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For instance, how does the proof for 7 being prime work?

We can start from 1 and work up to to 7 and show that 7 has exactly two factors, namely 1 and 7. But, how do we rigorously establish that no number greater than 7 can be a factor of 7?

The definition of factor is as follows: For all n, For all x element of N, x is a factor of n iff There exists k element of N in such a way that n=kx

So, the question can be rephrased as a proof that For all n, For all x element of N, x is greater than n => For all k element of N, n is not equal to kx

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  • $\begingroup$ Don't use the term 'natural numbers' (which is ambiguous) and instead call them positive or nonnegative integers (which has clear meaning). Your statement is wrong if you include $0$ in $\Bbb N$. $\endgroup$
    – user26486
    Jun 3, 2015 at 11:13

3 Answers 3

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Hint: if $m$ and $n$ are positive integers, then $mn \ge n$.

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Let $n$ be a natural number and let $d \mid n$ (where $d\in \mathbb{N}$). Assume $d>n$. Note that by definition of divisibility, $n=kd$ with $n,k,d \in \mathbb{N}$. Then $d>kd$ implying $1>k$. There are no natural numbers less than $1$. So $d\leq n$.

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If you are talking about positive integer numbers and factors, then, if $a=db$ then $$a-b=db-b=b(d-1)\ge0$$ This implies that $a\ge b$.

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