Well the following post isn't really a question but actually a verification of a proof of a problem. I would be highly grateful if you would just check my proof. Here goes the problem and the proof:-
Let $N $ $\epsilon $ $ \mathbb{Z}^+$. Now, we know that if N is composite then $m$ divides $N$ where $1\le m \le \sqrt{N}$. I wanted to see that if $m_1$, $m_2$ $\epsilon $ $ \mathbb{Z}^+$ such that $m_1\le \sqrt{N} and $m_1 \nmid N$, then $m_2\nmid N$. Here $m_1$ is any natural number $\le \sqrt{N}$ and $m_2$ is any natural number $> \sqrt{N}$ but $.
PROOF:- Let $m_1$, $m_2$ $\epsilon $ $ \mathbb{Z}^+$ such that $m_1\le \sqrt{N} and $m_1 \nmid N$. Let us assume that $m_2|N$. $\Rightarrow$ $N$ = $\alpha m_2$ and $N$ = $\beta m_1 + \gamma$ where $0<\gamma. From the above equations, we can say that:- $\alpha m_2$ = $\beta m_1 + \gamma$ where $0<\gamma.$\Rightarrow$ $\beta m_1 = \alpha m_2 - \gamma$ where $0<\gamma or $\beta m_1 = \alpha^{\prime} m_2 + \gamma^{\prime}$ where $0<\gamma^{\prime}. $\Rightarrow$ $m_2\nmid\beta m_1$ or $m_2\nmid(N-\gamma)$ $\Rightarrow$ $(N-{\gamma})= \alpha^{\prime\prime} m_2 + \beta^{\prime}$ where $0 < \beta^{\prime} or $N= \alpha^{\prime\prime} m_2 + \beta^{\prime} + \gamma^{\prime}$ or $N= \alpha^{\prime\prime\prime} m_2 + \beta^{\prime\prime} $ where $0< \beta^{\prime\prime} < m_2$ $\Rightarrow$ $m_2 \nmid N$. But this is contradiction to our assumption. Hence $N$ is a prime number.
Note:- Here $\mathbb{Z}^+$ is the set all non zero positive integers only.