What Really Happens If the Sun Disappears?

To demonstrate that the speed of gravity is no faster than light, physicists love to point out that if the sun suddenly vanished, it would take approximately eight minutes for the information to reach earth, and over five hours for the information to reach Pluto. It is assumed that earth and Pluto would remain under the influence of the sun's gravity for eight minutes or over five hours respectively. After all, it is currently understood that information, including gravitational information cannot exceed light speed.

But here is the irony: to make the point that nothing is faster than light, the sun (poof!) disappears faster than light! Further, such a thought experiment only takes into account an observer on earth (or Pluto). Allegedly, both gravitational and electromagnetic information reach the observer simultaneously, so said observer (Alice) observes nothing out of the ordinary. She sees the sun vanish and notices that the sun's gravity has vanished along with it. However, another observer (Bob) has parked his spaceship telescope halfway between the sun and the orb where Alice is located (see drawing below).

The orb is distance r from the sun and Bob's telescope is distance r from the sun and the orb. Information from the sun and the orb reach Bob simultaneously within a time of r/c seconds, where c is light speed. Now, let's suspend disbelief and pretend the sun vanishes instantaneously (see drawing below).

Because information from the sun and the orb reach Bob simultaneously, Bob must wait another r/c seconds to observe what Alice observes. Suppose he doesn't wait. Suppose he plugs in the numbers he observes when he first receives information from the sun and the orb. Here is what Bob's math reveals:

At 1) we have Einstein's field equation. If the sun vanishes, the stress-energy tensor (T) value would be severely reduced, so at 2) T has a limit of zero. A little algebra gives us a value that is substantially greater than the gravitational constant. Both 2) and 3) reveal that Newton's constant is not constant, i.e., has a much greater value than expected for a time period of r/c, the time it takes for the information to reach Alice.

If we assume that gravitational information is necessary and that it propagates at light speed, then we must abandon the idea that Newton's constant is constant for all observers. Clearly it is not if gravitational information is necessary to cause gravity.

On the other hand, if we assume gravity is more analogous with the equivalence-principal thought experiment, where the dropped pen seems to fall to the spaceship's floor but the floor really accelerates to the pen, then gravitational information is not necessary to cause gravity. Further, the math shows that Newton's constant remains constant, because there is no r/c time delay--the spaceship does not send a signal to the pen--it simply accelerates to the pen, creating the "persistent illusion" of gravity, where reality, according to Einstein, is a "persistent illusion."

Of course nothing I have written here addresses gravitational waves or how gravity is supposed to work outside an equivalence-principle spaceship. I address those concerns in "The Beautiful Destruction of the Graviton" and "Quantizing Gravity without the Graviton."--which you can find at my profile page at Accademia.edu.

OK, then ... if the sun should magically disappear what really happens? If Newton's constant is really a constant, the impact on the solar system will be immediate but this will not violate the light-speed limit because no gravitational information needs to propagate from the sun.

P = NP? A Solution to the Clay Mathematics Institute's Sample NP Problem

ABSTRACT:

This paper offers a solution to the sample NP problem that can be found at the Clay Mathematics Institute's website. A python program is included along with commentary on whether or not P = NP.

At their P vs. NP page, the Clay Mathematics Institute (CMI) provides the following NP problem example:

"Suppose that you are organizing housing accommodations for a group of four hundred university students. Space is limited and only one hundred of the students will receive places in the dormitory. To complicate matters, the Dean has provided you with a list of pairs of incompatible students, and requested that no pair from this list appear in your final choice. This is an example of what computer scientists call an NP-problem, since it is easy to check if a given choice of one hundred students proposed by a coworker is satisfactory (i.e., no pair taken from your coworker's list also appears on the list from the Dean's office), however the task of generating such a list from scratch seems to be so hard as to be completely impractical. Indeed, the total number of ways of choosing one hundred students from the four hundred applicants is greater than the number of atoms in the known universe! Thus no future civilization could ever hope to build a supercomputer capable of solving the problem by brute force; that is, by checking every possible combination of 100 students."

To make this problem even more difficult and more explicit, let's assume there is only one set of 100 students who qualify for places in the dormitory, and, let's assume we can randomly select any two students from this set--and the pair will not appear on the dean's list of pairs of incompatible students. Thus, no pair from this set will match any pair on the dean's list. The following math shows that this ideal set is buried in a haystack of approximately 10^96 (not 10^200) sets of 100 students!

