## robert 46's equivocations

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**1**of**1**### robert 46's equivocations

This is a thread to collect the many incorrect assertions and/or vague implications Robert 46 makes in order to support his wild deductions. I will likely use it to simply refer to the response to the incorrect points he makes, instead of endlessly repeating the explanations for why he is wrong, only to be ignored.

And I'm not looking for - and will not respond to again - argumentative replies, like objecting to whether the index of a mathematical summation's formula should start with n=0 or n=1, when it is clearly implied by the example.

Since this thread is a collection, it seems appropriate for the first one to be about the definition of a SET in mathematics. All it is, is a collection of distinct abstract objects. That is, for any abstract object you can define, there is a description and/or function that tells you if that object is, or is not, in the set S. The objects that fit are called elements of the set.

We can express the description and/or function in many ways. Like “S is the even integers between 2 and 10, inclusive,” “S is 2N for N=1 to 5,” or “S={2,4,6,8,10}.” All that is required is that the expression define the binary relationship (Is A Member)/(Is Not A Member). In particular, there is no order, and no concept of duplication, associated with sets. So “12-2N for N=1 to 5” and {8,4,6,2,10} are also descriptions of this same set S.

Robert 46 seems to think (this is one of his vague implications) that being able to physically list the elements is a requirement for a set. That if you can’t write the elements down, in a specific order, it is not a set. He uses several words for the inability to do this – “you can’t find all the elements,” “you can’t complete the list,” “the elements don’t all exist,” etc. These are all equivocations on his part, that he uses to deny facts that are obvious to any logical observer.

One of those facts is that the description can apply endlessly. “All positive integers” is a description of a set I’ll call N that defines any abstract object as a member, or not a member. More formally, one could define it as the iterative test “The number 1 is in N and, if the number n is in N if n-1 is in N.” Although it may take a while to apply this explicitly, or require logic to exclude some objects (like -1), it can always provide the “yes/no” answer for any object, that is required to define a set.

+++++

The axiom needed in Cantor’s Diagonal Proof is called the Axiom of Infinity. Informally, it says that a set can be defined by an endless description. While this seems obvious from the correct definition of a set found above, Robert 46 denies it based on his incorrect definition of a set. He calls it “unrealistic” and compares it to belief in unicorns.

What he ignores, is that it is supposed to be an abstract concept. As such, his requirement for “realism,” by which he means “the ability to find all of these elements,” is irrelevant. It is not comparable - as he claims - to the belief that unicorns actually exist; but to the truth that the abstract concept "unicorn" exists. Which he proves, by referring to it.

[Edit #1, 3/26/2016] Clarify the purpose.

And I'm not looking for - and will not respond to again - argumentative replies, like objecting to whether the index of a mathematical summation's formula should start with n=0 or n=1, when it is clearly implied by the example.

Since this thread is a collection, it seems appropriate for the first one to be about the definition of a SET in mathematics. All it is, is a collection of distinct abstract objects. That is, for any abstract object you can define, there is a description and/or function that tells you if that object is, or is not, in the set S. The objects that fit are called elements of the set.

We can express the description and/or function in many ways. Like “S is the even integers between 2 and 10, inclusive,” “S is 2N for N=1 to 5,” or “S={2,4,6,8,10}.” All that is required is that the expression define the binary relationship (Is A Member)/(Is Not A Member). In particular, there is no order, and no concept of duplication, associated with sets. So “12-2N for N=1 to 5” and {8,4,6,2,10} are also descriptions of this same set S.

Robert 46 seems to think (this is one of his vague implications) that being able to physically list the elements is a requirement for a set. That if you can’t write the elements down, in a specific order, it is not a set. He uses several words for the inability to do this – “you can’t find all the elements,” “you can’t complete the list,” “the elements don’t all exist,” etc. These are all equivocations on his part, that he uses to deny facts that are obvious to any logical observer.

