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Gave my shot at a proof of the above statement. It goes like this:

Proof: Let $n$ by the Fundamental Theorem of Arithmetics, we express it as a sequence of primes $n=p_1,^{\alpha_1} \cdot \cdot \cdot p_i^{\alpha_i} \cdot \cdot \cdot p_s^{\alpha_s}$ such that $\forall_i \{ \alpha_i \geq 1\} \land s \geq 1 $.

Then if $n | ab$ we can say $\exists c \in \mathbb{Z}$ st $nc = ab$. Let, $a = p_1^{\alpha_1} \cdot \cdot \cdot p_j^{\alpha_j} $ and $b = p_{j+1}^{\alpha_{j+1}} \cdot \cdot \cdot p_s^{\alpha_s} \cdot \cdot \cdot p_k^{k}$. It is clear that $n \nmid b$ and $n \nmid a$.

Im not sure if this cover all cases, any comments are appreciated as I am getting on proves after a long time.

Bill Dubuque
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Martin Carrasco Castaneda
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  • $j$ would be $1 < j < k$ but, I now see the problem. If $s=1$ then the "slice" I was trying to remove would not be remove but stay in $a$, then $b=1$ and $n=a$ when the idea is that if $a < n$ and $b < n$ then necessarily $n \nmid a$ and $n \nmid b$ – Martin Carrasco Castaneda Apr 10 '24 at 12:22