Below we derive equation 4 which shows that not only k increases as n increases, but the exponent increases as well:

A brute-force algorithm will undoubtedly exceed polynomial time (P). However, we can convert this NP problem to a P problem if we replace student names with numbers ranging from 0 to 399, then pair with each student number an incompatibility score. We do this by examining how many times each student number appears on the dean's list of student pairs. For example, if the list shows [[1,2], [1,3] ...], student 1 appears twice and receives a score of 2. Students 2 and 3 only appear once, so they receive a score of 1. Having the lowest incompatibily scores, students 2 and 3 will appear on the final 100 list along with 98 other low-scoring students.

The final list of 100 students can be checked against the dean's list in polynomial time. The process of checking the dean's list to give each student a score can also be performed in polynomial time. This can be mathematically shown. Let's start by calculating the size of the dean's list of pairs:

The first term on the right side of equation 5 is the total number of student pairs possible with no repeats. We subtract from that the total possible pairs (with no repeats) from the final 100 list. Thus, the size and complexity of the problem has been reduced from approximately 10^96 arrangements of 100 students to less than 100,000 arrangements of student pairs. Below is the relevant math showing the problem is solved in polynomial time:

The term on the right side of equation 11 does not have the number of students (n) as an exponent. The variable k' increases as n increases but the exponent stays fixed at 2--unlike equation 4 where the exponent increases as n increases.

Now, on paper it looks as though the NP problem can be reduced to a P problem, but what about in practice? Below is a python program that solves the CMI's sample NP problem:

Here is the final output:

Lines 1 to 21 are devoted to creating a random dean's list. The remaining program solves the puzzle by giving each student number an incompatibility score. The 100 students with the lowest scores make the final cut. As you can see, there is only 45 lines of code, most of which are used merely to set up the problem. Only the final 22 lines of code are used to solve the problem. The total run time on my personal computer is about one minute. Now, does this mean that P = NP?

At the very least it can be shown that an NP problem can be reduced to a P problem. But is this always the case? One thing I have noticed is that some NP problems have more clues than others. The CMI's example has 74,850 clues, i.e., the student pairs on the dean's list. These clues tell which pairs don't belong to the set of pairs of students on the final 100 list. So there's no need to guess which of the approximately 10^200 sets of 100 students is the right answer. We can use the power of deduction.

Now, imagine this same NP problem without any dean's list, without any clues. The dean simply tells you whether or not you have the right answer. Even the great Sherlock Holmes would be forced to test up to approximately 10^96 sets of 100 students. His powers of deduction would be useless. So even if we can make NP equal to P, we can always make the NP problem harder--the hardest problem has no clues. Thus it seems reasonable to conclude that P doesn't equal NP where there are no clues. Where there are clues, there is opportunity!

References:

1. Cook, Stephen. The P Versus NP Problem. Clay Mathematics Institute

2. Stewart, Ian. Ian Stewart on MInesweeper. Clay Mathematics Institute

3. P vs NP Problem. Clay Mathematics Institute

A Solution to the Continuum Hypothesis

ABSTRACT:

Re: the Continuum Hypothesis: Is there any set which has more members than the set of natural numbers (N), but fewer members than the set of real numbers (R)? The short answer is no. The long answer, that includes mathematical proof, shall be set forth in this paper.

Imagine a finite set of natural numbers (N') with n' members. Imagine another finite set of natural numbers (N) with 2^n' members. Also imagine a set of real numbers (R) with c members:

Even though N and N' are both sets of natural numbers, they do not bijectively map to each other. However, there is a bijective mapping between N and the power set of N' (P(N')), since P(N') also contains 2^n' members:

Now let's take n' to its infinite limit. R has c members, where c = 2^n'. N and R have the same cardinality:

N and R have the same cardinality? This is not consistent with Cantor's theorem which states: ""Let f be a map from set N' to its power set P(N'). Then f: N'-->P(N') is not surjective. As a consequence, the cardinality of N' is less than the cardinality of P(N') holds for any set N'."

