# Number Theory

## This Prime Generating Product generates successive prime factors

Any integer greater than 1 is called a prime number if and only if its positive factors are 1 and the number p itself. The basic ideology involved in this post is flawed and the post has now been moved to Archives. Prime Generating Formulas We all know how hard it is to predict a formula for prime numbers! They have extremely uncertain patterns at various number ranges. Some prime

## A Possible Proof of Collatz Conjecture

Our reader Eswar Chellappa has sent his work on the solution of ‘3X+1’ problem, also called Collatz Conjecture. He had been working on the proof of Collatz Conjecture off and on for almost ten years. The Collatz Conjecture can be quoted as follow: Let $\phi : \mathbb{N} \to \mathbb{N}^+$ be a function defined  such that: \phi(x):= \begin{cases} \frac{x}{2}, & \text{if } x \text{ is even } \\

## My mobile number is a prime number

My personal mobile number 9565804301 is a prime number. What is a prime number? Any integer p greater than 1 is called a prime number if and only if its positive factors are 1 and the number p itself. In other words, the natural numbers which are completely divisible by 1 and themselves only and have no other factors, are called prime numbers. 2 , 3,5,7,11,13…  etc. are prime numbers [or just Primes].

## Complete Elementary Analysis of Nested radicals

This is a continuation of the series of summer projects sponsored by department of science and technology, government of India. In this project work, I have worked to collect and expand what Ramanujan did with Nested Radicals and summarized all important facts into the one article. In the article, there are formulas, formulas and only formulas — I think this is exactly what Ramanujan is known

## Smart Fallacies: i=1, 1= 2 and 1= 3

This mathematical fallacy is due to a simple assumption, that $-1=\dfrac{-1}{1}=\dfrac{1}{-1}$ . Proceeding with $\dfrac{-1}{1}=\dfrac{1}{-1}$ and taking square-roots of both sides, we get: $\dfrac{\sqrt{-1}}{\sqrt{1}}=\dfrac{\sqrt{1}}{\sqrt{-1}}$ Now, as the Euler’s constant $i= \sqrt{-1}$ and $\sqrt{1}=1$ , we can have $\dfrac{i}{1}=\dfrac{1}{i} \ldots \{1 \}$ $\Rightarrow i^2=1 \ldots \{2 \}$ . This is complete contradiction to the fact that $i^2=-1$ . Again,