One of those facts is that the description can apply endlessly. “All positive integers” is a description of a set I’ll call N that defines any abstract object as a member, or not a member. More formally, one could define it as the iterative test “The number 1 is in N and, if the number n is in N if n-1 is in N.” Although it may take a while to apply this explicitly, or require logic to exclude some objects (like -1), it can always provide the “yes/no” answer for any object, that is required to define a set.

+++++

The axiom needed in Cantor’s Diagonal Proof is called the Axiom of Infinity. Informally, it says that a set can be defined by an endless description. While this seems obvious from the correct definition of a set found above, Robert 46 denies it based on his incorrect definition of a set. He calls it “unrealistic” and compares it to belief in unicorns.

What he ignores, is that it is supposed to be an abstract concept. As such, his requirement for “realism,” by which he means “the ability to find all of these elements,” is irrelevant. It is not comparable - as he claims - to the belief that unicorns actually exist; but to the truth that the abstract concept "unicorn" exists. Which he proves, by referring to it.

[Edit #1, 3/26/2016] Clarify the purpose.

Last edited by JeffJo on Sat Mar 26, 2016 9:20 am, edited 1 time in total.

- JeffJo
- Intellectual
**Posts:**2246**Joined:**Tue Mar 10, 2009 11:01 am

### Re: robert 46's equivocations

It can be proven that the square root of two is found buy the following infinite summation:

If so, there must be integers M and N such that (N/M)^2=2.

There aren't.

- 1/2 + 3/8 + 15/64 + ... + (2n+1)!/2^(3n+1)/(n!)^2 + ...

robert 46 wrote:... the sum of rational terms is rational ...

If so, there must be integers M and N such that (N/M)^2=2.

There aren't.

- JeffJo
- Intellectual
**Posts:**2246**Joined:**Tue Mar 10, 2009 11:01 am

### Re: robert 46's equivocations

JeffJo wrote:It can be proven that the square root of two is found buy the following infinite summation:1/2 + 3/8 + 15/64 + ... + (2n+1)!/2^(3n+1)/(n!)^2 + ...

For that series, n would be a counting number ( a positive integer), so it should be

1/2 + 3/8 + 15/64 + ... + (2n - 1)!/2^(3n - 2)/((n - 1)!)^2 + ...

instead.

- phobos rising
- Intellectual
**Posts:**114**Joined:**Sun May 24, 2009 11:29 am

### Probability Problems

TCP1) You know that Ann has exactly two children, and you also know a gender that applies to at least one of them, but not if it applies to both. What are the chances that Ann has two children of the same gender?

TCP2) You know that Beth has exactly two children, and you also know that at least one of them is, say, a boy. But not if both are the same gender. What are the chances that Beth has two children of the same gender?

TCP3) You know that Cindy has exactly two children, and you also know that at least one of is a boy. But not if both are boys. What are the chances that Cindy has two boys?

GSP1) On Dan Hall’s game show, you are offered the choice of three doors. One has a car behind it, and the other two have goats. You pick one, but Dan opens a different one instead to show a goat. What are your chances of getting the car if you are allowed to switch to the remaining door?

GSP2) On Ed Hall’s game show, you are offered the same choice. You pick a door; say Door #1. But Ed opens a different one instead to show a goat; say, Door #3. What are your chances of getting the car if you are allowed to switch to the remaining door (#2 in this example)?

GSP3) On Fred Hall’s game show, you are again offered this choice. You pick Door #1. But Fred opens a Door #3 instead showing a goat. What are your chances of getting the car if you are allowed to switch to the Door #2?

+++++

The answer must be the same within either set of three (more below): 1/2 for the TCP, and 2/3 for the GSP.

Robert makes all kinds of equivocations about these probability problems, trying to justify different approaches to the originals (TCP3 and GSP2, respectively) that he chose first (1/3 and 2/3/respectively), and then justified by methods he made up.

