41–50 of 74 posts
I remember when negative zero was labelled a hoax. (-: * https://en.wikipedia.org/w/index.php?title=-0&diff=25603603&... * https://en.wikipedia.org/wiki/Wikipedia:Articles_for_deletio...
Although it smells like a design flaw (which negabinary conveniently doesn't have)
i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1
Can anyone provide some examples of practical applications of this?
― Aldous Huxley
Taken one step further, could you have irrational bases? i^1 = i i^2 = -1 i^3 = -i i^4 = 1
Earlier quoted context omitted.
I don't think anyone ever imagined a use for imaginary numbers either, but those turned out to be quite useful for reducing dimensionality. Towards the bottom of the article it states that Donald Knuth proposed imaginary base numerical systems. So this may eventually find a use, likely with higher-dimensional math.
It will be really funny if there is some bit of new, undiscovered physics, that went undiscovered for so long just because it's best described using some esoteric maths like complex base numbers. Highly unlikely, but fun to think about.
Earlier quoted context omitted.
It will be really funny if there is some bit of new, undiscovered physics, that went undiscovered for so long just because it's best described using some esoteric maths like complex base numbers. Highly unlikely, but fun to think about.
Are there any examples where the representation of a number matters in physics?
Actually, another neat example is analog vs digital computation. your models can get very different answers. I recall some Mandelbrot guy talking about that.
Earlier quoted context omitted.
Well, the number of digits tell you the sign of the number, right? It could save a sign bit I think the ideal wikipedia page for this kind of thing has a 'Properties' section from which to start thinking about this
The bit saved for the sign is lost because you need more bits to represent a number.
Earlier quoted context omitted.
XVI ⨉ XXIII = XVI ⨉ (XVI + I) = (XVI ⨉ XVI) + XVI = CCLVI + XVI = CCLXXII
I didn't say that it wasn't computeable, only that I bet you couldn't do it in your head. Note that your calculation for some reason (EDIT: ah, maybe because my Arabic-numeral problem has 16 ⨉ 17?) replaces XXIII = 23 by XVII = 17, and then black-boxes the calculation XVI ⨉ XVI = CCLVI (which I at least wouldn't know without converting).
1 6
1 1 6
7 7 42
1, 6+7, 42 = 100 + 130 + 42 = 272
X V I
X C L X
X C L X
I X V I
I X V I
I X V I
CC LL(=C) XXXXX(=L) VVV(=XV) III = CCCLXVIII = 16*23=368
There's probably a way to formalize this with an entropy argument: that roman numerals are inefficient encoding. Because given some n-length string of numerals, firstly many are invalid encodings, and secondly among the valid numerals there isn't a uniform distribution from strings to integers. Something like that.
Earlier quoted context omitted.
I'm going to have to take this riddle bait. My solution: Move 1: weigh 6 coins vs 6 coins. Isolate the heavier set for the next move. Move 2: weigh 3 coins vs 3 coins. Isolate the heavier set for the next move. Move 3: weigh any two of the 3 remaining coins. If one is heavier, that is the counterfeit. Otherwise (if coins are equal in weight) the remaining coin is counterfeit.
You failed because: > One is counterfeit and is either heavier or lighter It turned out the counterfeit was lighter and you discarded it after the first weighing.