Pythagorean Theorem proof based on Law of Sines independent of the Pythagorean identity

2023.3.30: news.math/
Pythagorean Theorem proof based on Law of Sines
independent of the Pythagorean identity:

the American Mathematical Society:

https://meetings.ams.org/math/spring2023se/meetingapp.cgi/Paper/23621

. for the 2000 years since trigonometry was discovered

it's always been assumed that any alleged proof of 

Pythagoras’s Theorem based on trigonometry

must be circular

-- that all the fundamental formulae of trigonometry

are themselves based upon

the truth of the Pythagorean Theorem.

But Ne’Kiya D Jackson and Calcea Rujean Johnson

from St. Mary's Academy High School

present a new proof of Pythagoras’s Theorem

openstax.org free college textbooks

 2022.7.2: news.adds/edu/
openstax.org/free college textbooks

I list about 80% of the openstax books

that were of the most interest to me

and organized some in the order that

I read them during college,

or will read them in the future.

@joboaler youcubed.org getting all kids inspired by math

 2022.5.9: news.adds/math/edu/
@joboaler youcubed.org getting all kids inspired by math:

summary:

Stanford Mathematics Education Professor 

https://ed.stanford.edu/faculty/joboaler

Jo Boaler, promotes inclusive math training

that shows all kids they have a brain for math.

. her resources include the website youcubed.org

that includes a free online student course.

https://www.youcubed.org/online-student-course/

. she has several videos explaining her way,

along with many books.

https://smile.amazon.com/Jo-Boaler/e/B001IODJ5U

common core math's subtraction

10.20: web.math/common core math/has new subtraction:
businessinsider:
. the new way is realizing a subtraction problem
is asking you to measure the distance between 2 numbers;
You do that, in turn, by measuring the distance between
landmarks (easy, round numbers).

phys.org 2015:
we all do arithmetic like this in our heads all the time.
Say you are buying a scone at a bakery for breakfast
and the total price is US$2.60.
You hand the cashier a $10 bill. How much change do you get?
Now, you do not perform the standard algorithm in your head.
You first note that you'd need another
40 cents to get to the next dollar, making $3,
and then you'd need $7 to get up to $10,
so your change is $7.40.

called the counting up subtraction method.
from the smaller number
count up to the nearest 10, 100, etc;
to that add the largest number's smaller digits.

0.999... is a hyperreal not equal to one

12.8: co.quora/math/
here's why I have a problem with (0.999...) = 1:
. if the number (0.999...) = (1 - 1/infinity) = 1;
then the set [0,1) = {0, ... 1- 1/infinity} = {0, ... 1} = [0..1];
but then we have [0,1) = [0,1] so did we want to mean that?

12.17: wiki:
The equality of 0.999... and 1 is closely related to
the absence of nonzero infinitesimals in the real number system,
the most commonly used system in mathematical analysis.
Some alternative number systems, such as the hyperreals,
do contain nonzero infinitesimals;
and then the symbol "0.999..." admits the interpretation
of falling infinitesimally short of 1.
The equality 0.999... = 1 has long been accepted by mathematicians
because they are concerned with real numbers not hyperreals.
12.27: me:
0.999... is not a real number; it's a hyperreal;
because it is equal to 1 -1/infinity (the infinitesimal);
making it infinitely close to 1 but not real;
that's why 0.999... can't be equal to 1;
1 is a real; 0.999... is a hyperreal.

probability function over naturals

4.3: 12.13: news.math/probability function over naturals:
askamathematician.com 2010:
. can we define a probability distribution on the set of integers
(rather than the real numbers between 0 and 1)
such that they each occur with equal probability
(i.e. does a uniform distribution on the integers exist)?
askamathematician states:
". it can not be the case that
each of the infinity of integers
has the same probability
and the probability density function integrates to 1"

. but intuitively you can see
that the value is non-zero:
it is the infinitesimal defined by 1/infinity .

euler's number derived from pyramid of additions

2.9: math/euler's number derived from pyramid of additions:
. "( if s#n is the product of the terms in the nth row;
) it shows a pyramid of numbers
but doesn't say how the numbers are generated .
. so there are rows with a list of terms,
and the product means
multiply all numbers in row n to get s#n .
s#(n-1)* s#(n+1) /s#n ^2 = e .
say n=7, s#n = 162000,
s#(n-1) = 2500
s#(n+1) = 26471025 .
(26471025 * 2500) / (162000* 162000) = 2.52
e = 2.7
. I can see how the pyramid is formed:
put 1's down the sides,
then for each adjacent pair in a row,
eg, for 1 2 1
see the pairs (1 2) (2 1)
add the pair and put the result between them
on the row below: 1 3 3 1 .

differentiate parameters under the integral sign

16: math/differentiate parameters under the integral sign:
"Surely You're Joking, Mr. Feynman!", p72:

That book [Woods`Advanced Calculus]
also showed how to
differentiate parameters under the integral sign
--it's a certain operation. It turns out
that's not taught very much in the universities;
they don't emphasize it.
But I caught on how to use that method,
and I used that one damn tool again and again.
So because I was self-taught using that book,
I had peculiar methods of doing integrals.

free online books

4.20: adds/lib/free books list

5.29: summary:

. I found a list of free academic books online,

and this is my reading list todo . :)

a fraction calculator's interface without over.key

1.22: adde/math/handling fractions:
. the Kochan`Fractions Calculator has an over.key,
for allowing a fraction to be entered without eval'ing it;
another way to handle this
is to enter complex expressions:
nothing is eval'd until you hit enter,
so, expecting fractions, the only division happens when
there is the form a/b / c/d .
. also, the policy could be that
div's would be eval'd only when
they represent exact fractions:
eg, 2/5 is 2/5, while 1/2 is 0.5 [1.31:
that does present a problem with the base being decimal, 2*5
but the storage's base being binary?
you could have an exact fraction like 1/5 = 0.2 in base 10,
but that is a irrational number in base 2 .]
...
. even if it did eval each operator a.s.a.p.,
it could simply reduce and stay exact:
14/10 = 7/5 = (1+2/5).

modulus vs remainder

2.7, 2.16: Using the mod() function with negative numbers
"modulo" as a relation:
[pointing in the same direction on a clock]
any two numbers a and b are congruent modulo m
if (a - b) is a multiple of m.
. math's idea of "integer division":
x . . . . : 2.7, -2.7
floor(x) .: 2.0, -3.0
ceiling(x): 3.0, -2.0 .

. for both mod (modulus) and rem (remainder),
they are related to div by:
A = ( A DIV B ) * B + A % B
where % is either { rem, mod };
. {mod, rem} are similar in that
they are both consistent with a div function;
but mod's div truncates towards -∞ (negative infinity);
whereas, rem's div truncates towards zero .
. mod(-n, d) -- vs rem -- is the complement
of mod(n, d); eg,
MOD(-340,60)= 20
MOD(340,60)= 40
(40 and 20 are complementary modulo 60;
ie, 40+20 = 60).
. truncating toward -∞ (negative infinity)
means that if n (the numerator) is negative;
then the usual integer div needs to be decremented:
div = int(n/d)-1
-- so that the truncation is consistent by
always reducing the value instead of
changing it willy-nilly towards nil
(that'd be adding value when truncating negatives
while subtracting value when truncating positives).

Ada's "mod" (modulus) and "rem" (remainder):
. notice that while Ada supports both {mod, rem}
it has only rem-consistent div (truncate toward zero)
ie, observing the identity (-A)/B = -(A/B) = A/(-B)
for A,B in positives .
by contrast, Python truncates toward -infinity .
. here is Python's mod-consistent div ( % means mod )
. 123 / 10 = 12,  123 % 10 = 3
-123 / 10 = -13, -123 % 10 = 7
. 123 /-10 = -13, 123 % -10 = -7
-123 / -10 = 12, -123 % -10 = -3

translation from ada to c:
. ada's {rem, mod} is defined for (n,d) in Z
(integers, numerator and denominator can be negatives);
let c`% = abs(n) % abs(d):
-- (%) is c's symbol for remainder function --
then depending on the original signs of n,d,
use the following table to know whether to
{complement, negate} c`% .
-- complement (~) means abs(modulus) - x;
so for modulus = 5, the complement
of 1..4
is 4..1, respectively .
for rem:
. (n rem d)`sign = n`sign
details:
. when n,d are both positive,
or only d(modulus) is negative:
eg, 1...4 rem -5 = 1..4
or 1...4 rem 5 = 1..4
--> c`% .
. when n,d are both negative,
or only n is negative:
eg, -1...-4 rem -5 = -1 .. -4;
or -1...-4 rem 5 = -1..-4
--> -c`% .
for mod:
. (n mod d)`sign = d`sign;
if only n or only d is negative,
then complement .
details:
. when n,d are both positive:
eg, 1...4 mod 5 = 1..4
--> c`% .
. when n,d are both negative:
eg, -1...-4 mod -5 = -1 .. -4
--> -c`% .
. when only d(modulus) is negative:
eg, 1...4 mod -5 = -4..-1
--> -~c`% .
. when only n is negative:
eg, -1...-4 mod 5 = 4..1
--> ~c`% .

unary operators not always taking precedence over binary

2.7: news.adda/math/unary not always taking precedence:
. negation has the same precedence as
multiplication and division;
because, negation means mult by -1.
So -a^b should be -1*a^b = -(a^b).
details:
. programmers were accustomed to the C language,
in which unary operators such as negation
have higher precedence than any binary operator;
(and there was no exponent operator in C
to cause them to think twice about the matter).
so, when programmers use an exponent operator,
they may have wished to remain consistent with C;
however, for centuries,
the polynomial -x^2
has meant -1*x^2 = -(x^2)
not (-x)^2 = x^2 .

. look at the HP48G User Guide/order of operations:
priority#1:
Prefix functions (such as sin, ln, ...)
and Postfix functions (such as ! (factorial)).
--[. many could say negation is a prefix -();
2.16: nevertheless,
notice the way math has superscripted powers
(rather than using an operator);
as if it was an extension of the symbol's name
like the way subscripts actually are,
and thus intuitively having higher precedence
than any operation applied to the name .]
priority#2: Power (^) and square root.
priority#3: Negation (-), multiplication, and division.
--[. here is the 2nd place -() fits;
but, only because of its equvalence to -1*();
many think it's obvious that the negative
is part of the number's value .]
priority#4: Addition and subtraction.

. clarity should take precedence over correctness;
so, the system needs to ask new users
-- at least those who use the form (-x^n):

"( how would you eval -2^2 ?
{ 4, -4 } ??
. -1*2^2 is definitely = -1(2^2) = -4 .
whereas (-2)^2 = (-2)(-2) = 4 . )

. furthermore, when exporting adda`binary,
or allowing copies to text
always write it unambiguously { -(2^2), (-2)^2 }.