For example, he says TCP3 represents a “one-time occurrence,” and deduces from that assertion that it was predetermined that (A) Cindy must have either two boys, or a boy and a girl, and (B) that you would know about a boy, but could not know about the possible girl. But, in typical equivocation, in GSP3 he says you need to know if these door numbers were predetermined before you can deduce if GSP3 is a “one-time occurrence.”

And what is really strange, is he seems to expect people to believe him when he does this.

+++++

There is no practical differences between what robert 46 calls (at least, I think these are what he calls them – his labels evidence further equivocation) a repeatable experiment (TCP1 and GSP1), an example of a repeatable experiment (TCP2 and GSP2), and a “one-time occurrence” (TCP3 and GSP3). All a one-time occurrence is, is a specific example; and all an example is, is a reduction of a repeatable experiment to any one of a set of equivalent possibilities.

The robust solution all these problems can be found only by using a simple relationship known in probability as Bayes Theorem. It comes from the definition that the conditional probability of event A happening, given that you observed event B, is the initial (meaning before any information is learned) probability of A and B happening together, divided by the initial probability of B happening regardless of A. That is, Pr(A|B) = Pr(A&B)/Pr(B). For example, if you roll a red die and a green die, the probability of rolling a sum of Five is 4/36=1/9. But the probability of Five, given that the green die was Three, is Pr(Sum=5|Green=3) = Pr(Green=3&Red=2)/Pr(Green=3) = (1/36)/(1/6) = 1/6. Bayes Theorem is derived by noting that A&B, and B&A, represent the same event. So Pr(A&B) = Pr(A|B)*Pr(B) = Pr(B|A)*Pr(A) = Pr(B&A).

Rearranging, Pr(A|B) = Pr(B|A)*Pr(A)/Pr(B). This is useful because sometimes, the reverse conditional probability is easier to assess. The above problems are all examples.

In GSP3, the car is initially equally likely to be behind any door; that is, Pr(Car=1) = Pr(Car=2) = Pr(Car=3) = 1/3. Fred Hall is initially equally likely[1] to open either Door #2 or Door #3; that is, Pr(Open=2) = Pr(Open=3) = 1/2. But these cases must be divided unequally, depending on where the car is:

The answer to TCP1 can only be 1/2. We have no information that can make any family arrangement more, or less, likely than any other. Two of the four possibilities have only one gender, so the answer is 1/2.[2] And, like I said, it can't vary within the set - in fact, the common answer to TCP3 of "1/3" is an example of Bertrand's Box Paradox since it differs from TCP1.

To get the answer to TCP3 directly:

+++++

[1] We must make some assumptions about the rules of this game to make this statement. We must assume it is possible for Fred Hall to open a door, reveal a goat, and offer a switch; he did, after all. It isn’t as clear why we can’t assume the rules allow anything else. The reason is twofold: first, we have no way to assume relative probabilities between any other set of rules and this one. So the question cannot be answered if we allow any other rules. Second, such rules are not inconsistent with game shows, so it is reasonable.

Robert 46 equivocates by intentionally misrepresenting these reasons.

[2] Another thing robert 46 equivocates about, is claiming that my solution identifies a single child as "the" boy in TCP3. This is false, as is clearly seen in the case where there are two of the same gender. It is true that, in mixed families, a specific child is implied by the known gender, but it is not true that you know which child that is by the gender, or somehow "picked" that child and "reported" the gender.

[3] Anything else assumes that our knowledge – independent of this probability question because it is based on initial probabilities – is biased toward boys. That if we recall incomplete gender information about a family, we always recall boys and never recall girls. Robert 46 equivocates about this assumption as well: he says it is the opposite of the one made about the GSP. It clearly isn’t: first, we can assume the relative probabilities are equal; and second, the assumption of unbiased knowledge is what is consistent here.

TCP2) You know that Beth has exactly two children, and you also know that at least one of them is, say, a boy. But not if both are the same gender. What are the chances that Beth has two children of the same gender?

