that was made by a statistician which is an aluminum ten-sided prism with the sides numbered (on the end of the prism) from 0 to 9. It is for generating random decimal numbers.
Re #103: Yeah, but what's the ANSWER? :)
Okay, let's roll out the old HS geometry and take a closer look at that tetrahedron. It's regular with edges of length 1, so all 4 faces are equilateral triangles with sides of length 1. Bisect the 3 angles in the base triangle and run the lines through the opposite sides. The result just oozes symmetries, similarities, and 30-60-90 triangles. Note that by easy symmetry arguments, the 3 bisectors meet at the center directly under the vertex. Using the old 1|1/2|sqtr(3)/2 sides rule for those 30-60-90 trianges gives us a distance of 1/sqrt(3) from a corner of the base to the center. This distance and the (unknown) height are legs of a right triangle with an edge as hypotenuse (1). Pythagoras says the height is sqrt(2/3), so (per #103), the answer is 1+sqrt(2/3). On to a probability problem. It's not hard, but I've seen grad. students flub it and I'm told that the government of India once made a major policy decision based on a wrong answer to it. Assume that the gender ratio at birth is 50-50 and that all births are independent events. (In other words, having babies is like flipping a fair coin - heads it's a boy, tails it's a girl.) The following policy is proposed: Each couple may have two children. If both are girls, then they may have a third, if that's a girl, a 4th is permitted, etc. (Two very different ways you could interpret this sexism! The intent was to guarantee everyone a son who wasn't an only child.) Assume that couples will have as many children as permitted. Ignore divorce, birth out of wedlock, fertility running out before a son is born, etc. What will the gender ratio of children born under this policy be?
Okay, I'll take a crack at it.
If you stop at two kids, then the ratio is 50/50. However, 25% of these
families will be two girls, and thus can have another kid. Of these, 50% will
have yet another girl, and of these 50% yet another... Thus we have:
M M 1/4
M F 1/4
F M 1/4
F F M 1/8
F F F M 1/16
F F F F M 1/32
.........
(n) F M 2^(-n-1)
In the first two cases, the ratio is one female to three males, multiplied
by 1/2. The rest can be generalized as n females to one male, multiplied by
(1/2)^(n+1). Expressing it as a sum:
R = 1/6 + sum(i=1 to inf) { i / 2^(i+1) }.
The series works out as 1/6 + 1/4 + 1/4 + 3/16 + 1/8 + 5/64 + 3/64 + ...
By experiment, the series seems to converge at 1.166666667 or 7/6. Shouldn't
be too much of a disaster in itself.
But the real problem is the population *growth*, the increase per generation.
This is best state, per couple, as
2/4 + 2/4 + 2/4 + 3/8 + 4/16 + 5/32 + ...
or
1/2 + 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + ...
or
1/2 + sum(i=1 to inf) { i / 2^i }.
This series converges, by experiment, at 2.5 children per male-female pair,
or 1.25 multiplication of the population every generation.
Where did you get that 1/6 from, Drew?
The first two cases are taken as the first term in the series - actually not really a part of the series itself, but a special case. The female to male ratio among these first two cases is 1:3, and they take up half the outcomes, for a product of 1/6. I suspect my answer may not be right. I'll have to think this one over.
There is a simple proof that the series sum(i=1 to inf) (i / 2^i) = 2 based on the method of differentiating the series...to wit Start with y= sum[i=0,inf](x^i) = 1/(1-x) in the region 0<x<1 differentiate w.r.t. x - (you can differentiate each term of the series separately). so it yields dy/dx = sum[i=0,inf] (i * x^(i-1) ) = (1 - x)^(-2) so if we substitute x=2^(-1) (another name for 0.5) we have this sum[i=0,inf] (i / 2^(i-1)) = 4 now divide each side by 2 and note that the i=0 term is 0, so drop it sum[i=1,inf] (i / 2^i ) = 2 QED H O W E V E R ..... If you wish to determine the gender ratio you do not need to evaluate a series. It is 1:1 because each birth is independent. I don't see any need to argue further. For skeptics, if you really must do it by the series approach, then evaluate the expected number of males per family and the expected number of females. Both evaluate to 1.25 so the ratio really is 1:1 case probability E(M) E(F) MM 1/4 2/4 0/4 MF 1/4 1/4 1/4 FM 1/4 1/4 1/4 FFM 1/8 1/8 2/8 FFFM 1/16 1/16 3/16 FFFFM 1/32 1/32 4/32 The E(M) series is 1/4 + sum[i=1,inf](1/2^i) = 1/4 + 1 The E(F) series is 1/4 + sum[i=1,inf](i/2^(i+1)) = 1/4 + 1
Another way to find the expected number of girls G (which doesn't require
any calculus) is:
G = 0/4 + 1/4 + 1/4 + 2/8 + 3/16 + 4/32 + ...
