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From: Ross Finlayson <ross.a.finlayson@gmail.com>
Date: Sun, 12 May 2024 11:21:23 -0700
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On 05/12/2024 10:46 AM, Jim Burns wrote:
> On 5/11/2024 9:17 PM, Ross Finlayson wrote:
>> On 05/11/2024 04:47 PM, Jim Burns wrote:
>>> On 5/11/2024 7:11 PM, Ross Finlayson wrote:
>
>>>> The case is that induction goes through,
>>>> an inviolable law you call it:
>>>> does it go all the way through?
>>>> Does it complete?
>>>
>>> It is complete.
>>> There is no completing.activity,
>>> so I wouldn't say it completes.
>>>
>>> Compare to right triangles:
>>> Are all the squares of two shorter sides
>>> summed to the square of the longest side?
>>>
>>> That's a tricky question to answer because
>>> there is no summing done.
>>> That relationship between the sides
>>> is simply something true about right triangles.
>>>
>>> And it is complete == it is true for each.
>>>
>>> We don't typically ask the tricky question
>>> about right triangles.
>>> We ask the tricky question about cisfinite induction
>>> because we imagine it as a process,
>>> which we don't for right triangles.
>>>
>>> Cisfinite induction is NOT a process.
>>> Cisfinite induction is an argument,
>>> completely correct or completely incorrect.
>
>> What I recall of the context of the Pythagorean theorem,
>
> Let's refresh our memories.
>
>   ͨₐ🭢🭕🭞🭜🭘ᵇ  =  ͨₐ🭢🭕ͩ   +  ͩₐ🭞🭜🭘ᵇ
>
> The right triangle 🞃cab is split into
> two right triangles ◥cda ◤adb
> by segment a͞d perpendicular to b͞c
>
> 🞃cab ◥cda ◤adb are _similar_
> [1]
> Corresponding sides have equal ratios.
>
> ∠acb = ∠dca
> ∟cab = ∟cda
> 🞃cab ≚ ◥cda
> c͡b/c͡a = c͡a/c͡d
>
> ∠cba = ∠abd
> ∟cab = ∟adb
> 🞃cab ≚ ◤adb
> c͡b/b͡a = b͡a/d͡b
>
> c͡b⋅c͡d = c͡a²
> c͡b⋅d͡b = b͡a²
> c͡b⋅(c͡d+d͡b) = c͡b² = c͡a² + b͡a²
> QED
>
> [1] needs its own proof,
> but that can be done, too.
>
>> What I recall of the context of the Pythagorean theorem,
>> was that after algebra already was trigonometry, and
>> the definitions of the trigonometric functions, for
>> sine and cosine and tangent, about the opposite and
>> adjacent and hypotenuse, then as of a right triangle
>> with its hypotenuse the radius of a unit circle, that
>> the right angle is as with regards to the abscissa
>> and ordinates or where the lines drop or slide to
>> the x or y axis of the usual X-Y coordinate setting
>> of a circle centered at the origin, it was of the
>> secondary school's first three years of geometry,
>> algebra, and trigonometry, or along those lines.
>>
>> So, we computed a bunch of ready things about
>> those often with the Pythagorean theorem,
>> which is as an addition-formula, mostly about
>> 30-60-90 triangles, and, isosceles triangles,
>> or 45-45-90, then those got used throughout
>> precalculus and a couple years of calculus
>> or high school.
>
> I agree that the Pythagorean theorem
> gets used in a lot of different ways.
>
> How we know that the Pythagorean theorem
> is a fact about each right triangle
> has important similarities to
> how we know that we cisfinitely.induced claims
> are facts about each natural number.
>
>> So anyways one time I see a diagram about
>> Pythagorean triples, those being tuples of
>> three integers that have a^2 + b^2 = c^2,
>> and what they'd done was right triangle,
>> then draw a square as of the square alongside
>> it, and counting the boxes of the squares of
>> a b c it's that the boxes of the squares of a
>> and b equals the boxes of the square of c.
>
> Actually,
> that works in the opposite direction.
> We know that 3:4:5 is a right triangle
> because of the Pythagorean theorem.
>
>> If that's not a proof of the Pythagorean theorem
>> and least it's graphically intuitive for some values,
>> where of course there are hundreds of known
>> proofs of the Pythagorean theorem, since the
>> time of Pythagoras as some even have as from
>> greater antiquity, then it reminds of things
>> like Rodriguez formula, Vieta's formulas,
>> Nicomachus' theorem and formulas,
>> Pascal triangle and bonomial theorem,
>> all what are sorts of addition formulas,
>> like an addition formula of the product
>> of exponents as the sum of the powers.
>>
>> So, that Pythagorean triples exist, and it results
>> that the rightness of a triangle with sides length
>> the Pythagorean triple can be established without
>> invoking the Pythagorean theorem, doesn't so
>> much make it so the other way around, from
>> induction over Pythagorean triples, without
>> showing as how all right triangles are somehow
>> as some congruence to what is some Pythagorean
>> triple,
>
> We shouldn't want to show that
> each right triangle has a Pythagorean triple,
> because we know that isn't true.
> Famously, Pythagoras executed one of his disciples
> for proving that the right isosceles triangle 1:1:√̅2
> has no Pythagorean triple.
>
>> of the equivalence class of all the triples
>> and all the congruences to triangles with a
>> unit length longest side, establishing infinite
>> expressions, and closures, of completion,
>> to make a case for the Pythagorean theorem
>> as via induction from an explication after
>> the enumeration of Pythagorean triples,
>> which via inspection have a^2+b^2 = c^2,
>> as for that it results congruences that
>> "go to" any given dimensions of a right
>> triangle.
>>
>> About the cisfinite and transfinite induction,
>> and I know it's not the languages fault that
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