https://www.ugrad.math.ubc.ca/coursedoc/math101/notes/moreApps/separable.html

And these people have never seen the possibility to start "the other
way around" ??

Chris

Yes I meant "straight-line" distance in Euclidean plane. That was the point of drawing the triangle.

Sorry for the ambiguity...

You arrive at x² + y² = C in what may be a reasonably consistent way. (I have doubts that it is possible to derive the whole of calculus without the distance formula. But this is not important here.) As I said, the mentioning of a right triangle with sides x, y, c and the derivation of the slope of the normal to the radius-vector are both irrelevant to what you did till then.

Now, at this point you could have drawn a right triangle but there would not be any way to establish a connection between C and c. You then apply the trick of picking up the initial condition on the y-axis. But this is equivalent to an a ariori assumption that you have a circle for a solution.

You mean an "a priori assumption" ? How so? The premise is that at the point where x = 0 on the coordinate plane, y = c (as illustrated in the triangle) and that this gives a particular solution to the differential equation.

But why? Just because you said "let's consider a curve with c constant"? How do you know that this curve comes through as the solution of your equation?

As I said before, you simply constructed a vector field of slopes and formed a differential equation for envelopes of those slopes. Indeed you got x² + y² = C as a solution to this equation. All the rest is plain hand waiving. Where y = 0 you can introduce c so that c² = C and get x² + y² = c². You need the Pythagorean theorem to claim that this c relates anyhow to the radius of a circle.

The calculus speaks for itself...but OK, I am but a lowly high school teacher... I suppose my argument and our discourse can be summed up as this: consider a circle of fixed radius c, whose equation is formed by using calculus, which arguably may have roots in the Pythagorean theorem somewhere, which has a particular solution a^2 + b^2 = c^2.

In no way am I an authority on Calculus, and I understand that I am taking a huge step in assuming that the Calculus I use in this "proof" is powerful enough to stand on its own without the Pythagorean Theorem. But I still think we could have arrived at "distance as an equation" this way if history had turned out differently.

Thank you gentlemen for your feedback. I think this is a good enough exercise to show to a Calculus class and hope that it does not break any mathematical law to do so. I have read about Descartes and his use of the the Pythagorean Theorem but would love to have some reference to the link between the Pythagorean Theorem and the invention of calculus...

Calculus has othing to do with my objections, although you may want to dig deeper and make sure that its development does not need the PT.

Simply put, you did get that equation x² + y² = c². So what? c is just a constant that came from your initial conditions. Can you argue that it has anything to do with the radius of a circle?

But the distance is the Euclidean distance whose definition stems from the PT. So, if you assume that the latter is unknown, you can't claim that the solution of the differential equation is a circle. I doubt whether you can define a circle in the common sense without a reference to the Euclidean distance and, hence, the PT.

Let's also assume that elsewhere on this circle are points (x,y) and a right triangle can be drawn using (0,0), (x,0), and (x,y). This might also take some evidence to show that it is indeed a right triangle using the parallel postulate (and some would say this is equivalent to the PT).

but taking this 'liberty' I know that the curve comes through as the solution curve through a slope field of orthogonal slopes to the line through (0,0) and (x,y) (i do not think the PT is needed for this... See https://whyslopes.com/Analytic-Geometry-Functions/analGeo09b_Perpendicular_Lines.html

I then use calculus to establish the relation between x, y, and c.
>Pythagorean theorem to claim that this c relates anyhow to
>the radius of a circle.

1) parallel postulate
2) 'the underpinnings of calculus'
3) the mention of slope implies distance exists via the PT

if these were removed, would the proof THEN be valid?

If a^2+b^2 is not the square of the distance from (a,b) to the origin then the points (0,c) and (c,0) are not on the circle. However the circle is defined to exist as one centered at the origin with radius c. So by indirect reasoning, I think it is still valid.

>I wish to add your derivation as an "invalid" proof to

I am either stubborn or ignorant or both but I
still do not see it as invalid unless by way of the underpinnings of calculus.

Perhaps in your writeup, I shall finally get it and concede. You are welcome to add it to your page.

in any event, to quote Wayne Bishop from mathforum, it is nice to know that the mathematics is consistent and the demonstration shows the relationship between the circle and the theorem in a somewhat different approach.

Why would I ignore anything? I'll put up there anything you want.

I have a collectin of invalid proofs. I think that yours belongs to that collection. There is nothing personal. I do not do that to offend you.

not offended really.. Edison took over 1000 attempts to make a light bulb. It's just that I wrote a refutation in post 21 and you did not comment on it. I tried to refute a statement you made earlier.

