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June 26th, 2014, 07:14 PM   #21
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Quote:
 Originally Posted by peterwilliam I am quite impressed to see the views here of all experts.
Ok, I see that the audience warmed up. Let's get on with the show.

Question, what is the logical flaw in this?
The 1x2 triangle is the blue right triangle. We are trying to trisect ∠d.

June 27th, 2014, 05:04 AM   #22
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Quote:
 Originally Posted by long_quach Ideally, who would you want the judges to be?
Professional geometers -- those with the expertise required to judge the claim. Within that group my only real concern would be to avoid the possibility of loons.

Quote:
 Originally Posted by long_quach How do you know what cards I'm holding? And do you think I don't know what cards you are holding?
I don't really care what your cards look like.

June 27th, 2014, 05:38 AM   #23
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Quote:
 Originally Posted by long_quach Question, what is the logical flaw in this?
The principal flaw is that it has been proved impossible to trisect an angle with ruler and straightedge.

Specifically, you'lll need to convince me that b+e=180 and that the red triangle is a) a triangle; and b) isosceles.

What this is missing (as a proof) is any method, so I can't say exactly what's wrong, because I don't know how the figure has been constructed.

June 27th, 2014, 06:22 AM   #24
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Quote:
 Originally Posted by v8archie The principal flaw is that it has been proved impossible to trisect an angle with ruler and straightedge.
Right, you need something else like a linkage, origami, or neusis. The claim is that this proof uses neusis which makes it at least plausible. I didn't examine it though.

June 27th, 2014, 01:48 PM   #25
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Quote:
 Originally Posted by v8archie The principal flaw is that it has been proved impossible to trisect an angle with ruler and straightedge. Specifically, you'lll need to convince me that b+e=180 . . .
b + e = 180° always. A line is 180°.

Quote:
 Originally Posted by v8archie and that the red triangle is a) a triangle
It's a triangle. A triangle has 3 sides.

Quote:
 Originally Posted by v8archie and b) isosceles.
Bingo! a = a, if and only if the 2 red segments are equal. This is not proven.

Quote:
 Originally Posted by v8archie What this is missing (as a proof) is any method, so I can't say exactly what's wrong, because I don't know how the figure has been constructed.
This is the method of its construction. I dial a "string" until the distance to the Y axis = 1. Is this valid why or why not? And don't regurgitate words like neusis, marked ruler, already proven, etc . . . REASON through the logic.

 June 27th, 2014, 03:57 PM #26 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra I have no idea what that picture is about.
June 27th, 2014, 10:56 PM   #27
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The string is attached to point C.
It moves and intersect the X axis, Y=0. My mistake in the previous post.

KD>1
HF<1

When GE=1, the triangle is isosceles.

Quote:
 Originally Posted by v8archie and b) isosceles.

June 29th, 2014, 08:34 AM   #28
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Quote:
 Originally Posted by v8archie I've read in the past that older scientists are more prone to this [self deception], than younger ones. SO old discredited theories only usually die with the scientists that cling to them.
If you know where the article is then I would like to read it.

June 29th, 2014, 09:13 AM   #29
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Quote:
Originally Posted by bob
I have now become a very radical logician and reject many of the axioms that most logicians accept.
Quote:
 Originally Posted by Evgeny Many? May I ask which axioms?
F -> T = T
T -> T = T
F -> F = T
T -> F = F

I don't believe -> is truth functional. Only & and v and ~ are truth-functional, meaning their syntax affects the transformation of the symbol T or F. The sole purpose of -> is one-way replacement. If we believe that -> is truth-functional, then we are obligated to believe that:

If there are round-squares then New York is a large city

is a true statement and it it's not. Something is wrong, and my solution is use -> solely for one-way replacement.

p v ~p = T

It's not always true.

Either round-squares are hexagons or round-squares are not hexagons.

Both statements are false.

p & ~p = F

We have to make a distinction between an atomic sentence and a molecular sentence.

p is atomic means the symbols that sentence p is composed of are indefinable.
p is molecular means the symbols that sentence p is composed of are definable.