TCP3) You know that Cindy has exactly two children, and you also know that at least one of is a boy. But not if both are boys. What are the chances that Cindy has two boys?

GSP1) On Dan Hall’s game show, you are offered the choice of three doors. One has a car behind it, and the other two have goats. You pick one, but Dan opens a different one instead to show a goat. What are your chances of getting the car if you are allowed to switch to the remaining door?

GSP2) On Ed Hall’s game show, you are offered the same choice. You pick a door; say Door #1. But Ed opens a different one instead to show a goat; say, Door #3. What are your chances of getting the car if you are allowed to switch to the remaining door (#2 in this example)?

GSP3) On Fred Hall’s game show, you are again offered this choice. You pick Door #1. But Fred opens a Door #3 instead showing a goat. What are your chances of getting the car if you are allowed to switch to the Door #2?

+++++

The answer must be the same within either set of three (more below): 1/2 for the TCP, and 2/3 for the GSP.

Robert makes all kinds of equivocations about these probability problems, trying to justify different approaches to the originals (TCP3 and GSP2, respectively) that he chose first (1/3 and 2/3/respectively), and then justified by methods he made up.

For example, he says TCP3 represents a “one-time occurrence,” and deduces from that assertion that it was predetermined that (A) Cindy must have either two boys, or a boy and a girl, and (B) that you would know about a boy, but could not know about the possible girl. But, in typical equivocation, in GSP3 he says you need to know if these door numbers were predetermined before you can deduce if GSP3 is a “one-time occurrence.”

And what is really strange, is he seems to expect people to believe him when he does this.

+++++

There is no practical differences between what robert 46 calls (at least, I think these are what he calls them – his labels evidence further equivocation) a repeatable experiment (TCP1 and GSP1), an example of a repeatable experiment (TCP2 and GSP2), and a “one-time occurrence” (TCP3 and GSP3). All a one-time occurrence is, is a specific example; and all an example is, is a reduction of a repeatable experiment to any one of a set of equivalent possibilities.

The robust solution all these problems can be found only by using a simple relationship known in probability as Bayes Theorem. It comes from the definition that the conditional probability of event A happening, given that you observed event B, is the initial (meaning before any information is learned) probability of A and B happening together, divided by the initial probability of B happening regardless of A. That is, Pr(A|B) = Pr(A&B)/Pr(B). For example, if you roll a red die and a green die, the probability of rolling a sum of Five is 4/36=1/9. But the probability of Five, given that the green die was Three, is Pr(Sum=5|Green=3) = Pr(Green=3&Red=2)/Pr(Green=3) = (1/36)/(1/6) = 1/6. Bayes Theorem is derived by noting that A&B, and B&A, represent the same event. So Pr(A&B) = Pr(A|B)*Pr(B) = Pr(B|A)*Pr(A) = Pr(B&A).

Rearranging, Pr(A|B) = Pr(B|A)*Pr(A)/Pr(B). This is useful because sometimes, the reverse conditional probability is easier to assess. The above problems are all examples.

In GSP3, the car is initially equally likely to be behind any door; that is, Pr(Car=1) = Pr(Car=2) = Pr(Car=3) = 1/3. Fred Hall is initially equally likely[1] to open either Door #2 or Door #3; that is, Pr(Open=2) = Pr(Open=3) = 1/2. But these cases must be divided unequally, depending on where the car is:

- He can’t reveal the car, so Pr(Open=3|Car=3)=0.
- He must open a goat door, so Pr(Open=3|Car=2)=Pr(Car=2)=1/3.
- If Car=1, he must choose between two doors, so Pr(Open=3|Car=1)=Pr(Car=1)/2=1/6.

