WildTrig2: Quadrance via Pythagoras and Archimedes

[DaveKotschessa]

I'm a U.S. student and wondering if I can pursue this Trigonometry as part of my undergraduate research. Any advice on how I might go about that?

[njwildberger]

@DaveKotschessa I would look for an interested advisor at your institution, show him/her my book and ask if you could do some kind of project from it. Even better, you could ask that they consider offering a course in Rational Trigonometry and Universal Geometry. Combined with Universal Hyperbolic Geometry, this could be a very nice course, I have taught something in this direction here at UNSW to 3rd year students.

[Toxie207]

I really like this form of trigonometry. It has taken me months, but I am getting my head round it now, despite 20 years of having classic trigonometry hammered into my brain! I am a head of a Mathematics department at a High school in the UK and would love to introduce these concepts of Rational Trigonometry. Unfortunately, the exam process really requires my students to use classic trigonometry, as the questions often involve trig proofs using the classic trig functions. Dogma prevails! :(

[njwildberger]

@Toxie207 Thanks for the comment. I am very glad that you are taking the time to learn this approach. Perhaps you can think about offering it as an extra topic to some of your better students?

[horstmueller1000]

It is simple to get the symmetric formular - just expand 2( Q1^2 + Q2^2 + Q3^2) into Q1^2 + Q2^2 + Q3^2 + Q1^2 + Q2^2 + Q3^2 - Wow !!! it remains to prove that 2( Q1Q2 + Q2Q3 + Q1Q3) = Q1^2 + Q2^2 + Q3^2 - UFF !  We know that (Q3-Q2-Q1)^2 = 4 Q1Q2. So now calculate the left hand side again to obtain (Q3-Q2-Q1)^2 = Q1^2 + Q2^2 + Q3^2 + terms. Meaning that Q1^2 + Q2^2 + Q3^2 = (Q3-Q2-Q1)^2 - terms. Now combine this: 2( Q1Q2 + Q2Q3 + Q1Q3) = triple quad formular - the terms. That's it.

[GGinFra]

Really valuble lecture, but I still feel it's not abstract enough.

I think, It's worth to mention here that acutally we are dealing with some interesting binary functions( NxN -> N).

[asnothe]

I'm having trouble with the pythagorean theorem proof. How can we conclude that the white area at the end is Q3, when we haven't yet proved that is in fact a square shape (perpendicular lines)? Do we need to use the fact that both *angles* in a triangle, and *angles* along a straight line, total 180 degrees? I'm really enjoying these series of Prof Wildberger's.

[Ohcacful]

Lol. Took me 15 minutes to prove the symmetrical relation and I'm a mathematics major.

[njwildberger]

@Ohcacful Don't be dismayed. Many maths students have insufficient experience working and manipulating with algebraic expressions. It is a key stumbling block to progress in maths, so take every opportunity to practice. Rational Trigonometry provides quite a few chances to improve your skills, as you shall see as you work through the series. 

[freezzertime]

this is how math should be taught in schools! Why isn't it?

[freezzertime]

you're pyth. thm. was excellent!!! *tears*

[mskimwild]

type in cats part 1 by ali the musical. It is just as exciting and interesting as Norman Wildbergers videos. WATCH CATS PART 2 BY ALI AS WELL BY THE WAY!!!!!

[GlorifiedTruth]

I finally got the symmetrical version proof! It only took me a zillion hours. I had to incorporate the triple quad formula, which we accept as proven, to get an identity that I used to create a true statement in the symmetrical version. It required, as Dr. Wildberger mentioned, expanding both. Now I'm ready to go on with the rest. Whoo hoo!

[isilder]

I don't think that no greek person ever just compared the two length directly...

Heck the easy way to make those right angles ? oh make a triangle 3,4,5... OH that would take measuring with a rule ??? of course...

That leaves a statement like "the greeks didnt measure length, they compared the size of the squares" to be propaganda.. just because we have Pythagarous telling us about squares doesn't mean they didnt do a lot of measuring.

[besimpel]

Hi njwildberger,

I have now after several try found out how I can rearrange the triple quad formula to the symmetric version. It helped a lot when I brought all the terms to one side of the equation. Thanks.

[ValentinKostadinov]

Awesome! Thank you!

[liverwurt4]

Wait, I'm confused about Heron's theorem.

If 4Q1*Q2 = (Q1+Q2+Q3)^2, then wouldn't 16(area)^2 = 0?

[njwildberger]

Hi liverwurt4, Yes that would be the case of a degenerate triangle consisting of three collinear points.

[njwildberger]

... and I presume there should be a negative sign in front of your Q3.

[FarrenfollowsBen]

Very cool, thanks.

[DeepWithinU29]

I like what you're wearing =P

[zebbox]

tried to derive the "symmetrical version" of the triple quad but got stuck - can anyone help?

[maths486]

nice proof of the pythagorean theorem! :)

[GR1o6180339887498948]

The ancients rule!!! 5 stars

[sprenner]

I just don't get the lemma. Why does Q squared = Q1xQ2?

