(On a computer impeded by broken keyboard now--please excuse necessarily-brief response.)
(Letters seven, seven-plus-one, and (crucially!) lowercase-quote 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 plus-prime 1) times-prime (a plus-prime 1) = a times-prime a.
Suppose one wants to multiply out (a plus-prime 1) times-prime (a plus-prime 1) (and assume my instinct is correct about ability to distribute) we immediately run into still-open question about definition of multiplication by and of one. Will respond later at time possessed of full I/O.
Notation: I use " +' ", " -' ", " x' ", and " /' " to represent the four arithmetic operations.
I interpret your definition of +' as: a +' b = append b items to an a-list and count interstices. (N.B. it's an operator definition, and that the identity element for +' is 1.)
If -' is defined as a -' b = remove b items from an a-list 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.)
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 old-fashioned reading of 2 x 4 as "twice four," or "two fours." This definition has the interesting, and possible undesirable consequence that 1 is a left-identity but not a right-identity: 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.)
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 items-and-spaces, 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
Using the most-recent definition of " x' ", and assuming that sqrt'(a) is the inverse of a x' a, we see that
Horribly sorry--I mis-read 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.
It's not possible, assuming that friendship is symmetric and not optionally-reflexive. 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.
Finding its volume was a fairly straightforward calculus problem. Letting the circular base lie in the x-y plane and the linear ridge lie along the y-axis:
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 3D-modeling program claims the volume is
Anyone else find the volume of a solid that looks like this?