= 1/4 + 1/4 + 2/8 + 3/16 + 4/32 + ...
==> 2G = 2/4 + 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + ...
==> 2G-G = 1/4 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ...
==> G = 1/4 + (1/2)/(1-1/2)
= 1/4 + 1
= 5/4
Drew's interesting observation about the increase in population is quite
right, each generation will be 25% bigger than the last if everyone has
children. That would be kind of a problem.
(Compared to the actual growth rates, 25% per generation [less early mortality, infertility, etc.] would have been wonderfully low, and India would be in much better shape today.) Steve's "H O W E V E R..." in #111 is the correct and "best" solution. There are simpler "add things up" solutions, but this is one of those problems where the student's approach tells the teacher more about his/her understanding than anything else. (The instructor threw this problem at a class of PhD-track math (not stats) grad students I was in. The initial class consensus was for a wrong answer [and they almost agreed on which one].) On to a more familiar mathematical topic - Sets. Apples and oranges are elements of the set of fruits, Mr. Figston's 3rd grade is a subset of the set of students at Washington Elementary School, the intersection of the set of wet thing with the set of dry things is the empty set, and all that fun. One very popular set is the Universal Set, which contains EVERYTHING - real, abstract, imagined or undiscovered, simple or complex, it's all there. But the idea that such a set can exist suffers from a fatal logical flaw. It is not that the Universal Set can't contain itself as a subset - but that's a good hint on where to start looking.
Did you mean to say "contain itself as a subset" or "contain itself as an element"?
resp:112 Hey, that's cool, Mark. I learned how to solve that silly series using differentiation. Your way is much better.
Would the Universal Set, then try to contain 'null' and 'infinity' at the same time?
The Universal Set (U for short), by "definition", contains null, infinity, and anything else you can think up. Yourself included. Re #114: U must do both. I believe that a contradiction can be derived either way, but the "nontraditional" one requires minimal knowledge of power sets.
Well, every set contains itself as a subset, so I figured you must have meant "contain itself as an element". The most familiar contradiction based on the notion of sets containing themselves as elements is the Russell Paradox, which goes as follows: Let S be the set of all sets that are not elements of themselves. Then if S is an element of itself, then by definition of S, S is not an element of itself. Conversely, if S is not an element of itself, then again by definition of S, S is an element of itself.
Mike shaves all men that do not shave themselves. Does Mike shave himself?
re #119: Your statement of Russell's "Barber" paradox is insufficient to indicate the paradox since it says nothing about whether or not the barber shaves some of those who shave themselves (which would only include the barber..) or whether or not the restriction applies to the barber (e.g. is the barber a woman?) You need a restriction more like "the barber shaves all those and *only* those who don't shave themselves.."
Ok, fine. But if you are going to be particular, remember that the defining relative pronoun is *that*, so it has to be stated as "the barber shaves all those and only those *that* don't shave themselves."
Re: #118 - my meaning was that a contradiction can be obtained by looking either at sets which are elements of U or at sets which are subsets of U (the power sets are used in the latter). I did't want to give too big a hint. This Statement Is False. (Is the above statement a paradox? Is it not a paradox only for "poetic" reasons - the contradiction is too poorly hidden, insufficiently interesting, etc.? What is a paradox?)
Hey, I know Barry & Sally Childs-Helton, they both have Phds, they are a Paradox.
re 122: two places to moor ships. Speaking of Russel's paradox, Greg (my roommate, flem) was just reading his autobiography and mentioned it to me last night. Interesting coincidance.
Isn't a paradox something that appears to be true, even though it is not?
Er, not exactly. I would describe a paradox as an assertion that seems to be contradictory. Almost the opposite of your definition.
I agree with Steve. A paradox appears to be internally contradictory. Paradoxes may have resolutions, or they may not. Something that "appears to be true, even though it is not" is just an error. I think there is an expression for it, such as "plausible but mistaken".
Chalk one up to fuzzy-headedness. *sigh*
paradox is self-contradictory.
Re #129: A paradox is something that *appears* so be self-contradictory.
That's what I said in #127....... ;)
Apparently, he missed it the first time. :-)
You have several choices: