Spiral of Theodorus

This is a short, animated visual proof demonstrating how to construct square roots of any positive integer using the Spiral of Theodorus
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#manim #math​​ #mathshorts​ #mathvideo​
#construction #geometry #mtbos​ #manim​ #animation​ #theorem​ #pww​ #proofwithoutwords​ #visualproof​ #proof​ #iteachmath #spiral #theodorus #squareroot
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[@CatOnACell]

no, but this will be a great tool for drawing seashells in the future.

[@mentallyderanged888]

Seashells look more like the Fibonacci Spiral which is also easier to construct

[@PC_Simo]

@CatOnACell My thoughts, exactly 🎯!

[@PC_Simo]

@@mentallyderanged888 I’m pretty sure that seashells (at least nautilus shells) are nowhere near golden, in terms of their featured spiral. Only that they’re approximately logarithmic. See the Mathologer-video: ”Visual Infinite Descent”, and follow the link, mentioned therein, for more. 🤔

[@ian-williamfountain608]

@@mentallyderanged888 i was thinking the same thing.

[@dang-x3n0t1ct]

​@@mentallyderanged888 seashell are logarithmic spirals not Fibonacci

[@wyattstevens8574]

I don't think I'd be able to construct sqrt(200), except as 10sqrt(2).

[@MathVisualProofs]

Much easier way for sure :)

[@hallfiry]

I guess you'd get less error by using a 2-14 right triangle (or a 5-15 right triangle and a little bit of Thales)

[@r2_bee2]

I was thinking about the same 🫣

[@wyattstevens8574]

@@hallfiry 5-15? That would produce 5sqrt(10) instead!

[@hallfiry]

@@wyattstevens8574 Nope, you use 15 as the hypotenuse and construct yourself a right triangle over that with 5 as one of the short sides. 15²-5²=200, so the other short side will be sqrt(200)

[@plathanos159]

The lore behind that first triangle is quite... "irrational"

[@ShivSingh-io5eh]

Ba dum tss 🥁🥁

[@willcooper8028]

damn he really wanted to know if I think I could construct this by hand

[@EvdogMusic]

You gonna tell him or leave him hanging?

[@Rev_Erser]

you look like Aliensrock

[@willcooper8028]

@@Rev_Erser no he looks like me

[@anshuraut4091]

Yeah you can construct it by hand, And a ruler (for straight lines)

[@giovanigomes3257]

Nice! finally something new to put on every image besides the golden ratio

[@ZzSlumberzZ]

I remember learning this 9th class but couldn't fully understand it back then

[@abhidababy6746]

... Cbse board?

[@Redtigerr]

​@@abhidababy6746yep

[@CyrusThePython]

I smell CBSE

[@olivetho]

In case anyone here is in the same boat: This happens because of the Pythagorean theorem - because the relationship between the length of the hypotenuse (​ _c_ ) and those of the other two sides ( _a_ and _b_ ) is given by the formula _a²+b²=c²_ , we can express the length of the hypotenuse directly by taking the square root of both sides of the equation: _c=√(a²+b²)_ . Now, if we take _a_ to be the square root of some positive integer _n_ , and _b_ to be 1 (as is the case in the video); we can fill in the expression for _c_ we got earlier. This results in the equation: _c=√([√n]²+1²)_ - notice that we are squaring both a square root (its inverse operation which cancels it out) and the number 1 (1 raised to any power is 1), so we can simplify it to receive the expression: _c=√(n+1)_ - which is exactly the relationship described in the video.

[@rip_kakarotto]

I SMELL CBSE

[@maggi_tael]

my teacher made us draw an entire page of this thing, thanks for reminding me of this traumatic experience

[@JTCF]

Sounds like a cool way to compute the square roots. Actually, I wonder how computers do that in the first... New rabbit hole, here I go!

[@duckymomo7935]

Back then computation wasn’t as straightforward in geometry-greeks

[@mrocto329]

Most computers use Newtons method I assume. If you know calculus, you probably know that a derivative gives you a rate of change. This rate of change corresponds to the slope of the tangent line at a point (Think of it like deconstructing a curvy line into many tiny straight lines.) You can use this fact that a derivative is a tangent line to solve equations of the form f(x) = 0 by starting at an arbitrary point on the graph and repeatedly drawing tangent lines and finding the point they intersect the horizontal axis to approximate the solution x of the equation. In our case, if we want e.g. sqrt(2) we are really trying to solve the equation x = sqrt(2) x^2 = 2 x^2 - 2 = 0 and we can apply newtons method starting at X = 1 to find the values step 1 -> 1.5 step 2 -> 1.416 step 3 -> 1.41421 and we already found the first 5 digits after the decimal point with 3 steps.

[@Paul-sj5db]

​@@mrocto329Good explanation. As computers have limited precision they'll just stop when they hit that limit. Some applications that require super fast computation, e.g. games, will sacrifice precision for speed. Newton's method requires lots of division and that's an expensive operation. I remember a long time ago trying to write a really fast circle drawing program in 6502 assembler. It required square roots. Division was really hard so I instead went for a process that took advantage of the fact that n^2 is the sum of the odd numbers from 1 to 2n-1. I had a loop that repeatedly subtracted odd numbers until the result would be less than zero, then rounded up or down as appropriate. It was accurate enough and fast enough. A binary lookup table would have been even faster.

[@stephenwithaph1566]

I believe square roots have an infinite sum series, and stop after the series stops affecting the last-most digit displayed. This is how most computer systems calculate Linear Transformations in different cores, each core with a different transformation, then all summed up after they're all done. Each core can calculate the square root series for different values (indeces) of n , then add up the result and repeat!

[@Velereonics]

there's a really good book on euclidean geometry that's just all constructions beginning with a line is that which subtends the distance between two points or some shit like that. But it does the Pythagorean theorem and it does square roots and it does all sorts of crazy stuff with just constructions. basically validates math

[@haasjeoverkonijn6961]

Clearest explanation ever

[@user-bv7fh1kd6s]

There is a much simpler and non-recursive way to construct sqrt(n) using the fact that sqrt(n)=sqrt(n*1) which is the geometric mean of n,1. The geometric mean of two numbers a,b can be seen as a perpendicular to a diameter of a circle with length n+1 when the perpendicular stops when it touches the circle. In other words, you can first construct n+1, which is a pretty simple task, then bisect the segment to get the center of the circle. Then you can draw the circle, draw a perpendicular line 1 units from the end of the segment and voila your sqrt(n) is just the length of that perpendicular segment.

[@Savio25]

what a beauty.

[@quinn7894]

I like the pacing of this short. Very contrary to the seemingly rushed speech and lack of breaks of other shorts

[@herbertbader7558]

then visual representation of the spiral motivates the conjecture, that the difference of the radius between the loops remain constant. Then one could draw the spiral with a pencil limited by a thread winded up around a cylinder with radius=1 in the center which is rolling off by drawing. The difference between loops therefore is constantly 2*pi.

[@KalliJ13]

I can't figure out the point of using the compass, since you don't show using it to find the perpendicular of your √ line. You can make this construction with just a right-angle triangle ruler for your straight edge.

[@MathVisualProofs]

All the lines must be length 1.

[@KalliJ13]

@@MathVisualProofs If my straight edge doesn't have markings on it, yeah, a compass would help. But any point on the circumference of that circle will be length 1 from the center, so how does having it help you make sure your new line is perpendicular to the √ line?

[@MathVisualProofs]

@@KalliJ13 Ah, still have to also construct the perpendicular line. Didn't want to show that full construction here :)

[@noxteryn]

Straightedge and Compass means you can only draw straight lines and radians. Right-angles can be drawn using these two tools. There is no right-angle tool allowed.

[@chrismartin9849]

No right angle tool allowed. Thats a luxury. Straight lines and circles only. The seeds of all angles.

[@Integral_dy]

Is he ever gonna stop surprising us?

[@zedmelon]

Constructing this by hand would be difficult. Even a tiny early error would compound with every iteration. Very very cool though. It could be manually kept _near_ correct by measuring lines with a ruler (and making adjustments if needed) when the root is also an integer, e.g. 3/9, 6/36, 11/121...

[@balboa9439]

Great way to train and check the precision of your drawing skills with pencil, compass and ruler. (when square number pops up you measure length of the constructed line)

[@franticzenster8140]

It actually starts from sq rt of 1 ;)

[@sileudies]

We were given a question on this in a maths - we were told that all the outside lengths were 1, and told to find the 9th hypotenuse (which, after working it out, was sqrt(9) which is 3!) Didnt realise it was an actual mathematical thing tho lol

[@TehJumpingJawa]

Now slap a z-coordinate for extra funkiness

[@theguyshetellsunottoworryabout]

I used to do this by hand in secondary school

[@andrebartels1690]

That's dope shit.

[@cheeseheadfiddle]

As the pattern evolves, I’d like to see the relationships between the perfect squares, 4,9, 16, 25, 36 etc. Or how the primes relate. I’ve drawn this before. Very satisfying.

[@user-bv7fh1kd6s]

I don't think that primes would give you an interesting pattern here, but perfect squares might.

[@alien_0_0_7]

I did it jn 7th or 8th standard to locate √2 or any other irrational number on number line.

[@wyattstevens8574]

You learned this in (USA) junior high? Neat way to find them!

[@user-bv7fh1kd6s]

Well this is not true. Not every real number can be constructed this way, only numbers of the form sqrt(n). Although all constructible numbers (en.wikipedia.org/wiki/Constructible_number) can be drawn by a ruler and compass. But it is not true that you can locate any irrational number since some numbers are non-constructible, like pi or e or all transcendental numbers or ever third roots of rational numbers. In fact it is known that there are more non-constructible numbers than constructable numbers. So what you have probably did is locate constructable numbers, which are dense in the reals so they may have given you an illusion that they cover all of the real number line.

[@Deedoo_r]

its also great for drawing top-down arial shots from the center of a spiral staircase

[@jsmunroe]

And its not even a logarithmic spiral, that is what is so beautiful and strange about this.

[@Tommy_007]

So you are saying that anything that is not a logarithmic spiral is beautiful and strange?

[@Pizhdak]

The urge to extend this spiral into a smooth curve and find a polar eaquation for it ↗️↗️↗️

[@pascalrockenstiehl6972]

Wait yes: 1 could have just drawn 2 lines with an 90° angle with lenght 10*1 each. The connecting line would be 10*sqrt(2) = sqrt(100)*sqrt(2) = sqrt(100*2) = sqrt(200) Or am I missing something?

[@RobaGuy]

Wait so… if you do this to infinity, you would have a hypotenuse of length root(infinity) = infinity, as n -> infinity, and a side of length root( n - 1 ) which is still infinity as n -> infinity, but have a line segment of 1 because of the circle drawn with radius 1

[@nicolabombardelli2874]

Un triangolo isoscele di base 1 e altezza infinita ha due lati di lunghezza infinita e DUE angoli retti.

[@TheGBZard]

Reminds me of the shells found in snails or the extinct ammonites

[@maricelty7744]

As you zoom out more. The spiral begins to follow (1/θ)r = 1

[@user-uv2sg6yh1t]

So that's what JoJo told us about!

[@Suo_kongque]

I heard spiral out. The TOOL fan in me has been awoken.

[@RonanDmfx]

my 7th grade math book's cover has this, when i found out about it, i found it very amazing

[@ikepigott]

The challenge here would be to do everything by hand all the way up to square root of 100, and then checking that against your unit 10 to see how good your measurements and calibrations are.

[@happinesstan]

If you spiral in, instead of out, you will find the secret to infinity.

[@user-gj1kf5lw6p]

The first line is sqrt(1).

[@potentiallyunaffiliated4285]

"Is that an Archimedean spiral?" -Peter Parker

[@X-SPONGED]

"Spiral of Theodorus" sounds like some maguffin from a new Indiana Jones movie

[@samuelmidzi8240]

Looks like the golden design rule (snail shell)

[@KanjiasDev]

Btw this is also an amazing visualization of the concept behind Pythagorean theorem ✌️

[@satatgautam2152]

We are taught about this in 9th grade. It is called square root spiral.

[@maulliksiktia9100]

CBSE wale ho na ?

