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# Division algorithm in Wolfram Alpha

Posted 11 months ago
 About 4 years ago WolframAlpha was using Euclidean division algorithm. The result of dividing -11 by 5 was the quotient equals -3 and the remainder equals 4. Which was absolutely correct according to Euclidean definition.Nowadays, the result is -2 and -1 respectively. Which is correct in the Modulo world, but not in the Euclidean one. The remainder, according to a vast majority of sources, is 0 ≤ r < |b|, so it simply can't be negative. On top of it, the remainder is what is left after we found smallest multiple of the divisor, right? The smallest multiple of 5 in respect to -11 is -15, because -10 is actually greater than -11. Why was it changed to yield an incorrect result?
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Posted 11 months ago
 You're right, 'incorrect' is too strong of a word to use here. Let's say it used to be strictly Euclidean and now it is not. The input was as simple as -11/5 (I tried a fraction as well, same result). The output section I'm talking about is 'Quotient and remainder'. In the past I used this example to explain to kids division algorithm on integers through subtraction, as a demo I was using Alpha and it was producing output based on Euclidean division.
Posted 11 months ago
 I see. Thanks for the clarification. When I click "Step-by-step solution" I read To compute -11 / 5 compute 11 / 5 and replace the negative sign at the end. So, this seems less about using a different variant of Euclidean division and more about re-envisioning the entire explanation. I don't have the pro version, so I can't see the rest of the explanation. I don't see anything on the page that refers explicitly to Euclidean division, so I don't think you can fault the information provided, but I can understand being frustrated with it changing.
Posted 11 months ago
 I don't think we can answer specifically without knowing the exact input you used and which part of the result you're referring to. You can't really say that it's incorrect in this unqualified sense, because we can use any constraint on the remainder that has the right "width". It also might be the case that the result was doing division with positive 11 and just adding the negative sign back in at the end.