0
$\begingroup$

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.

  • 0
    You have not stated the problem. Despite (or more probably because) of the many symbols that you use, there is a lack of clarity. You presumably mean that if $N$ is composite, then there is an $m$ such that $\dots$. Surely we should not allow $m=1$. Are $m_1$ and $m_2$ *any* numbers such that $m_1\le \sqrt{N} and such that $m_1$ does not divide $N$? Then how are we to conclude anything about $m_2$? Are you only trying to prove that if $N$ has no divisors $d$ with $1 then $N$ is prime?2012-03-20
  • 0
    I'm sorry, I don't understand what you are trying to prove. Set $N=91, m_1=2, m_2=13$ - whether 91 is divisible by 13 is not determined by the fact that it is odd.2012-03-20
  • 0
    @ Andre Nicolas. You are right about my thought process. That's exactly what I thought. I can see it in many examples but tried to prove it here.2012-03-20
  • 0
    @ Mark Bennet. If we set $N$ = 91, then $\sqrt{91} = 9.53.......$. Now, $7$ does divides $91$ and well this shows that $91$ is composite. Also, I don't think that this proof depends on the fact that a number is odd or even.2012-03-20
  • 0
    Andrew, you then need to specify that your condition is true for all integers less than the square root. And it would help at the beginning if you would state that you are trying to prove than N is prime. André has given an answer which reflects my best guess as to what you mean.2012-03-20
  • 0
    @ Mark Bennet. I am really sorry for my stupid comment. I completely see your point and the flaw in my proof.2012-03-20
  • 0
    Andrew - I've posted some embarrassingly silly things in my time, but if I hadn't, I wouldn't have learned ...2012-03-20

2 Answers 2