TreueEckhardt2

The quadratic equation is definitely something that would be part of a high school algebra class.

I have a question about something similar to the twin prime conjecture. If you have two consecutive numbers, is there a limit to how smooth they can be? Here is the basic definition: If the largest prime factor of a number is less than or equal to the nth root of that number, then the number is nth root smooth. I am curious about pairs of consecutive numbers that are both nth root smooth for large values of n. Let’s call them twin nth root smooth numbers. For example, 2400 and 2401 are consecutive numbers. The largest prime factor of 2400 is 5, and 5 is less than the 4th root of 2400, so 2400 is 4th root smooth. The largest prime factor of 2401 is 7, and 7 is equal to the 4th root of 2401, so 2401 is 4th root smooth. Thus, 2400 and 2401 are twin 4th root smooth numbers. I have found an example of twin 10th root smooth numbers. Let A=2^210-1 and B=2^210. A is equal to the following 64-digit number: 1645 504557 321206 042154 969182 557350 504982 735865 633579 863348 609023 The largest prime factor of A is 1,564,921. I found this by using the calculator at the following website: https://www.alpertron.com.ar/ECM.HTM Since the largest prime factor of A is smaller than the 10th root of A, A is 10th root smooth. The largest prime factor of B is 2, and since 2 is smaller than the 10th root of B, B is 10th root smooth. So A and B are twin 10th root smooth numbers. My question is this: Is there a largest value of n for which twin nth root smooth numbers exist? I haven’t been able to find any examples twin nth root smooth numbers for values of n larger than 10, but I don’t suppose that means a lot.

I think I have found an answer to the question. For a random positive integer, the probability that its largest prime factor is greater than the square root of the integer is ln2. I found the answer on the following web page: https://mathworld.wolfram.com/GreatestPrimeFactor.html

Do most composite numbers have a large prime factor? First, I’ll define what I mean by a “large” prime factor. Let N be a number. If a prime factor of N is greater than the square root of N, then that factor is a large prime factor of N. As an example, 11 is a large prime factor of 22, because 11 is greater than the square root of 22, and so 22 has a large prime factor On the other hand, 3 is not a large prime factor of 12 because 3 is less than the square root of 12, and so 12 does not have a large prime factor. Below is a list of composite numbers with large prime factors: 6, 10, 14, 15, 20, 21, 22, 26, 28, 33, 34, 35, 38, 39, 42, 44, 46, 51, 52, 55, 57, 58, 62, 65, 66, 68, 69, 74, … It seems that, as numbers increase, a greater and greater percentage of them have large prime factors. I say that that seems to be true, because I have sampled some groups of big numbers, and most of them had large prime factors. Of course, that isn’t proof and as far as I know it could also be wrong. If we check all of the numbers up to 330, the majority of counting numbers are composite numbers with large prime factors. If I understand it correctly, then what I’m asking about is similar to the question answered by the Prime Number Theorem. According to the Prime Number Theorem, for a very large number N, the probability that a random integer not greater than N is prime is equal to 1/log(N). Because the prime numbers are distributed in this way, and 1/log(N) can be arbitrarily close to zero, the composite numbers can be seen as essentially the same as all integers, for very large values of N. For very large numbers, my question is the same as asking what percentage of all integers have a large prime factor. My question is, “For a very large number N, what is the probability that a random integer less than N has a large prime factor?” “Is this probability greater than 0.5?” I’m hoping there might be some kind of answer to this in the same way that the Prime Number Theorem answers the question about the distribution of prime numbers.

Thank you for your response. When you say, "This thread can be analyzed under algebra and number theory," Do you mean that I should post this topic on a different website that deals with algebra and number theory? Or maybe I should post to the general forum instead of this subforum?

Do most composite numbers have a large prime factor? First, I’ll define what I mean by a “large” prime factor. Let N be a number. If a prime factor of N is greater than the square root of N, then that factor is a large prime factor of N. As an example, 11 is a large prime factor of 22, because 11 is greater than the square root of 22, and so 22 has a large prime factor On the other hand, 3 is not a large prime factor of 12 because 3 is less than the square root of 12, and so 12 does not have a large prime factor. Below is a list of composite numbers with large prime factors: 6, 10, 14, 15, 20, 21, 22, 26, 28, 33, 34, 35, 38, 39, 42, 44, 46, 51, 52, 55, 57, 58, 62, 65, 66, 68, 69, 74, … It seems that, as numbers increase, a greater and greater percentage of them have large prime factors. I say that that seems to be true, because I have sampled some groups of big numbers, and most of them had large prime factors. Of course, that isn’t proof and as far as I know it could also be wrong. If we check all of the numbers up to 330, the majority of counting numbers are composite numbers with large prime factors. If I understand it correctly, then what I’m asking about is similar to the question answered by the Prime Number Theorem. According to the Prime Number Theorem, for a very large number N, the probability that a random integer not greater than N is prime is equal to 1/log(N). Because the prime numbers are distributed in this way, and 1/log(N) can be arbitrarily close to zero, the composite numbers can be seen as essentially the same as all integers, for very large values of N. For very large numbers, my question is the same as asking what percentage of all integers have a large prime factor. My question is, “For a very large number N, what is the probability that a random integer less than N has a large prime factor?” “Is this probability greater than 0.5?” I’m hoping there might be some kind of answer to this in the same way that the Prime Number Theorem answers the question about the distribution of prime numbers.

