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 March 15th, 2016, 11:20 AM #1 Newbie   Joined: Mar 2016 From: Belarus Posts: 4 Thanks: 0 Help prove or disprove my hypothesis Help prove or disprove my hypothesis: Theorem (Axiom) There is an ordered set of natural numbers, which is given by the function f (x) (x - a natural number). The operation of addition is defined on this set only if the successor function S (x) associated with the operation of addition in the form of: f (n) + f (n + 1) = f (k) (1) Or f (k) + f (n) = f (n + 1) (2) k, n - integers If the set has a pair of consecutive numbers that satisfy the condition (1) or (2), the set has an infinite number of triples that are solution of the equation a + b = c (3) If the set does not have a pair of consecutive numbers that satisfy the condition (1) or (2), the set has no members who are solution of the equation (1). On this set the operation of addition is not defined, because it is not connected with the following function. Corollary. It is easy to show that the sets that are defined function f (x) = x ^ n (n> 2) does not satisfy the condition (1) or (2). Therefore, the equation a ^ n + b ^ n = c ^ n (n> 2) has no solution in nonzero whole numbers. March 15th, 2016, 01:27 PM   #2
Math Team

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Quote:
 Originally Posted by AndreyShvets The operation of addition is defined on this set only if the successor function S (x) associated with the operation of addition in the form of:
That's not a sentence... Also, what's $S(x)$? I'm not familiar with the "successor function". March 15th, 2016, 07:53 PM #3 Newbie   Joined: Mar 2016 From: Belarus Posts: 4 Thanks: 0 I talked about it: https://en.wikipedia.org/wiki/Successor_function March 16th, 2016, 12:42 AM #4 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 Okay, so $S(x) = x + 1$. Now what does that sentence mean? March 16th, 2016, 01:54 AM #5 Newbie   Joined: Mar 2016 From: Belarus Posts: 4 Thanks: 0 Refinement (translation problems) ... Theorem (or axiom) Is an ordered set of integers , is given by the function f (x) (x - is an integer). the addition operation is defined on this set, only if there is n, such that the function sequence S (x) associated with the operation of addition: f (n) + f (n + 1) = f (k) (1) Or f (k) + f (n) = f (n + 1) (2) k, n, f (n), f (n+1), f (k) - an integer nonzero number If the set has a pair of consecutive numbers that satisfy the condition (1) or (2), the set has an infinite number of triples that are solution of the equation (the addition operation) a + b = c (3) If the set does not have a pair of consecutive numbers that satisfy the condition (1) or (2), the set has no members who are solution of the equation (3). On this set the operation of addition is not defined, because it is not connected with the following function. Corollary. It is easy to show that the sets that are defined function f (x) = x ^ n (n> 2) does not satisfy the condition (1) or (2). Therefore, the equation a ^ n + b ^ n = c ^ n (n> 2) has no solution in nonzero whole numbers. Last edited by AndreyShvets; March 16th, 2016 at 02:18 AM. Tags disprove, hypothesis, prove ### content

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