The question says it all. It is obviously true, but how do you prove it with actual mathematical symbols?
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1What is your definition of $\bmod ?\ \ $ – Bill Dubuque Feb 07 '19 at 00:40
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Below I have already accepted the answer – Math Bob Feb 07 '19 at 00:49
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Why the downvote? There is actually no more context to give, and based on the answer below, it is a one step proof, so I wouldn't have any attempts unless I have no idea how to prove this. – Math Bob Feb 07 '19 at 00:54
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1@MathBob Wasn't me, but I suspect it's because the (not exactly standard) meaning of the terms used in your question aren't found in your question. They're not even found in the comments; they had to be speculated by the answerer. – Theo Bendit Feb 07 '19 at 01:21
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This question is very similar to the recent one asked at Use WOP to prove that for integers $a, m > 0$ there are integers $q, r \ge 0$, with $r \in {0,1,2,\dots, m-1}$ such that $a=qm+r$. – John Omielan Feb 07 '19 at 03:03
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I suspect this is your definition of operation 'mod' $$ x\text{ mod }y = \min\{ x-k\cdot y : k\in \mathbb{Z}\,\land\, x-k\cdot y\geq0\} $$ Directly from definition, $x\text{ mod }y\in\mathbb{Z}_+ = \mathbb{N}$.
Note that the minimum of that set exists, because for example $$ x\in \{ x-k\cdot y : k\in \mathbb{Z}\,\land\, x-k\cdot y\geq0\}$$ and natural numbers are a well-ordered set (every non-empty subset of them has a minimum).
Jakobian
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One may want to use well-ordering to show that there is a min to that set ... if one wanted to be pedantic, which perhaps the OP wants to be. – kcrisman Feb 07 '19 at 00:36
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I have accepted your answer at least I tried. It says I can up vote in four minutes:) – Math Bob Feb 07 '19 at 00:40
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