Constructibility 1: Compass & Straightedge

[@tobiasgertz7800]

Had they explained it half as well in high school, I would have found my love for geometry much sooner. My life may have taken a different path. The finessing between geometry and algebra is strong here as well. Having a geometric intuition about algebra makes all the difference in the world concerning your enjoyment of the subject material.

[@anjaninator]

Thank you for this constructibility series!

[@pieter-jan26]

I wish I could've had this lecture 10 years ago. Enlightening stuff!

[@evanbarnes9984]

Oh man! I don't know how this just happened, but seeing you write the equation of a circle somehow finally forced my brain to make the connection that it's just the Pythagorean theorem. I was never uncomfortable with either the Pythagorean theorem or the formula for a circle, but had never quite linked the two so directly in my head before. It's funny how that can happen!

[@omarmaswadeh5434]

Let me say, this was awesome!

[@trizarabia]

Very informative and clear, thanks!

[@alastairbateman6365]

Sorry but your wrong. What you have done circa 25/31 minutes is NOT due to Descartes but to Euclid book 2 proposition 14. It is in effect the Quarter Square Rule otherwise book 2 proposition 8. It is a proof that the QSR is also a proof for the Pythagoras theorem as we can easily construct a triangle (a+b)/2, (a-b)/2, (ab)^1/2 on the diagram. The upshot is that we can construct a square of any line of any length including pi. If you put pi +/-1 into the QSR equation you get pi for the square and pi^1/2 for the side of the square. Hence we have squared the circle as I and Manoj Khabal have shown in our videos.

[@MathatAndrews]

Since pi cannot be constructed from the rationals with the basic algebraic operations, it cannot be constructed. We cover this as the series progresses.

[@alikaperdue]

@28:55 - an easier way to find root a is: make the same construction, but join the 3 points on the perimeter of the semi-circle, making 2 similar triangles inside the circle. Giving the vertical line a distance of x, we have the relation: 1/x = x/a or x^2 = a or x = root a

[@rhke6789]

great video succinct and clear

[@AlessandroZir]

very useful, thanks!!! ❤️❤️

[@rhke6789]

multiplication as explained cannot be understood. Better: 1. draw a line length 1, mark it, and extend until it is length a, at the zero point, make a line right angle up and perpendicular with length b. 2. complete the triangle by adding 3rd side, hypotenuse. 3. From end of a, put a line parallel to the hypotenuse to close the larger similar triangle. 4. Call the vertical length of the larger triangle p. Now you can see b:1 as p:a or b/1=p/a or ba=p.

[@unluckyponnuart]

Another one question Cos (2π/33) is constructible or not?

[@marionhelminger4151]

Is there any source for the comment by René Descartes at 31:29?

[@softwaresirppi]

"If the greeks had known this, they wouldn''t have made so many fatt books" such a ballsy thing to say, Descartes!!

[@JéssicaMendes-k1r]

This is great

[@responsiblparty]

I tried the multiplication as detailed and it didn’t come out right.

[@softwaresirppi]

UNDERRATED

[@unluckyponnuart]

√5 is constructible or not?

[@MathatAndrews]

Yes, you can construct the square root of any integer (or rational number). We give an argument for this 25 minutes in.

[@unluckyponnuart]

@@MathatAndrews thank you so much sir

[@latetodagame1892]

Area= a x b

[@samisiddiqi5411]

36:00

[@jshellenberger7876]

Just went by?

[@jshellenberger7876]

OPM jobs isn’t usajobs isnt LinkedIn isn’t Facebook #POW

[@jshellenberger7876]

Business Lounge lmao