I just don't know where to start, I tried induction but that didn't work. I wrote a and m are greater than 0 and and we can say a is congruent to r(modm) but im notsure if thats correct. Any hints?
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This question is very similar to the more recent one asked at How to prove if x ∈ ℕ and y ∈ ℕ, then (x mod y) ∈ ℕ?. – John Omielan Feb 07 '19 at 02:57
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Let $a$ and $m$ elements of $\mathbb{Z}$ and $A=\{a-mk \mid k\in \mathbb{Z}\}\cap\mathbb{N}$. $A$ is a non empty subset of $\mathbb{N}$, then contain a least element, $r$ say. Since $r\in A$, there must be $q\in \mathbb{Z}$ so that $a-mq=r$. We now prove that $r<|m|$.
Let us assume that $r\geq |m|$, then $0\leq r-|m|=a-m(q+\epsilon), \epsilon=\pm 1 \in A$, contradiction because $r$ is minimal. This prove existence of $(q,r)$
DINEDINE
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