The answer to TCP1 can only be 1/2. We have no information that can make any family arrangement more, or less, likely than any other. Two of the four possibilities have only one gender, so the answer is 1/2.[2] And, like I said, it can't vary within the set - in fact, the common answer to TCP3 of "1/3" is an example of Bertrand's Box Paradox since it differs from TCP1.

To get the answer to TCP3 directly:

- Pr(Boys=0)=Pr(Boys=2)=1/4
- Pr(Boys=1)=1/2.
- In any family, of any size, where we know only one non-specific gender, it must be equally likely we would know about “boy” or “girl”[2]. Pr(Know=Boy)=1/2.
- But we can only know “boy” if there are only boys, Pr(Know=Boy|Boys=2)=1.
- So Pr(Boys=2|Know=Boy) = Pr(Know=Boy|Boys=2)*Pr(Boys=2)/Pr(Know=Boy)
- Pr(Boys=2|Know=Boy) = (1)*(1/4)/(1/2) = 1/2.

+++++

[1] We must make some assumptions about the rules of this game to make this statement. We must assume it is possible for Fred Hall to open a door, reveal a goat, and offer a switch; he did, after all. It isn’t as clear why we can’t assume the rules allow anything else. The reason is twofold: first, we have no way to assume relative probabilities between any other set of rules and this one. So the question cannot be answered if we allow any other rules. Second, such rules are not inconsistent with game shows, so it is reasonable.

Robert 46 equivocates by intentionally misrepresenting these reasons.

[2] Another thing robert 46 equivocates about, is claiming that my solution identifies a single child as "the" boy in TCP3. This is false, as is clearly seen in the case where there are two of the same gender. It is true that, in mixed families, a specific child is implied by the known gender, but it is not true that you know which child that is by the gender, or somehow "picked" that child and "reported" the gender.

[3] Anything else assumes that our knowledge – independent of this probability question because it is based on initial probabilities – is biased toward boys. That if we recall incomplete gender information about a family, we always recall boys and never recall girls. Robert 46 equivocates about this assumption as well: he says it is the opposite of the one made about the GSP. It clearly isn’t: first, we can assume the relative probabilities are equal; and second, the assumption of unbiased knowledge is what is consistent here.

- JeffJo
- Intellectual
**Posts:**2246**Joined:**Tue Mar 10, 2009 11:01 am

### Re: robert 46's equivocations

"Robert 46 seems to think [..] that being able to physically list the elements is a requirement for a set."

Robert apparently is a finitist or an ultrafinitist. Nothing wrong with a little constructivism taking a jab at the 'purely theoretical, less practical' use of mainstream math.

"It can be proven that the square root of two is found buy the following infinite summation:

1/2 + 3/8 + 15/64 + ... + (2n+1)!/2^(3n+1)/(n!)^2 + ..."

which again is purely theoretical, because we can't, in the real world, sum infinitely many times. A constructivist would simply instead use a fast converging series, like Xn+1 = Xn - (Xn^2-2)/(2Xn), X1 = 2/3, which is the Newton-Raphson method for solving x*x-2=0, terminating it, at some N.

Robert apparently is a finitist or an ultrafinitist. Nothing wrong with a little constructivism taking a jab at the 'purely theoretical, less practical' use of mainstream math.

"It can be proven that the square root of two is found buy the following infinite summation:

1/2 + 3/8 + 15/64 + ... + (2n+1)!/2^(3n+1)/(n!)^2 + ..."

which again is purely theoretical, because we can't, in the real world, sum infinitely many times. A constructivist would simply instead use a fast converging series, like Xn+1 = Xn - (Xn^2-2)/(2Xn), X1 = 2/3, which is the Newton-Raphson method for solving x*x-2=0, terminating it, at some N.

- Gofer
- Intellectual
**Posts:**276**Joined:**Mon May 09, 2016 8:24 am

### Re: robert 46's equivocations

(Note that quotes work properly in this thread.)

1+1=2

1+1/2+1/4+...+2^N+... = 2

Construct a right, isosceles triangle. The length of the hypotenuse is sqrt(2).