[@satatgautam2152]

@@maulliksiktia9100 yes

[@bonami.]

this was asked in my 8th grade math exam lmao

[@James-sc8jz]

Theodorus means "GODS GIFT"

[@mironoleszczuk5660]

I thought it meant "the smelly one", as in "The odoures" 😅

[@pendragon7600]

Yes. It's easy to construct square roots, not just of integers, but of anything in the field you're working over. Start at the origin. Draw a line segment of length 1+x. Draw a circle of radius (1+x)/2 at the midpoint of the segment. The line drawn perpendicularly between 1 along the segment and its intersection with the circle is of length sqrt(x)

[@jargontrueseer]

That last bit sounded pained, like "do you really think you can do what I did?? do you really think it's that easy?!?" XD

[@boriswilsoncreations]

The way he asks with such an intensity if I'm able to construct this by hand is somehow intimidating

[@lukasbirkmann3240]

U givin me flashbacks to a problem we had to solve In School a while back. We had to figure out the perimeter of any given shape, N.

[@andrewmcgill3178]

Wowie! Thanks

[@skypepper577]

Easy for me, I am really good in spiraling

[@RockyNitro5470]

Mr. Nautilus

[@CrazyDiamondo]

Looks like infinite spiral staircase

[@logx-ow1us]

This was a bonus on my math test in like 4th grade;-;

[@reinerzufall1292]

I mean as long as you're accurate enough, making sure you're drawing line perfectly perpendicular and on length 1 I don't see why not. I actually did this a lot in school when I was bored

[@TheOriginalDeaf]

It looks like a fossil seashell, it's like the math is embedded in nature 😮❤

[@xitheris1758]

That's exactly how it works, actually.

[@alireda646]

Its a graphical method to get square root 2 perpendicular lines with lengths 1, sqrt (x-1 ) Third line length will always be square root of x 🎉 Thus before calculators peoples did all numbers in tables makes it easier for the generations comes after

[@channel-wt3ff]

I remember using this at school to find a root of a prime number on the number line.

[@arcanine_enjoyer]

The staircase that the characters have to go up (it's for comedic relief)

[@user-ru9rp1pf6m]

Мне нравится что угля становятся всё круглее и круглее))

[@BlackbodyEconomics]

What immediately struck me as a further question to consider is if any of the radial lines from the center overlap exactly and if there is a pattern to it. The second go around it seemed that there were a fair number of 2nd level radial lines bisecting first level triangles and 3rd level radial lines trisecting the first level (though it doesn't seem very consistent). If you don't investigate that - I will :P

[@BlackbodyEconomics]

Nope - let me correct that. As a new round starts, it seems to start trisecting the round before it, then it eventually moves toward bisection. Then it starts over again with the next round. There's gotta be a pattern in there for sure. Also - couldn't you also go inward from the first right triangle? Just perform the operation backwards. Which begs the question: are there other algorithms that don't include a right triangle? If you started with an equillateral, isosceles, or some other random triangle - would you eventually run into a right triangle if you spiraled in the direction it's most likely to inhabit?

[@fudgylumpsforlife]

Bro really said √4 💀

[@MathVisualProofs]

my favorite way to say 2 for sure.

[@augustday9483]

Looks like a nautilus shell.

[@TruthOfZ0]

As N goes to infinity does the length of the inner spiral equal to the latest external spiral ..what is the radius of that external spiral/circle ??

[@user-awrssadk]

سؤال جيد اعتقد يساوي طول منحنى الدالة جذر n

[@TruthOfZ0]

@@user-awrssadk yeah but after 360 degrees of rotation is the inner and latest spiral circle close enough to minimize the gap ? or it still increases geometrical? to infinity...and last after a full rotation 360 degrees in which N we end up? is it steady a fixed number or it increases to with every full rotation? it takes more steps to fill to 360??

[@user-awrssadk]

يزيد مع كل دورة جديدة إلى ما لا نهاية على ما اعتقد ، ولا ينتهي في الدورة الداخلية ابداً

[@alejrandom6592]

Yeah because sqrt(200) is just 10*sqrt(2) which is the diagonal of a square of side length 10

[@ikes_Pikes]

wow

[@ItzRoxyAlpha]

Haha this is at my school i had to draw this and the roots

[@terrariariley1643]

I just simplify roots and then use newtons method to approximate whatever the root was simplified to

[@EmmaSquire-ks9nu]

But what is the angle of the tangent lines drawn from the circle? [Edit: oh I see, you are making right triangles the whole way

[@Teledoor24]

the perfect whorl...

