A killer question from Japan. Is tan 1° a rational number?

[@rowantodd827]

He was sentenced to death? That doesn't seem like the most rational way to handle the situation.

[@itsphoenixingtime]

It was radical though.

[@alifsheikh4237]

I dont think that this is a natural thing to do

[@Mike-H_UK]

It was his own fault for being transcendental towards Pythagoras!

[@giork2828]

Should have rationalized his decisions ref. his discovery

[@maxhagenauer24]

Only numbers can be rational or irrational, not actions.

[@marcusscience23]

Allegedly, Pythagoras hated 2 things: irrational numbers and beans. The former he killed for, the latter he died for.

[@CosmicHase]

An irrational number of beans

[@Nelster]

Did he bake beans into a pi?

[@zsoltnagy5654]

Yes, (tan(1°) is rational) ⇒ (tan(2°) is rational). BUT it's not really, that _"(tan(2°) is rational) ⇒ (tan(3°) is rational)"_ but rather *(tan(2°) is rational AND tan(1°) is rational) ⇒ (tan(3°) is rational).* This is a very important distinction here, since one might prove with the naive version _"(tan(1°) is rational) ⇒ (tan(2°) is rational) ⇒ ... ⇒ tan(60°) is rational",_ that _"tan(60°)=√3 is not rational ⇒ ... _*_tan(45°)=1 is not rational_*_ ⇒ ... ⇒ tan(2°) is not rational ⇒ tan(1°) is not rational"._ The correct version is rather the following: *(tan(1°) is rational) ⇒ (tan(1°) is rational AND tan(2°) is rational) ⇒ (tan(1°) is rational AND tan(3°) is rational) ⇒ ... ⇒ (tan(1°) is rational AND tan(45°) is rational) ⇒ ... ⇒ (tan(1°) is rational AND tan(60°) is rational) ⇒ (tan(60°) is rational)* such that (or if and only if) *(tan(60°)=√3 is not rational) ⇒ (tan(1°) is not rational OR tan(60°)=√3 is not rational) ⇒ ... ⇒ (tan(1°) is not rational OR tan(45°)=1 is not rational) ⇒ ... ⇒ (tan(1°) is not rational OR tan(2°) is not rational) ⇒ (tan(1°) is not rational)* (by contraposition and De Morgan's law) So be carefull with this otherwise one might come to the wrong and false idea of proving with the naive version, that _"tan(45°)=1 is not rational"!_

[@minerscale]

That's a very good point, the *only* thing this proves is that tan 1° is irrational since it's what we assumed for the contradiction.

[@asdbanz316]

their problem if they assume it Also tan 1 => tan 2 => tan 3 => ... => tan 60 is an entire chain with dependencies on all previous steps If you start with tan 3 without relying on 1 and 2, you create new independent assumption and it goes like tan 3 => tan 6 => tan 9 => ... btw I just realized that all factors of 60 will make irrational tangents

[@trumpetbob15]

@@minerscale But we only got an irrational because we went to 60 degrees; why that one? If we had instead only gone from 1 to 45, we end at a rational number and therefore, no contradiction, everything is rational. I'm not sure this proof actually does show that Tan 1 degree is irrational.

[@theupson]

@@trumpetbob15 no. youre confusing the converse with the contrapositive. edit: if tan(1 degree) is rational, then tan(n degrees) is rational for ALL natural numbers n, by strong induction using the sum formula for tangent. therefore, finding a single natural value of n yielding an irrational tangent does the job

[@trumpetbob15]

@@theupson Yeah, I'm totally confused with this one.

[@jackychanmaths]

tan(pi/6) is already well-known to be sqrt(3)/3 which is irrational so it is not necessary to go to pi/3

[@BC-ng8yk]

Also tan(pi/4)=1🤓😱🤯 pi/4=45 dgr < 60 . Not a valid proof!!!!!

[@Mike-H_UK]

Totally true, but once you introduce the concept of the incrementing angle, it is no harder to use 60 degrees than 30 degrees, even if it is not quite as minimalist.

[@Geek37664]

He should stopped at π/12

[@ingiford175]

@@Mike-H_UK Yes, but you do go though a rational point as 45 degrees (pi/4 is rational) which may confuse the issue

[@atrus3823]

Since he’s really using induction here, he doesn’t actually go all the way to 60. Really once you’ve shown the pattern continues, he could go to any whole number of degrees and it’s no more effort.