I don't think he fully understands what that means; but yes. Which is one reason why he can't have disproved Cantor Diagonalization. The Axiom of Infinity is a postulate in Cantor's set theory. If he dismisses that postulate, he can't disprove a theorem in that theory. Any more than a Euclidean geometrist could disprove that there are more, or less, than exactly one parallel line through a single point in a non-Euclidean geometry.Gofer wrote:Robert apparently is a finitist or an ultrafinitist.

In his version, there is. The inability to construct a result by one method does not prove that the you can't by another. And if the two can be shown to be equivalent, then that inability is meaningless.Nothing wrong with a little constructivism taking a jab at the 'purely theoretical, less practical' use of mainstream math.

1+1=2

1+1/2+1/4+...+2^N+... = 2

Construct a right, isosceles triangle. The length of the hypotenuse is sqrt(2).

- JeffJo
- Intellectual
**Posts:**2246**Joined:**Tue Mar 10, 2009 11:01 am

### Re: robert 46's equivocations

JeffJo wrote:It is not comparable - as he claims - to the belief that unicorns actually exist; but to the truth that the abstract concept "unicorn" exists.

B.t.w., f.y.i., this is an incorrect use of semicolon, which instead should be a comma. And there's no reason to put "as he claims" within hyphens, as it does fit with the main sentence. Just some friendly advice - as I've seen you doing this before!

In Cantor's proof, is it really necessary that the set, N, of naturals is infinite? Isn't S just put in a 1-to-1 correspondence with N, which is really just a function? So it seems that the infiniteness of N is irrelevant to Cantor's proof, despite you stating that "The axiom needed in Cantor’s Diagonal Proof is called the Axiom of Infinity"; but I can't be bothered to check for the details.

- Gofer
- Intellectual
**Posts:**276**Joined:**Mon May 09, 2016 8:24 am

### Re: robert 46's equivocations

Gofer wrote:JeffJo wrote:It is not comparable - as he claims - to the belief that unicorns actually exist; but to the truth that the abstract concept "unicorn" exists.

B.t.w., f.y.i., this is an incorrect use of semicolon, which instead should be a comma.

The semicolon separates two closely related independent clauses,[1] which is what a semicolon is supposed to do. Yes, a comma could have been used (like I am about to), but would not have separated the clauses as I intended.

And there's no reason to put "as he claims" within hyphens, as it does fit with the main sentence. Just some friendly advice - as I've seen you doing this before!

No, you are purposely picking nits because you do not appreciate the errors I have pointed out in your statements. And again, I could have used comas in each place, but it would have made a run-on sentence.

Whether or not it is necessary for the proof, N is infinite. So what is the point of the question?In Cantor's proof, is it really necessary that the set, N, of naturals is infinite?

But yes, it is critical. If N is not infinite, then the string defined by inverting the nth character of the nth string, for all n, is not infinite in length. Which means it is not a member of T.

And before you ask, the strings have to be infinite to have the property of being uncountable.

No, not "just". A 1:1 correspondence is a function with the unique property that the inverse is also a function.Isn't S just put in a 1-to-1 correspondence with N, which is really just a function?

You'll have to actually explain why you conclude that, so I can explain to you exactly where you misunderstand the proof.So it seems that the infiniteness of N is irrelevant to Cantor's proof, ...

Yes, I have noticed that about you.I can't be bothered to check for the details.

The axiom is required for the proof to show that D(n) is a completed string.

+++++

[1] Edit: You should use a semicolon between independent clauses joined by a coordinating conjunction (and, but, or, nor, yet) if the clauses are already punctuated with commas or if the clauses are lengthy. Look it up.

Last edited by JeffJo on Wed Jul 05, 2017 12:32 pm, edited 1 time in total.

- JeffJo
- Intellectual
**Posts:**2246**Joined:**Tue Mar 10, 2009 11:01 am

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