Clearly P(N') has more members than N'. We can prove this by counting the members of P(N'). We start with 1 and count all the way to 2^n'... but this implies the existence of a bigger set of natural numbers (N) with 2^n' members. Albeit, the power set of N (P(N)) has more members than N, but these "more members" imply the existence of an even larger set of natural numbers and so on and so on to infinity. Speaking of infinity, the cardinality of R (c) equals aleph-1 which is believed to be greater than aleph-0:

If aleph-1 is greater than aleph-0, then the reciprocal of aleph-1 is less than the reciprocal of aleph-0 (see 5 below). However, if the absolute value of the reciprocal of aleph-0 equals zero, then the absolute value of aleph-1's reciprocal is less than zero (see 6). But no number has an absolute value less than zero, so aleph-1 can't be greater than aleph-0 (see 7, 8, 9).

When comparing infinite sets, it appears that N can have the same cardinality as P(N) and R. Cantor, of course, set forth a brilliant counter-argument in support of his theorem. His argument (or a contemporary version of it) begins with a complete list of members of the power set:

Each number is classified as either "selfish" or "non-selfish." A number is selfish if it is a member of a subset and that same number is the natural number paired with the subset; otherwise, it is non-selfish. In the list above, rows 1 and 3 are examples where 1 and 3 are selfish. They are each members of their respective sets as well as natural numbers paired with those sets. Rows 2 and 4 are examples of 2 and 4 being non-selfish numbers. Now, by definition of the power set, there allegedly should be a subset S that contains all non-selfish numbers. Here's where Cantor creates the template for Russel's paradox: the natural number s can't be a member of S or S would not be the set of all non-selfish numbers. Also, the natural number s can't be a non-selfish number either, since it would be a member of S. A third option is a selfish number that is not a member of S. Unfortunately, such a number is a natural number paired with a different subset. Thus, Cantor argues that there is no possible natural number s that can pair with S. Therefore, there is no bijective mapping between N and P(N).

Yes, a brilliant argument! But Cantor had to move the goalpost to make it. If we move the goalpost back to its original position we see that a bijective mapping between two sets only requires that the sets have the same cardinality, i.e., the same number of members. It does not require the numbers and subsets to have certain attributes like "selfishness" or "non-selfishness." As an illustration, suppose we have a subset S={1, 2, 3, ..., n} that is the only subset that hasn't been paired with a natural number. Subset S is defined as the set of all non-selfish numbers. If we stay true to that definition, we will never find a number to pair it with, notwithstanding the fact that number 3 hasn't been paired with any subset. We break down and decide to map 3 to S. Now there is a complete bijective mapping between N and P(N) (where N and P(N) each have infinite members)--so why should anyone care that S is no longer "the set of all unselfish numbers"?

Let's now address the other proofs (including the famous diagonalization proof) that seem to support the claim that R and N don't have the same cardinality. These proofs start with a list of, say, all the real numbers. Each real number is paired with a natural number. (It is assumed that all natural numbers are listed.) A new real number is then produced that is not on the list. This supposedly contradicts the assumption that the real numbers are countable. If the real numbers are countable, the new real number would be on the list--so the argument goes. However, if the real numbers are countable, then it should be possible to produce a new counting number (natural number) that is not on the list for every new real number that is not on the list. Here's how it is possible:

Translate the natural numbers and/or Cantor ordinals to ASCII or unicode. This will produce natural numbers that will bijectively map to the real numbers. Now, use any proof that will produce a real number that is not on the list. This new real number will map to a unique natural number that is also not on the list. Below is an example of a possible translation scheme:

The following is the bijective mapping scheme:

Note that omega+1 circled in red is translated to 9694349. This is a unique natural number that is not on the list (that includes omega). All the natural numbers listed (translated to unicode) have a left-leading digit of 4 or 5. The unlisted natural number has a left-leading digit of 9, and, unlike omega (969) has a 43. It can be mapped to the unlisted real number. Also note that every order of infinity can be translated into a unique, finite natural number. One concern is the unicode numbers can become quite large compared to untranslated natural numbers. Will the larger unicode numbers reach infinity sooner than their smaller counterparts as the count increases, causing a more incomplete list? The following shows a comparison between n and much larger n^n. If n is the largest number short of infinity, is n^n infinity or beyond?

It appears that n^n will always be finite as long as n is finite and n^n won't be infinite until n is infinite. Thus the unicode natural numbers can map to the same infinite (or finite) list as the smaller untranslated natural numbers. This is good news! There is simply no finite or infinite number that can't be translated into a unique natural number. For every unlisted real number there is a corresponding unlisted natural number. To falsify this conclusion, all one needs to do is show that a Cantor ordinal or combination of Cantor ordinals and natural numbers can't be translated into a unique unicode natural number.