For example,
1. Some men go to the beach and some men do not go to beach.

Logicians have an explanation for why that is not of the form:
p & ~p

But I believe their explanation refers to an axiom which is not basic, it's a consequence of some other basic axiom. The basic axiom I use to show that
1 is not false is that you cannot compute the truth value of a composite sentence (i.e., composed of several conjunctive atoms) unless it is symbolized in atomic form. So let lowercase letters stand for atomic sentences and molecular sentences stand for capital letters then:

P & ~P = T v F
p & ~p = F

It's more complicated than that but you get the idea. There are 3 examples of axioms I reject and there are probably about 10 or 15 more than that. If you're interested in seeing the others then send me a PM because I would be happy to discuss it with you.

Quote:
 Originally Posted by Evegeny What do you mean by processing the symbols on a computer: numerical calculations?
Very good question and I do not believe Church and Turing solved the problem back in the 30's, let's go ahead and start working on a solution. This relates to the distinction between pure and applied math.

Machine x processes symbol y means x rewrites y as z on another line and z means the same thing as y.

E.g. Say you have line:

~(x & y)

Machine x processes symbols ~(x & y) means x rewrites ~(x & y) as ~x v ~y on another line and ~(x & y) means the same thing as ~x v ~y.

Now, what happens when you don't want equivalence of meaning but you want to know the answer, as in:

2 + 2

Then you just write the symbol as:

2 + 2 = ?

And the linguistic being (because an alien could also do the same thing) will understand that

2 + 2 = ? has the same meaning as 4.

In other words,

(2 + 2 = ?) <-> 4

Let's do an example of symbols that are not calculable (I admit that process was not a very good word to have used in the first place. Calculate is more precise).

I'm really against dialetheism, the belief that some contradictions are true. Remember I do not believe that: "Some men go to the beach and some men do not go to the beach is a contradiction". So here is a string of symbols by Graham Priest, taken form here:
Dialetheism (Stanford Encyclopedia of Philosophy)
"Hence, if a sufficient case can be made out for a contradiction, it will be rational to believe it."

What I would like to do is build a machine which can calculate whether or not that sentence is true or false. But it cannot be calculated at the moment because such symbols like "sufficient" and "rational" are very hard to define and if you cannot define a symbol then you cannot calculate with it. But I do believe that sentence is calculable just not until we get a definition for sufficient and rational. I have hypotheses for the definitions of these words but I've learned the hard way that a hypothesis is changed more than 95% of the time when you get down to the hard work of testing it.

Hope that helps.

June 29th, 2014, 10:17 AM   #30
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Quote:
 Originally Posted by bobsmith76 I don't believe -> is truth functional. Only & and v and ~ are truth-functional, meaning their syntax affects the transformation of the symbol T or F. The sole purpose of -> is one-way replacement.
It sounds like you just want the symbol to mean something different from its standard meaning, not that you actually disagree with standard math. Simply replace Q -> R with ~(Q) ^ (R) whenever you see it and you'll be golden.

Quote:
 Originally Posted by bobsmith76 If we believe that -> is truth-functional, then we are obligated to believe that: If there are round-squares then New York is a large city is a true statement and it it's not.
I'm not sure how to read you here. If you mean for "round-squares" to be some undefined term like "sjfujerui" then either they exist or not and either way you're good (remembering to first transform the statement into "There are no round-squares or else New York is a large city" to avoid your notation issues, as above). If you instead mean it to be the set of squares which are also round, where roundness is some property squares lack, then the statement is "There are no round-squares, or else New York is a large city" that is "(For all squares s, s is not round) or (New York is a large city)" which is also fine.

Aside from the minor translation need to convert normal math into bobsmith-speak this isn't a big deal.

More interesting to me would be a description of this new operator, ->_{bobsmith}, and why you think it's interesting. (You refer to it as "one-way replacement", not quite sure what you mean by that.)

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