Spiral of Theodorus Constructs Square Roots (visual proof; straightedge and compass) 

Thanks!

Glad it helped!

He didn’t use a visual proof that I know. He used the standard argument. This was used to construct square roots using straightedge and compass.

Thank you so much 😀

:)

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Thanks!

Thank you!

@@MathVisualProofs But the question is can you construct 3rd roots?

@@SuperYoonHo What do you think?

@@MathVisualProofs Probably impossible because Fermat told us that A63+b^3is never =C^3

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Thanks! And Thank you for watching :)

Thanks! And thanks for watching :)

@@MathVisualProofs the angle of decent on this is interesting.... almost a perfect orbit which then quickly goes catastrophic

@@KaliFissure :)

Sorry, sir but pi can't be expressed in the form of algebraics.

@@boruahmohendra4349 Who says so? π is a ratio! Just like the Golden ratio is… a ratio.

​​@@PASHKULI pi is a transcendental number, but phi isn't

@@simonsidorov8315 That is where the terminology is wrong about π.

Thanks! Glad you liked it.

Thanks!

Each time we use a circle of radius 1 to draw the triangle base length of 1.

@@MathVisualProofs yes, but I meant the spiral arms that are built by the end points of the construction, the square roots of 2, 3, 4, ... n, n+1, n+2 ... Is the distance of these end points to the center increasing linearly with the angle around the center?

I try to answer my own question: as the points spiral from the origin and around the origin, does their distance to the origin increase linearly with the angle they're moving around the origin? . . . For a step from n to n+1 (for large n), the distance to the origin increases from sqrt(n) to sqrt(n+1): the difference is about 1/(2*sqrt(n)). During the same step, the angle increases by about 1/sqrt(n). That means the increases in distance are about half of the increases in angle, and we should expect the points to line up on a nice linear spiral :-)