Math-Forums » Number Theory
by Murray Cantor
1M ago
Let's call a number N 'sub-perfect' if the sum of its divisors less than N is N-1. (There may be a different term for such numbers). It is easy to see that powers of 2 are sub-perfect. Does anyone know a proof or counter-example to the conjecture that all sub-perfect numbers are powers of 2 ..read more
Math-Forums » Number Theory
by syndixxx
2M ago
I am sorry for my English. This is a repost from one forum member from another mathematical forum. I'll try to translate it into English (as best I can). [this is a text from another person on the forum]: Since Petrov is being quoted here, he also has a much more interesting article devoted to prime numbers: “Petrov I.B. METAREMULTION (general superficial numerical study of an interesting prime number)” Author’s article, self-publishing, 2023, 5 pp. (attached to... Is Petrov’s “metaremultion” finite ..read more
Math-Forums » Number Theory
by nemo
4M ago
My idea of proving this theorem is presented in the attacheded document. May be it is not worth your time but I was facinated by the simplicity of the proposition and devoted a good amount of time and thinking ..read more
Math-Forums » Number Theory
by John.Anderson
5M ago
Hi think I may have found a new formula for the Riemann hypothesis. This is how you find the real side of the limit of zeta(n+bi): https://www.desmos.com/calculator/rcyapfsmuz Here is how you can find the imaginary side of the limit: https://www.desmos.com/calculator/uaociu20r4 The y=a(60) on the top is your final answer, you can increase its accuracy by setting the v to a larger number or by plugging in a higher number into that y=a(60). You can check the final answer... New formula for the Riemann Hypothesis ..read more
Math-Forums » Number Theory
by Logic
6M ago
Addition, subtraction, multiplication and exponentiation modulo n are no problem. Division modulo n (often) IS a problem since fractions are not allowed, only integers. Normally you would calculate the answer to a modular division by calculating the multiplicative inverse of the denominator and then multiply the numerator by that inverse. (everything mod n of course) This is perfectly okay and works fine. My question is: is there an algorithm that gives me the answer to a modular division... Modular Division Algorithm ..read more
Math-Forums » Number Theory
by eski
7M ago
Hey guys so I was wondering what you thought of this number Meum. I met it while talking to myself - ironically in a jail. I was trying to compute the shapes of fractal equations in my head and got a shiver down my spine. "Why does this number show up everywhere, who are you? "I am meum, 1.19758, I exponentiate to infinity perfectly. Years later I found the top level equations. Apparently the value doesn't vary ..read more
Math-Forums » Number Theory
by Seff
7M ago
Diagonaliztion as a process involves constructing a number that cannot possibly exist in an infinite list of numbers of a set such as the reals, then because that list was assumed to have a bijection with the naturals it concludes that a bijection is impossible. This conclusion however is flawed in that it is never tests if diagonalization will also create a new natural number not in the list of natural numbers that we can then use to continue the bijection. Say we have a list of all... The flaw in Cantor's Diagonalization Argument ..read more
Math-Forums » Number Theory
by Seff
7M ago
The natural numbers are analogous to an infinite line with a beginning and no end. We can continue that line into the integers and it becomes infinite in two directions. A line is defined by two points, we'll denote those two points as a set {x,y}. For the natural numbers x is 1 and y is infinitely large. For the integers x is infinitely small and y is infinitely large. As such I'll denote the naturals as {1,y} and the integers as {x,y}. When we expand into the rational numbers we are... Working on a theory of countable cardinals ..read more
Math-Forums » Number Theory
by Seff
7M ago
Claim: The real numbers are countable Proof using bijection Create a bijection between the natural numbers > 1 and the set of all multisets of prime numbers using the definition of prime factorization and index each correspondence using the natural numbers enumerated by n. N:n>1⇒ {M(p)} 1: 2 ⇒ {2} 2: 3 ⇒ {3} 3: 4 ⇒ {2,2} 4: … Using a bijection between the natural numbers and the prime numbers in sequence of least to greatest, create a... Proof of Countable Reals ..read more
Math-Forums » Number Theory
by Seff
7M ago
Let the set of natural numbers be denoted by N. Let the set of real numbers between 0 and 1 be denoted by R_01. Let f be a function that maps each natural number n to a real number between 0 and 1, where f(n) is obtained by reversing the digits of n and removing any trailing zeroes. If f(n) has a repeating decimal of the form 0.999..., then increment the digit in the next highest place by 1 and remove the repeating nines. Formally, we can define f as follows: f: N → R_01 f(n) = 0 if n = 0... A very simple bijection between reals and naturals ..read more

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