This is what I'm thinking might be the answer to my question: When you are talking about Pythagorean triples, you are usually talking about whole numbers. Pythagorean triples have a geometric meaning because they can be used for the sides of a right triangle, but the fact that they are whole numbers doesn't have anything to do with this geometric meaning. Is that right? I suppose you could say that 1+2=3 is an irrational Pythagorean equation. I think Pythagorean triples are interesting. When I first started thinking about complex Pythagorean triples, I was thinking of numbers of the type a+bi where 'a' and 'b' are both integers. I think you might use the term "complex integers", right? I think solutions to the Pythagorean equation that are complex integers are interesting, and I suppose they might have some meaning or significance, (although I don't know what that might be) but I'm guessing that they don't have any geometric meaning, because any three complex numbers x, y, and z, where x+y=z can be made to be the solution to a complex Pythagorean equation. The only thing that might make complex Pythagorean triples special is the fact that they are complex integers, and that doesn't have anything to do with Geometry. Any comments?

You're right, the resultant is the sum of two vectors. I don't know the term for the product of two complex number vectors. How about if I use the words "product vector"? I understand the next sentence. We're talking about two product vectors. Then I think I understand the first part of the next sentence. Since the two product vectors are both vectors themselves, when we add them, we get a "resultant in the proper sense", that is, the sum of two vectors. Now for the next part. If you add any two vectors together, you will get a third vector, the resultant, that will "close the triangle" in the complex plane. Is that what you were saying? That is true, but I don't think that really answers the question. If you take any two vectors and add them together and show that they equal another vector, you don't necessarily have a Pythagorean equation. Do the vectors in a complex Pythagorean equation have any geometrical meaning beyond the meaning that you get when you add any two complex numbers together? I've been thinking about my last post. If you take any complex number it does have a square root that is another complex number, so any equation x+y=z is necessarily a complex Pythagorean equation. Thinking about that I think that answers my question.

Whether or not you have the means of making drawings, thanks for your reply. Yes, what you said about stretching and rotation is right. So, using my example, when we multiply (3+18i) by itself, the original vector, which had an angle measure of about 80.5 degrees is rotated another 80.5 degrees, so the resultant has a measure of about 161 degrees. Also, the original vector had a magnitude of about 18.2, and the resultant is stretched out to a magnitude of about 333. So the vector (3+18i) does have a geometric meaning, and when you square a single complex number, the resulting square has a geometric meaning that is related to the original vector. A Pythagorean equation is something different though. We have three numbers. If the numbers are real then they have a geometric meaning relating to right triangles. But what if the numbers are not real? There are solutions to the equation, and they look interesting (to me at least) but do they mean anything geometrically?

Yes, I do mean squares. If we multiply by the complex conjugate then one side of the equation doesn't equal the other side, right? (169+36)+(36+484)=(9+324) is false. But if we square the two complex numbers on the one side and add them together, then they do equal the square of the complex number on the other side. (169-156i-36)+(36+264i-484)=(9+108i-324) is true. Unless, of course, I've made some mistake in the math. My question is, does it mean anything Geometrically?

Real Pythagorean triples are numbers like 3, 4, and 5. If a, b, and c are three positive real numbers and [math]a^{2}+b^{2}=c^{2}[/math], then a, b, and c can be used as the sides of a right triangle. So real Pythagorean triples have geometric meaning. But what if a, b, and c are not real numbers? For example: [math](-13+6i)^{2}+(6+22i)^{2}=(3+18i)^{2}[/math] If three complex numbers form a Pythagorean triple (and they are not real numbers) do they have a geometric meaning?

Maybe it would help if I talked a bit about what kind of feedback I'm looking for. It seems that this is quite a simple problem and it seems that someone else would have found a solution for it, but I haven't found any other mention of a solution for this problem. If someone has seen a different solution to this problem, I would appreciate if they could tell me where I could find it. Just so we can be clear about what I am looking for, let me restate the problem. Suppose someone makes a series of predictions that certain events have a certain probabilities of happening. The predicted probabilities are not close to one and not close to zero. Some time later, we see which events have happened and which have not happened. How do we judge whether the predictions were good predictions or bad predictions? I would love to hear comments on this topic.