Also I suppose you could add the proof using complex numbers to your list as well (although I would not attach my name to it...see the weblink on that topic)...

What if the distanace is (x^3 + y^3)^(1/3)? Circles in this and the Euclidean metric cross the axes at the same points.

>However the circle is defined to exist as one
>centered at the origin with radius c.

Circle is a circle in a certain metric!

https://www.cut-the-knot.org/do_you_know/far_near.shtml

>Does
>this have anything to do with spherical or hyperbolic
>surfaces?

Why? There are different metrics in the common plane.

>If so I suppose the problem with my argument is
>that it assumes the Euclidean plane.

Yes. I believe so.

>But if so how do all other proofs not assume this?

Just following Eulid's Elements. Some theorems hold regardless of how the distance is defined. In the modern terminology, a plane is Euclidean when you define distance in a certain way, viz., sqrt(x² + y²).

One iota of truth may have slipped into my mind as I write this... I think you may be saying that (0,c) is on the curve, and (c,0) is on the curve but the rest of the curve may be the circle while relation (x^3 + y^3)^(1/3) describes the distance c, the hypotenuse of the triangle. i.e. the point (x,y) just dangles in space in the first quadrant between the points (0,c) and (c,0) on some curve and that curve may or may not be the circle. And the only way to show that this curve where the point dangles in thin air IS the circle is to reference the PT?

Have I finally arrived on your side of the fence? If not, then I shall wait for your write up. Nevertheless I look forward to being published on your page again, and I hope if I have this finally right I can finish this article I am working on with some resolution as to why the proof is invalid.

Alex, I want to add to this tpost hat I appreciate the past few months you have entertained ideas I have posted on CTK Exchange. I have grown rather passionate about them, and you have been patient with me while I have tried to defend them.

Regards,

Yes.

>If not,
>then I shall wait for your write up. Nevertheless I look
>forward to being published on your page again, and I hope if
>I have this finally right I can finish this article I am
>working on with some resolution as to why the proof is
>invalid.
>

I'd be happy to see you explain that. This is quite along the lines of what you just wrote. I actually would be more comfortable to have your explanation. There is nothing wrong with making a mistake. Still, it is much more rewarding to have your name associate with a useful result even if it'started with an error than a naked mistake.

>Alex, I want to add to this tpost hat I appreciate the past
>few months you have entertained ideas I have posted on CTK
>Exchange. I have grown rather passionate about them, and
>you have been patient with me while I have tried to defend
>them.

Not always patient, no.

OK, how does this sound?

"The aforementioned argument is an invalid proof. The reason, in simple terms, is that the triangle may or may not be inscribed in the first quadrant of the circle. As a counterexample consider a length c where c is a distance given by x^3 + y^3 = c^3. This curve also comes through as the solution through a slope field where the slopes are orthogonal to something. Also, it contains the points (c,0) and (0,c). So in this way the circle equation is not attached to the distance. To prove that this value of c is not the distance for the hypotenuse of the triangle would take us back to the PT, making the argument circular.

To see further counterexamples, one could animate the curve given by |x|^n + |y|^n = c^n for different positive values of n.  To show that n = 2 comes through as the solution for the triangle requires more than I have argued, likely the PT itself.

That is correct but needs a little finess. You should say "the word distance in the implicit (or assumed) context" or something to this effect. There may be distance but not the one you had in mind. For the PT you need the distance that is defined via the PT.

Perhaps that same notation is fitting here? Let's say that the circle contains point (x,y) moving along its curve in the first quadrant. The terminal point of my vector drawn from the origin as the hypotenuse c of my right triangle should be referenced as (x1,y1). The flaw in the proof is that (x,y) and (x1,y1) are not necessary the same values, even at the same "slope" defined as y1/x1.

Or even further I could state that y1 - y and x1 - x are not necessarily zero.

Have I dwelled on this too long?

Very likely. I just do not remember the details. Right now I am doing something else:

https://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013350&sid=816288a78771e5fbb263f272c9aa2278#p2013350

Can't switch right away.

>Have I dwelled on this too long?

The thread is indeed rather long. You may want to start a new one, if need be.

So it is a useful result to show that my own proof is invalid. Perhaps this is an accomplishment in and of itself.

Thanks for allowing me to submit the writeup.

I may have already mentioned this, but I am posting again so I may receive email notification. Yes, I want you to mention my name and place a reference to my online document. Again thanks for thinking of me,

https://www.cut-the-knot.org/pythagoras/CalculusProof.shtml