|March 30th, 2017, 11:10 PM||#1|
Joined: Mar 2017
From: Nairobi, Kenya.
Math Focus: Number theory
Proof to Collatz Conjecture.
Consider any positive integer n from which the sequence is formed. n has a probability of 0.5 of either being odd or even. If even, we divide it by two. If odd, we multiply it by three, add one and then divide the result by two since the resulting number must be even. This is same as multiplying n by 1.5 and adding 0.5. The resulting integer, say m, will hence either be n/2 or 1.5n+0.5 and also has a probability of 0.5 of being either even or odd. The 0.5 that is added has considerable effect on outcome only if n is 1 (which can explain the repeating cycle when the sequence reaches 1). In this case we will neglect it.
Since n has equal chances of being odd or even, it therefore has equal chances of being divided by 2 or being multiplied by 1.5, and so is m and the rest of the outcomes. The factor by which the outcomes are divided by is greater than the factor they are multiplied with and hence, the sequence will converge.
|collatz, conjecture, proof|
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