Talk:Remainder is Primitive Recursive

Can be slightly simpler
If I’m not mistaken, then when using a convention $a\bmod 0 = a$ (more on justification of that at SE), it’s possible to omit $\operatorname{sgn}(m)$ from the final expression. Analogously it allows removing that same subexpression from the final result at Quotient is Primitive Recursive. (I don’t think it’s essential enough for the potential controversy of this convention (though I myself think it’s more useful than any other), and I’m inexperienced with the traditions here, so I won’t edit these pages, but maybe someone would decide it’s woth the candle?) --Arseniiv (talk) 20:20, 10 June 2019 (EDT)


 * The development of this subject on has not at this point defined $\bmod$ as a primitive recursive function, so it cannot be used in a proof of primitive recursion. --prime mover (talk) 01:28, 11 June 2019 (EDT)