[@speedsystem4582]

I had done it when I was in school on a big chart. I got it done till sqrt(400+something)

[@Croissantification]

I did , its hell

[@wolfxlover]

How do you do this by computer? I would like to program this ! :D

[@JacobBarrowlikecursive]

Average calamity mod boss attack:

[@itsiwhatitsi]

It’s time to make a ledder 👍

[@abehankens7456]

insane that it looks like a nautilus shell! and no i couldnt construct this by hand lol, i swore i would never construct another geometric figure by hand again after i finished geometry honors

[@smileyp4535]

How does the perpendicular line extend in length if it's only length 1? The circle is radius 1 so the line can't ever actually get bigger, can it?

[@rulojuka]

Not the simplest way of drawing sqrt(200) but definitely a beautiful one!

[@shadowydevilksgc3671]

Nautilus shell? 💀

[@wongmanwaihehe]

Well, I made one before I watched this video.

[@DanaTheLateBloomingFruitLoop]

Never learnt about this in school but I wish I did. just cause it's cool

[@EvolvedFinch]

Unfortunately it’s not posted anywhere, but in Blender 3D software I actually made a pretty good looking one via brute force with pre-made assets. Mine was difficult to create for the fact that each side protruded outward. Gonna save this video and try again using the math you provided. 👍🏻

[@MathVisualProofs]

I wish I knew how to use Blender :)

[@GenericInternetter]

@@MathVisualProofs it's not hard once you get a grasp of the basics, definitely worth a try

[@deanwinchester3602]

There’s actually a better way to do it in terms of efficiency. consider that you want to draw √n. you can write down the binary representation of n and then the problem would be showing √(2^a1 + 2^a2 + … + 2^ak), where k is about log2(n). wlog you can consider ai’s sorted in ascending order. then you can write down 2^ai = (2^(ai/2))^2 for even ai’s and 2^ai = (2^((ai-1)/2))^2 + (2^((ai-1)/2))^2 for odd ai’s, and you can calculate next term by doubling current term and using the same method in video to draw √(sum of the (term^2)s). the upper bound of number of steps would be 1.5log2(n) considering the binary representation of n is 11…1 in the worst case. so for example. If you want do draw √1000 then you would need at most 1.5log2(1000)

[@PedroCristian]

X^2=n generates a field extension of degree 2... Hence constructible..

[@JosephYohanGino]

I know the third triangles area is equal 2.

[@0news28]

I make one of these for a math class one year. I turned the spiral into a goat horn

[@Astroponicist]

I could make a spiral stair case in a house from this insane idea. It would be expensive. the house would have to be huge.

[@MathVisualProofs]

And the upper stairs would be quite thin and long I think. :)

[@stk8510]

The trick is not to draw a 1-circle but to draw a 1-line in the right angle 90°. Then usung Pythagoras. 1 sqared + sqareroot of x sqared = xä(x+1) sqared.... Simple trick. Please make a video that shows us the squareroot of 123456789

[@user-dk8mj4uo9c]

Wow, I have drawn this spirals for fun, cool to know that some mathematician also think they are funny

[@pranavgajakos6427]

It looks like fibaunacci series

[@Mnaughten601]

Could I do it by hand. Maybe. But I’d rather draw it with python, or LaTex.

[@limoncheto]

roblox obby staircases be like:

[@antflam7473]

Almost had a heart attack when I saw this lol. In geometry class we had to make a this for an extra credit. We also had to decorate it which was easy but it was very tedious and a little difficult at some points. But I needed the extra credit so that’s what I had to do

[@LevianDaWolforca]

And if a magic cat decide to trap you in a circle, and whenever you choose to move in a direction, the cat can decide to move you in the opposite direction. Just remember to follow the path of this spiral, you will alway make it out.

[@planktonfun1]

nautilus gear

[@plamenlorea6910]

Yes

[@GenericInternetter]

i'd be interested to know what is the angle between the first line and the sqrt(4) line. is it a radian?

[@Rando2101]

It's about 1.92 radians

[@ShivSingh-io5eh]

We studied this and made this spiral as a project in 9th grade, but we only had to make it upto √4 or √5 I think

[@Seraph2101]

That looks like a 4D staircase but 2D

[@tigerfifty]

Root 4 is just 2

[@maxborn7400]

what happens when the starting line segments are infinitesimal? What kind of spiral does it produce?