[njwildberger]

Suppose that the rectangle is m units wide and n units long. Then its area is mxn, and so Q squared will be (mxn)^2 .

But this is the same as (m^2)x(n^2), which is Q1xQ2.

[mediteight]

These proofs are wonderful - many thanks..

Would I be being too pedantic to suggest that this explanation of the lemma seems to be using the notion of distance (am I right that we're using quadrance to avoid any reference to distance?)..

I admit the lemma seems completely obvious but I can't prove it too myself without considering distances - is there any way of doing this?..

Thanks again!

[njwildberger]

Thanks for the nice comment.

Distance is not really being used in the lemma. You can think of the area as being composed of lots of little unit squares, and we are just counting those squares by multiplying the number along the base by the number along the height.

[mediteight]

Many thanks for the kind reply!

Neat yes - I think I follow..

just to check, do we have to allow the little squares to be infinitesimal? (If one of the sides of the rectangle were an irrational multiple of the other...)

Does this matter? (if I'm not just wrong) Would proving this lemma need any notion of limit?..

[njwildberger]

A good question.

You are quite right, and touching on a historically delicate point. If we restrict our attention to the rational grid plane, then limits are not needed. Otherwise we need limits to define our notion of area. In my MathFoundations series, you will find that I prefer to work purely in the rational number plane, for exactly this kind of reason.

[mediteight]

Thanks for the info - I'll take a look at MathFoundations..

I must admit, the more I look into the foundations of maths the more bewilderingly subtle it appears (tried looking at Category theory and the difference between a class and a set - mystifying :)

Just pondering, even in the rational grid plane would you still have irrational distances all over the place and so tricky areas - like the quadrance on the side of the unit square...(?)

Anyhow - thanks for all the videos and explanations.

[ironclownfish]

It took me some time to figure it out too. Just think of it this way:

Q = (one side of Q2 * one side of Q1)

since Q2 and Q1 are squares, the length of one side of them would be the square root of their area, so:

Q = sqrt (Q1) * sqrt (Q2)

then just square both sides

Q^2 = [sqrt (Q1) * sqrt (Q2)]^2

or

Q^2= Q1*Q2

[veritasdivina]

Think algebraically in this case and not geometrically. If you have rectangle with sides a and b, then Q=a*b,

further Q1=a*a, and Q2=b*b.

Then Q^2=(a*b)^2=(a^2)*(b^2)=(a*a)*­(b*b)=Q1*Q2

which proves the lemma.

[Daood416]

I was all good until he talked about the symmetries!

[Ecoi1]

I do not obtain the same equation for the symmetrical version, am I doing something wrong or what? Anyone mind explaining how he got that?

[MedwayMan1945]

The way I was told that the Greeks proved Pythagoras' Theorem is this:

Let X be the point of Q2 opposite A3.

Let Y be the point of Q3 opposite A2.

Drop a perpendicular from A3 through the hypotenuse so as to divide Q3 into two rectangles and call the rectangle on the left R1 and the one on the right R2.

Area of A1A2X is half area of Q2.

Area of A1A3Y is half area of R1.

Rotate A1A2X by 90 degrees anticlockwise about A1 which takes it to A1A3Y.

Thus area of Q2 = area of R1.

[MedwayMan1945]

continuation of my proof:

Similarly area of Q1 = area of R2.

Thus area of Q2 + area of Q1 = area of R1 + area of R2 = area of Q3.

[verbal007]

I'm a little confused here. In the proof, I see how 4Q^2 becomes 4Q1*Q2, but what happened to

Q3-Q1-Q2? Why did all the values inside the brackets invert?

[mattrock23]

The square of x is equal to the square of -x. Thus inverting the values does not change anything.

[DescryHearts]

Ok.. and what does all of this mean again? I mean... what good is it?

So we as human beings found a relationship between distances and areas.. but what good is it to us? Seriously... you teach how to do it.. but can we teach its actual meaning?

[jlbjjlbj]

Wow, what an idiotic monkey. You should just stick with eating bananas. What is the meaning of math? trig? To take precise measurements of objects as the basis to manipulate them for a desired purpose. To understand pretty much all things, to a certain degree. It's just a tool for people to use to do what humans always try to do, and seems to be our future, which is to form things around us for understanding, and overall security, whether it is for personal, or sociatal.

[sikory]

when i critically look to your proof of pythagoras, i don't see why Q3 is a square, but I know, of course, that it's true

[sixbillionmorons]

All four sides are tilted by the same amount, so they all remain at right angles to each other. The triangles the tilt them are identical, so the four tilts are the same and the four side lengths are the same.

[MattMcDon]

I believe you have misscalculated the symmetrical version equation. You distributed the square to the entire equation which is not right. Perhaps that is why the distance doesn't equal that equation, because it's wrong.

[sapoty]

Thanks for this gift! In view of Garrett Lisi's theory of everything being founded soley on geometry your revision of the foundations is very relevant.