With all the forgoing in mind, let's review the question again: "Is there any set which has more members than the set of natural numbers (N), but fewer members than the set of real numbers (R)?" No, because R and N have the same cardinality.

ADDENDUM:

Is Infinity Real?

By definition, no number is greater than infinity, and, infinity seems unobtainable. The largest number imaginable can always be made larger. If infinity is obtainable, what is the largest finite number just short of infinity? We can agree that an arbitrary line segment has an infinite number of points. We might infer that a longer line segment has a larger infinity of points--but it too has an infinite number of points. How is that possible? If we think of the points as infinite zeros added together, these infinite zeros can add up to any finite number n. The value of n can be determined by how zero is expressed:

Clearly infinity is obtainable when the point length is zero and we are not willing to traverse one point at a time. Instead, we choose to slide past those infinite points to reach a finite number. Also, there doesn't appear to be more than one infinity, since the same infinite sum of zeros leads to any number n. On a grander scale, an infinite number of members leads to any infinite set. Like finite numbers, different infinite sets appear to be different, but involve the same infinity.

Limits and Infinite Precision

If infinity does not exist, then no precise or approximate number would exist? Wait ... approximate numbers should exist, since they don't require infinite precision, right? Wrong. Each approximation is a precise value of itself. To illustrate, suppose we want to expand the exponent e^x using a Taylor series. Let's assume there is no such thing as infinity, so the expansion only has a limited number of terms:

We have an approximation of e^x, but how can we if we don't believe in infinity? Let's set the approximation equal to k, then we shall expand k:

Without an infinite number of terms, we can't have an approximation of e^x--only an approximation of an approximation of e^x. Plus, we can carry out this insanity further with a limited expansion of the approximation of the approximation ... and so on. To claim that we have an approximation is to claim we don't have the number we were shooting for, but we do have the precise value of a number that is close. If we come up short or overshoot a precise point on a number line, we land on a different point that has its own infinite precision. When we have an error margin, that margin is bound by two numbers. Whether they are deemed precise or approximate, they are precisely what they are. That precision implies a Taylor expansion with infinite terms. Even if we insist that a number can't be precisely had, we can at least slide past it along the number line. If there is no infinity, there are no numbers.

Circling back to the question, what is the largest finite number short of infinity? Using the line-segment analogy, there are infinite points between the start and finish. If we can't pin down the exact point that qualifies as the largest finite number, we can at least slide past it.

References:

1. Cantor, Georg. Grundlagen einer allgemeinen Mannigfaltigkeitslehre (Foundations of a General Theory of Sets). jamesrmeyer.com.

2. Cantor, Georg. Uber eine elemtare Frage de Mannigfaltigketslehre (On an Elementary Question of Set Theory). jamesmeyer.com. 3. Cantor's Theorem. Wikipedia

4. 2020. SP20:Lecture 9 Diagonalization. courses.cs.cornell.edu

5. Cantor's Diagonal Argument. Wikipedia

6. Cardinality of the Continuum. Wikipedia

7. Continuum Hypothesis. Wikipedia

8. Huge List of Unicode Symbols. vertex42.com

Why Cantor's Diagonalization Proof Comes Up Short

ABSTRACT:

This paper reveals why Cantor's diagonalization argument fails to prove what it purportedly proves and the logical absurdity of "uncountable sets" that are deemed larger than the set of natural numbers.

Cantor's diagonalization proof (CDP) is used to prove various things like Gödel's Incompleteness theorem (GIT) and the uncountability of real numbers (URN). In the case of GIT, it is assumed that you can have a complete list of provable statements. You use CDP to prove a statement that is not on the list, then conclude the statement is not a provable statement; otherwise, it would be part of the list of all provable statements. However, if your complete list is not really complete, then it makes sense that you can prove a new provable statement not on the list. In the case of proving the URN, you use proof by contradiction: assume the real numbers are countable and assume a complete list of them between, say, .000 ... and .111 ... The list is believed to be complete because an infinite number of counting numbers (1 to n) each allegedly have a one-to-one correspondence to each real number on the list. CDP is then used to show there is a real number that is not on the list (a contradiction). Therefore, according to the proof, there is no bijection between the real numbers and the counting numbers.

Below, we will work through an example proof using CDP and demonstrate why it comes up short. Let's begin with an nXn matrix of real numbers in base 2. We go along the diagonal and flip each bit from 0 to 1 or 1 to 0 to create a new number not on the list. If we think the number is on the list, say, at row r, we can check it and see that its bit at column r doesn't match. We can repeat this exercise for all rows and see that we truly have a new real number not on the list.

