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austinm added an answer to a question Drawing a long line

austinm added an answer to a question Filling a room
By the way, I wasn't just being coy here. The proof is a bit interestingcontains one or two knots that have to be thoughtthrough carefullyand is worth taking five minutes to lay out.

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austinm added an answer to a question Filling a room

austinm added an answer to a question Up

austinm added an answer to a question Chain conundrum
Disagree
I'm going to go moremathematical:

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austinm added an answer to a question The deconstruction and reconstruction of numbers pt. 2
(On a computer impeded by broken keyboard nowplease excuse necessarilybrief response.)
(Letters seven, sevenplusone, and (crucially!) lowercasequote do not work!)
Recall in posed system 1 is additive identity. Answer to your latest question is same as answer to #4 in OP.
I.e. 36 = (a plusprime 1) timesprime (a plusprime 1) = a timesprime a.
Suppose one wants to multiply out (a plusprime 1) timesprime (a plusprime 1) (and assume my instinct is correct about ability to distribute) we immediately run into stillopen question about definition of multiplication by and of one. Will respond later at time possessed of full I/O.

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austinm added an answer to a question The deconstruction and reconstruction of numbers pt. 2
Notation: I use " +' ", " ' ", " x' ", and " /' " to represent the four arithmetic operations.
I interpret your definition of +' as: a +' b = append b items to an alist and count interstices. (N.B. it's an operator definition, and that the identity element for +' is 1.)
1.
If ' is defined as a ' b = remove b items from an alist and count interstices (i.e. the logical complement of +'), then
If a ' b is defined as the inverse of the +' operation, such that a +' b ' b = a = a ' b +' b, then
(Using Common Notion 2: "If equals be added to equals, the wholes are equal.")
Note that the second comports with the notion that a ' a should equal the identity element: 1. (This is why I favor the second.)
2.
If x' is defined as repeated addition (via +'), such that 2 x' 4 = 4 +'4, then
Note that the ordering is that implied by the oldfashioned reading of 2 x 4 as "twice four," or "two fours." This definition has the interesting, and possible undesirable consequence that 1 is a leftidentity but not a rightidentity: 1 x' 4 = 4, but 4 x' 1 = 3. (All multiplications not containing a one commute.)
Lack of universal commutivity can be remedied by defining 1 x' 4 as 4 +' 0; i.e. multiplying by one is adding nothing to 4, which then requires the final counting of interstices. In this case 1 x' 4 = 3 = 4 x' 1, all multiplication commutes, but there is no multiplicative identity.
I'd be interested to hear opinions on which definition of 1 x' a is preferable. The second, containing some appeal, leads to a breakdown in distributivity when 1's are involved. The first, I believe, allows for distributivity in all cases. (I've not rigorously proven it, though.)
3.
Properly speaking, I'd say
However, we don't really want that result, do we? So, let's clean up our definitions by dropping the notion of itemsandspaces, and just go with
a +' b = a + b 1 (where " + " is the usual addition operator),
a x' b = a x b 1 (ditto).
Note that a +' 1 = a, and a +' b = b +' a. Good so far.
a x' b = b x' a, but we still have no (and have no hope of a) multiplicitave identity element. This has grave consequences for the idea of division, but we forge ahead blindly...
Defining division as the inverse of the multiplication operation, such that a x' b /' b = a = a /' b x' b, we can use the above to determine that
4.
Using the mostrecent definition of " x' ", and assuming that sqrt'(a) is the inverse of a x' a, we see that

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austinm added an answer to a question Leaning cubes of Pisa
I can make the separationfromoriginalfootprint greater than 1/24 (and spend an arbitrarilysmall portion of that gain to ensure it's not precarious)can you?

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austinm added an answer to a question Leaning cubes of Pisa
This question gets much more interesting if we go, say, 1.5 blocklengths outside of the original footprintshape of tower differs dramatically.

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austinm added an answer to a question Party with strangers
Horribly sorryI misread the original as "no two people in the group..." rather than "no more than two people in the group...." For the OP it's clearly possible; my first reply applies to the situation in Prime's note.

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austinm added an answer to a question Party with strangers
It's not possible, assuming that friendship is symmetric and not optionallyreflexive. That is, friendship is either reflexive or not, but it is that way for all people. (If one does not make those assumptions, it's both possible and uninteresting.)
Proof is inductive on number of people at party; inductive step uses the Pigeonhole Principle. There's a tiny subtlety in the inductive step.

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austinm added an answer to a question Put a cork in it [original thread]
Finding its volume was a fairly straightforward calculus problem. Letting the circular base lie in the xy plane and the linear ridge lie along the yaxis:
Of course, I had to go back to my CRC to look up some of the integrals, so I wouldn't be surprised if there were a small transcription error in there. Especially as my 3Dmodeling program claims the volume is
Anyone else find the volume of a solid that looks like this?
[edited to catch factorof2 error]

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