Difference between revisions of "Talk:410: Math Paper"
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Shouldn't it say something about the whole math licence, and that you don't actually need a licence to do math? [[Special:Contributions/108.162.231.228|108.162.231.228]] 21:01, 31 October 2013 (UTC)Synthetica | Shouldn't it say something about the whole math licence, and that you don't actually need a licence to do math? [[Special:Contributions/108.162.231.228|108.162.231.228]] 21:01, 31 October 2013 (UTC)Synthetica | ||
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| + | Despite what this comic implies, the divisor function is defined over the Gaussian integers. There still is a problem, though. If a divides b, then so does -a, along with ai and -ai. The divisors will inevitably sum to zero. You could get around this by ignoring all the numbers that aren't in a given quadrant. I personally like the idea of using ones where the real part is greater than the imaginary part (although that still does become a problem with multiples of 1+i). This way, a friend of a natural number will also be a natural number (though it's only the same as what you'd get normally if all the factors are three mod four). | ||
Revision as of 08:14, 17 November 2013
Shouldn't it say something about the whole math licence, and that you don't actually need a licence to do math? 108.162.231.228 21:01, 31 October 2013 (UTC)Synthetica
Despite what this comic implies, the divisor function is defined over the Gaussian integers. There still is a problem, though. If a divides b, then so does -a, along with ai and -ai. The divisors will inevitably sum to zero. You could get around this by ignoring all the numbers that aren't in a given quadrant. I personally like the idea of using ones where the real part is greater than the imaginary part (although that still does become a problem with multiples of 1+i). This way, a friend of a natural number will also be a natural number (though it's only the same as what you'd get normally if all the factors are three mod four).
