15 Jan, 14 tweets, 5 min read
@GWOMaths One might say that physicists study the symmetry of nature, while mathematicians study the nature of symmetry.

1/
@GWOMaths Observing symmetry in nature, such as noting the similarity between the symmetries of a snowflake and a hexagon, is readily comprehensible. What does it mean then to study "the nature of symmetry"?

2/
@GWOMaths Mathematicians define a "group", G, as a set of elements {a,b,c, ...} with a binary operation ⊡ and a distinguished element e (the identity of G) satisfying these specific properties:

3/
@GWOMaths For all a, b, c in G
1) Closure: a⊡b is in G.
2) Associativity: (a⊡b)⊡c = a⊡(b⊡c)
3) Identity: e⊡a = a⊡e = a.
4) Inverse: There exists an element a* such that a*⊡a = a⊡a* = e.

4/
@GWOMaths Just as functions on the integers, rationals, or reals are defined as mappings, mathematicians define a *group morphism* μ as a mapping from one group to another that preserves the group structure:

5/
@GWOMaths For groups G = [{a,b,c, ...}, ⊡, e] and H = [{α,β,γ, ...}, ⊠, ε]
then μ: G ⟼ H is a "group morphism" if for all elements of G:
μ(a⊡b) = μ(a)⊠μ(b). Note that for all a:
μ(a)⊠ε = μ(a) = μ(a⊡e) = μ(a)⊠μ(e)
and hence μ(e)=ε; Similarly it can be shown that
μ(a*) = μ(a)*.
@GWOMaths Thus the definition of such a *group morphism* preserves the group structure. When such a *morphism* is both *onto* (ie every element of H is mapped to by one or more elements of G) and *one-to-one* (only one element of G maps to each element in H) it is termed an *isomorphism*.
@GWOMaths For mathematical purposes, when there exists an *isomorphism* between two groups G = [{a,b,c, ...}, ⊡, e] and H = [{α,β,γ, ...}, ⊠, ε] then G and H are termed *isomorphic*, or *the same up to isomorphism*.
@GWOMaths Now all the finite groups can be classified in terms of various internal structures, first collecting those which are *the same up to isomorphism* and then collecting families with similar internal structure.
@GWOMaths When all the families of groups - Cyclic, Alternating, and assorted Lie Group Types - have been defined, there are remaining 26 groups that don't fit anywhere: the *sporadic groups*.
@GWOMaths Of these 26 *sporadic groups*, two stand out from the others in terms of their size:

- the *Baby Monster*, *B*, of size
2⁴¹ ⋅ 3¹³ ⋅ 5⁶ ⋅ 7² ⋅ 11 ⋅ 13 ⋅ 17 ⋅ 19 ⋅ 23 ⋅ 31 ⋅ 47; and
@GWOMaths - the *(Fischer–Griess) Monster), *M*, of size
2⁴⁶ ⋅ 3²⁰ ⋅ 5⁹ ⋅ 7⁶ ⋅ 11² ⋅ 13³ ⋅ 17 ⋅ 19 ⋅ 23 ⋅ 29 ⋅ 31 ⋅ 41 ⋅ 47 ⋅ 59 ⋅ 71.

Counting up the number of distinct primes in that last number gives us 15.
@GWOMaths Therefore today's answer is that:

The number of distinct prime factors
in the size, n, of the *Monster Group* M
is

15.

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# More from @pgeerkens

17 Jan
@johnrobb We have abandoned competence,
in the futile search for universal expertise.

"What?" I hear you say.
Clearly definitions and explanations are in order.

1/
@johnrobb Expertise, certainly as regards this discussion,
but perhaps also universally, should be regarded
as the knowledge of a very great deal - about very
very little.

2/
@johnrobb To become an expert in a field requires
years - nay, decades - of diligent study of that
solitary field. In diving deeper, the pool in turn
becomes ever smaller: a diving pool rather than a
swimming pool. Nothing is free.

3/
2 Dec 20
@pnjaban None of the above:
Field-trained former military paramedic.

I don't want anyone not trained specifically in emergency medicine anywhere near; and none of those civilian trained E.R. either, cause they're trained in CYA first, and I want someone who's serious about saving my life
@pnjaban When my Dad suffered acute hemechromatosis at ~34,
for which the ONLY treatment is *bleeding*,
and the only options "a whole lot" and "a whole lot more",
the attending physician was too embarrassed to prescribe the correct treatment.
@pnjaban The nurse (Irish of course, because they understand hemechromatosis) told me Mom:

Mrs. Geerkens:
unless we remove him from this hospital immediately.
I'll help.

The nurse lost her job;
but my Dad lived another 45 years.
2 Dec 20
From Albert H. Beiler's
Recreations in the Theory of Numbers:

47² = ?
96² = ?
113² = ?
179² = ?

47² = (47 + 3) ⋅ (47 - 3) + 3² = 50 ⋅ 44 + 9 = 2209

96² = (96+4) ⋅ (96-4) + 4² = 100 ⋅ 92 + 16 = 9216

113² = (113+3) ⋅ (113-3) + 3² = 116⋅110 + 9
= 12760 + 9 = 12,769

179² = (179+21) ⋅ (179-21) + 21²
= 200 ⋅ 158 + (20² + 20 + 21)
= 31600 + 441 = 32041
The trick is from noting that from difference of squares
a² - b² = (a + b) ⋅ (a - b)
one can obtain by rearrangement
a² = (a + b) ⋅ (a - b) + b².
22 Nov 20
@GWOMaths Hint follows
@GWOMaths Rochambeau is just a fancy name for Rock-Paper-Scissors.
@GWOMaths If you're having difficulty visualizing this,
consider the situation after Game 1 as the baseline.
6 Nov 20
@LarrySchweikart My take:
SCOTUS must not be seen to play favourites - yet has an obligation to ensure that corruption is voided.

So wherever widespread and continuous failure to observe due process (as per State law) is seen, void all ballots from that county / counting centre.

1/
@LarrySchweikart If the number of voided counties exceeds more than 1 or 2, then void the entire election for the State.

In the case of House and Senate representatives, require special elections.

2/
@LarrySchweikart For Presidential Electors, legislation now in effect gives the State legislature responsibility and authority to select the State's slate of Electors - as per the original Constitutional design.

3/
20 Mar 20
@GWOMaths Define
y ≡ x - 16
to allow factoring the term 2¹⁶ completely.

Then 2¹⁶ + 2¹⁹ + 2ˣ
= 2¹⁶ + 2¹⁹ + 2ʸ⁺¹⁶
= 2¹⁶ . (1 + 2³ + 2ʸ)
= 2¹⁶ . (9 + 2ʸ)

Now a solution, viz y=4, is readily obvious as yielding
= 2¹⁶ . (9+16)
= 2¹⁶ . 25
= (2⁸ . 5)²

∴ x = y + 16 = 20.
@GWOMaths However - is our solution unique?

Suppose
∃ a ∈ Z
such that
9 + 2ʸ ≡ a²

Then
2ʸ = a² - 9 = (a-3) . (a+3)
and both (a-3) and (a+3) must be powers of 2.

This only occurs for
a = 5 => 2ʸ = 16 => y = 4 => x = 20.

Our solution is unique.
@GWOMaths Finally - as a fun calculating observation,for those who have memorized powers of two at least to 2¹⁶:

(2⁸ . 5)²
= (2⁷ . 2 . 5)²
= (2⁷ . 10)²
= (128 . 10)²
= 1280²

= 2¹⁶ . 25
= 2¹⁴ . 4 . 25
= 2¹⁴ . 100
= 16,384 . 100
= 1,638,400.