This post deals with a mathematical formula for deciding how well people have made predictions. I would like to receive feedback about this formula and, if possible, test it out with participants in this group in the upcoming March Madness NCAA basketball tournament. In order to help everyone understand what I'm talking about, I'll start with an example. Suppose three people, Adams, Benson, and Carter are watching a basketball game between Iowa State and Connecticut. Adams thinks that Iowa State has a 60% chance of winning, Benson thinks that Iowa State has a 70% chance of winning, and Carter thinks that Iowa State has an 80% chance of winning. Suppose Iowa State wins. How do we know who has made the best prediction? The following game that is designed to answer this question. The participants in this game make a series of predictions about the outcome of future events. They predict the percentage chance that certain events will happen. Later, when it is determined whether or not the event actually did happen, each participant gains or loses a certain number of points based on their predicted percentage chance that the thing would happen. Below is an example of this game followed by the mathematical formulas on which this game is based. If you predict Then if they win, And if they that the team you win this lose, you lose has this chance number of this number of of winning points points 1% 202 1 2% 408 4 5% 1052 28 10% 2222 123 15% 3529 311 20% 5000 625 25% 6667 1111 33% 9851 2426 40% 13333 4444 45% 16364 6694 50% 20000 10000 55% 24444 14938 60% 30000 22500 66% 40000 40000 75% 60000 90000 80% 80000 160000 You should circle no more than one of the percentages in each of the lines below. You should circle the percentage chance that you think each of these teams has to win their game(s). South What is the probability that #1 seed Kentucky will win its first five games 1% 2% 5% 10% 15% 20% 25% 33% 40% 45% 50% 55% 60% 66% 75% 80% #9 seed Connecticut will win against #8 seed Iowa State 1% 2% 5% 10% 15% 20% 25% 33% 40% 45% 50% 55% 60% 66% 75% 80% #8 seed Iowa State will win its first two games. 1% 2% 5% 10% 15% 20% 25% 33% 40% 45% 50% 55% 60% 66% 75% 80% Now let's run through a betting example from beginning to end. First, a person named Adams predicts that there is a 40% chance that Kentucky will win its first five games, a 40% chance that Connecticut will win its game against Iowa state, and a 40% chance that Iowa State will win its first two games. Second suppose that Kentucky does win its first five games, but Connecticut loses to Iowa State, and Iowa State loses its second game. Since Kentucky won its first five games, Adams gains 13333 points, but since Connecticut lost, he loses 4444 for that bet, and he also loses 4444 points for his bet on Iowa State winning its first two games. For these three bets, Adams gains 13333 points and loses 8888 points, and he has a score of +4445. Now for the formulas: The formulas are based on the following idea: Suppose I ask you to make a bet. You are betting that X will happen. If X does happen, then you will win some points, but if X does not happen, you will lose that number of points squared. For example suppose I ask you to bet on whether a tossed coin will lands on heads. If you bet two points, then if the coin lands on heads, you get two points, but if it lands on tails you lose four points. Betting two points is clearly not a winning strategy, but you can also bet a number smaller than one. Suppose you bet 0.5 points. Then if the coin lands on heads you get 0.5 points, but if it lands on tails you only lose 0.25 points. That's a winning strategy for a coin toss. We want to find, not just a winning strategy for this game, but the optimal strategy. To do this, we need to write out a formula for the game. This formula is given below. s=pb-(1-p)b^2 In this formula, p is the probability that something will happen, b is the number of points that you bet that the thing will happen, and s is the average score that you would expect to receive if you made a large number of bets with a fixed probability. In this situation, we are assuming that you know the probability that the thing will happen, and you are trying to choose the best number of points to bet in order to get the highest possible score. So p is a constant, b is the independent variable, and s is the dependent variable. We find the optimum score by finding the maximum point of the graph of this equation. Since the graph of this equation is a parabola, the result is straightforward. It will have a maximum where b has the following value: b=p/2(1-p) So, if Adams thinks that Connecticut has a 0.4 chance of winning a basketball game, then he should bet approximately 0.33 points. If Connecticut does win, then he will gain 0.33 points. if they lose, he will lose 0.11 points. Now look at the table below. This is the table that we have been using for the basketball prediction game that I have just been describing. 1% 202 1 2% 408 4 5% 1052 28 10% 2222 123 15% 3529 311 For example, suppose a team has a 1% chance of winning. Then the amount that you should bet is 1/198, and the amount that you will lose if the team loses is 1/39204. I multiplied both of those numbers by 40,000 so that the numbers in your score would be whole numbers rather than fractions or decimals. I then multiplied all of the other values for 'b' by 40,000 and rounded them to the nearest whole number. I have tested this prediction game with some friends and family and also several times with my high school math classes. I would like to test it out in this forum and I am interested in receiving feedback on it. Please let me know if you are interested in participating.