Hearted Youtube comments on Mathologer (@Mathologer) channel.
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I've got a proof for the last problem distinct from ilonachan's very elegant sudoku-style proof, inasfar as grid-puzzle proofs can be distinct, giving the same fully general 2n*2m result.
I) Any solution must have an "Alternating Orientation" (equivalent solution under reflection over a diagonal or 90° rotation) where each row alternates between two colors, and the sequence of rows alternates between two such pairs of colors.
II)
a. For an even number of columns, an alternating orientation of a solution must have different colors for the first and last columns in each row.
b. For an even number of rows, an alternating orientation trivially has different colors between the first and last rows.
c. The colors in each of the four corners of an alternating orientation must be different.
III) Reflection along the diagonal or rotation by 90° preserves both the property of being a solution, and the span of the colors in the corners. (Trivial)
Proof of I:
Ia) If the first row alternated over a pair, the second row must contain the complement pair; and that row must therefore alternate, as, trivially, two adjacent cells cannot share a color, and there are only two colors in the row. Inductively, each successive row must also contain the complement pair to the previous row, and must itself alternate. This satisfies the definition of an alternating orientation.
Suppose there is a solution which does not have an alternating orientation.
By Ia, the first row does not alternate over a pair.
Without loss of generality, it can be shown that the following configurations remain:
i. abcd[...]
ii. abcb[...]
iii. abac[...]
Where [...] represents any sequence of successive terms, for grids of more than 4 columns.
We shall first assume configurations i and ii, together.
Both configuration i and ii start with the same three colors, and we may say they both satisfy of configuration 'abcx[...],' where 'x' is undefined.
As no steps of the remaining steps do not depend on cells in columns 3 or later for this orientation, configurations will be denoted with only the first three color values.
Cell (2, 2) is part of both the top-left 2x2 square, and the top-center 2x2 square. It must therefore not be the same color as the first two or second two cells in row 1. As all three colors are represented in those three cells, cell (2, 2) must be the 4th color.
Cells (1, 2) and (3, 2) are now determined, as they each have 3 colored neighbors.
In particular, row two is now in a configuration of 'cda.'
This configuration is isomorphic to configurations i. or ii. of row one. Therefore, each row can be determined inductively.
In particular, row three is also in a configuration 'abc,' which is the same as row one.
If we look at column 1, we find it must alternate between colors 'a' and 'c.' This means that if we reflect over the diagonal, we must have a solution where the first row, equivalent to the original solution's first column, must alternate over the pair of colors 'a' and 'c.'
This satisfies the condition of Ia), and so this reflected solution must be an alternating orientation.
This means the original solution had an alternating orientation, contradicting the premise for configurations i and ii.
We may use a similar argument for configuration iii, but shifting our perspective over by 1 column. We first remove column one from the solution, and observe that we now have a configuration isomorphic to configuration i or ii. As removing a column does not alter the property of being a solution, as it does not affect 2x2 grids not intersecting that column, the argumentation for configurations i/ii follow.
If we take the resultant alternating orientation, and then reflect this solution vertically so that the top row is now at the bottom, we may notice that our new top row is also alternating, as given by Ia. If we try to replace an appropriately reoriented row at the bottom corresponding to the column we removed here, the solution must still satisfy Ia, as the new top row is not changed.
The property of having an alternating orientation is trivially preserved under horizontal and vertical reflections. As such, this re-appended row must still be alternating, and our overall solution still has some alternating orientation.
I: QED
QED
(Sorry for the haphazard alterations to the proof of I, but I realized I needed to generalize it to get the full 2n*2m result, but that proved non-trivial.)
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Excellent work! And yes, you should be wary of discontinuous "logarithms" lurking in the dark :) The axiom of choice guarantees their existence (but that also explains why you can't find any of them - because they're non-constructive).
Essentially, the multiplicative group of positive real numbers forms a vector space over Q (taking vector addition to be multiplication and scalar multiplication to be exponentiation). So the axiom of choice guarantees that there is a Hamel basis for the multiplicative group of positive real numbers over Q, and we can assume e is part of that basis. Similarly, the additive group of all real numbers is a vector space over Q, so the axiom of choice guarantees us a Hamel basis here, too, which we can assume contains 1. (Note that we can prove the dimension of both of these spaces over Q is |R|.)
Then we can create a function which maps the Hamel basis from the multiplicative positive real numbers to the Hamel basis of the additive real numbers, choosing e to get mapped to 1, and choosing it so that some basis element d of the multiplicative positive reals is mapped to something other than ln(d). But a function defined from a basis of one vector space to elements of another vector space defines a unique linear transformation between the two vector spaces. Let's call this linear transformation L. We can then check by the definition of these vector spaces and the definition of linear transformations that it has all the algebraic properties of the natural logarithm.
If a and b are two positive real numbers, then L(ab) = L(a)+L(b) because vector addition in the domain is actual multiplication but vector addition in the codomain is actual addition. For any positive real number a and any rational number c, we have L(a^c) = cL(a) since a^c is scalar multiplication of a by c in the domain but scalar multiplication in the codomain is actual multiplication. We also have L(1) = 0 since 1 and 0 are the identities respectively, and L(e) = 1 by choice. So we have the "properties" of a logarithm that you were able to pin down without continuity. But we have L(d) = something other than ln(d), so L is not the same thing as the natural log.
(As an added bonus, I made sure L is bijective too!)
But your argument pretty much shows that ln(x) is completely pinned down by
ln(e) = 1, ln(ab) = ln(a)+ln(b) for all real positive numbers a and b, and continuity.
So the only conclusion here is that this nasty L function cannot be continuous. And since it has the other properties of logarithms, it is in some sense a discontinuous "natural log".
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I was thinking about that Pythagorean triple tree. If one associates going straight with a 1, going left with a 2 and going right with a 3, then one can encode every primitive Pythagorean triple (PPT) with a rational number between 0 and 1 in base 4 as follows:
0 corresponds to (3,4,5), 0.1 to (21,20,29), 0.2 to (5,8,17), 0.3 to (5,12,13), 0.12 to (77,36,85) and so forth. That way one gets a 1-1 correspondence with the finitely representable base 4 numbers between 0 (included) and 1 that do not contain the digit 0. The three families of PPTs correspond to the numbers 0.11111..., 0.22222 and 0.3333...
I wonder if one can get anything geometrically meaningful with that correspondence.
Can one interpret the periodic or irrational numbers as interesting infinite paths/families of PTTs in the tree?
What about interpreting numerical manipulations like multiplication of 0.1111 with to to get 0.2222 in terms of the associated PTTs?
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