Assuming we have used up all the counting numbers to make the complete list of reals from .000 ... to .111 ..., we are forced to conclude that the number of reals exceeds the number of counting numbers by at least one. Therefore, the reals are uncountable (contrary to an initial assumption, via proof by contradiction, that the reals are countable). However, a list of positive integers (counting numbers, including 0) can also be shown to be uncountable:

Using CDP we can create a positive integer that is not on the list! The good news is we can use this new integer to count the new real number, so could it be that the real numbers are countable after all? If your alarm bells are going off, I feel your pain. You may have noticed that the new integer is 0...11 which is 3 in base 10. How can the number 3 be absent from a complete list of positive integers?! Answer: the complete list is not really complete. An nXn matrix only provides a subset of the set of positive integers. This is why CDP can produce numbers that are not part of the list. The assumption that we have a "complete list" of members of a given set is a false premise. Here's why:

Each matrix column corresponds to a digit. Thus we have a list of numbers with n digits. Where there's n digits, there's B^n (B is the number base) numbers with n digits. For example, if B = 2, and there are 3 digits, there are eight possible numbers, not three. If we only list three numbers, there are five more numbers that are not on the list. Once included, though, no amount of diagonal gimmickry will create a three-digit number not part of the list. This is not only true for integers but also real numbers, since reals are just integers with decimal points. Below is a mathematical proof showing that a complete list has 2^n members, not n members:

If the goal is to have a complete list, then we need a (2^n)Xn matrix:

Note that 0...11 is now included on the list. Now, just for the sake of argument, let's assume the list of reals we started with is complete, i.e., countable. By some method m, we derive a new real number not on the list. Let's assume method m is an indisputable method. We conclude that the set of real numbers is larger than the set of counting numbers, and, cannot be counted. But this implies that we run out of counting numbers before we run out of reals, i.e., it is possible to have a higher "count" of real numbers than the the numbers that count them. If we can't count the extra real numbers, then it is meaningless to say we have more, or, a higher count of real numbers and a lower count of the numbers that do the counting. The following diagram provides a visual representation of this absurdity:

If we have a list of n real numbers and derive one more, what does "one more" mean if not n+1 real numbers? If "one more" means n+1, then for every new real number created, there is an additional natural number added as well. With this in mind, we can map the positive integers (including 0) to real numbers in a range of 0 to 1:

Notice that if we take n to the limit of infinity, the number of real numbers grows and so does the number of positive integers. To count the reals between 0 and 1 requires 2^n + 1 counting numbers as n moves to infinity. Suppose we expand the range of reals: 0 to 2^n:

We now need 2^(2n) + 1 (n-->infinity) natural numbers to count this expanded range. If we continue to map 0 to 0, we could map the even positive integers to the positive real numbers, and, the odd positive integers to the negative real numbers:

We now need 2^(2n+1) + 1 (n-->infinity) natural numbers to count real numbers from -2^n to 2^n.

The great thing about the concept of infinity is infinity isn't a finite number. If we think the natural number set N is too small and can't count a bigger real number set R, we can make N bigger without using up its infinite potential. Both sets are forever growing and neither is fixed in stone. Therefore, for every increase in set R there is a corresponding increase in set N. If not, then it is meaningless to claim there are more members in R when its extra members can't be counted.

In conclusion, Cantor's diagonalization result makes perfect sense once we are aware of the fact that the "complete list" is only a partial list of members of a set. This should cast doubt on whatever Cantor's diagonal argument allegedly proves. Additionally, all methods of proving the uncountability of real numbers, etc. fail to address the absurdity of claiming that a set R has more members than the set of counting numbers needed to confirm that R does in fact have more members. It is obvious that there are more real numbers between, say, 1 and 2 than natural numbers. Albeit, the phrase "more real numbers" implies a higher count, i.e., countability.

References:

1. 2020. SP20:Lecture 9 Diagonalization. courses.cs.cornell.edu

2. Cantor's Diagonal Argument. Wikipedia

3. Dr. Pevam. 2018. R is Uncountable. youtube.com

4. Zap, Elvis. 2010. The Cantor Set is Uncountable. youtube.com

5. Hartl, Werner (and a comment from Pendaran Roberts.) Cantors Diagonal Argument Fails. youtube.com