[@joachimkeinert3202]

He was not sentenced to death, he was suspended from the pythagorean community and lated perished at sea, what was then interpreted as a death penalty by god.

[@GenUrobutcher]

Bro was exiled to a unsurvivable Island and they call God killed him lol

[@NestorAbad]

Nice proof, thanks for sharing! Another easy but interesting proof of irrationality by contradiction that you could include is: "Let p and q be prime numbers. Show that log_p(q) is irrational." (where log_p denotes logarithm in base p)

[@Tuqtgref]

Hey what are you? a mathematician ? Just curious 🤔.

[@NestorAbad]

@@Tuqtgref I'm a math teacher at high school 😄

[@theupson]

*cough* p relatively prime to q, i think you mean. if log(p)/ log(q) = n1/n2 where n1 and n2 are natural numbers, then p^n2 = q^n1, which contradicts the uniqueness of prime factorization (itself a really excellent proof, my favorite example of a non-algebra-grind strong induction)

[@BryanLu0]

​@@theupsonWell, prime numbers are relatively prime to each other. A problem can hide information

[@NestorAbad]

@@theupson Indeed, the only thing you need is that p and q are not a power of the same number.

[@Nxck2440]

Spent 6 minutes explaining basic stuff and then finally the answer in only 2-3 lol Also you could stop at tan 30 = 1/sqrt(3), which is irrational by inspection.

[@user-matlee2477]

And tan 18 degrees as well. You can find out this value with a regular pentagon.

[@dennisdeng3045]

I’d say, if anything, he shouldn’t have gone past 45 degrees…

[@Planetyyyy]

​@@dennisdeng3045stopping at 45° would prove nothing, as tan(45°) is rational

[@IdeaSlug]

@@dennisdeng3045 you don't get a contradiction if you stop there

[@kyleliao4445]

tan(15º) = (sqrt6-sqrt2)/4 :eyes:

[@randomjin9392]

Suppose tan(𝜋/180) = p/q with integer p, q: 0 < p < q. We then get cos²(𝜋/180) = q²/(p²+q²) and sin²(𝜋/180) = p²/(p²+q²). Multiply these to get: sin(𝜋/90) = pq/(p²+q²) and cos(𝜋/90) = (q²-p²)/(p²+q²). We now have sin(𝜋/90) and cos(𝜋/90) expressed as rational numbers. This means that any multiple of 2, 3 or 5 of the cosine and sine of 𝜋/90 will be rational - simply because double/triple/quintuple/etc angle formulas only involve polynomial operations on sine and cosine, so will always map a rational number to a rational one. Since 𝜋/3 = 2∙3∙5(𝜋/90), by our construction both cos(𝜋/3) and sin(𝜋/3) must be rational which isn't the case. Hence tan(𝜋/180) is irrational.

[@unholycrusader69]

How did you type that "π"?

[@evreatic3438]

sin(π/90) = 2pq/(p²+q²) You were off by a factor of 2, though it doesn't affect the proof.

[@randomjin9392]

@@evreatic3438 Good one. I thought to fix it, but .. let's leave it to check if anyone is actually reading it and understanding what's going on ;)

[@Deficard]

​@@unholycrusader69youtube can't do that. on keyboard, you're eqipped with π. but there's such things as unicode. basically an sprite for text in computer. for "𝛑", i had to copy the unicode U+1D6D1. i can easily access it through an app that lists unicode

[@wiwaxiasilver827]

@@randomjin9392 Wow, very neat use of the double angle formula :) How do you know the triple or quintuple angle formulas though? Edit: wow, I just realized that those necessarily follow from the angle addition formulas lol XD

[@ashutoshpendse4273]

"I don't want to go off a tangent" and you came back to the original problem which is in fact about a tangent ratio! A nice wordplay!

[@PerMortensen]

After six minutes of filler...

[@Traw-ve7qf]

8:36 if we continue to tan(45)=1 that is rational

[@n00bxl71]

Yes, but that doesn't prove anything. tan1 being rational implies tan(k) is rational. But tan1 being irrational doesn't imply that tan(k) is irrational. tan45 is rational regardless of whether or not tan1 is rational. The way cofunctions in trig are defined shows that cos(x)=sin(90-x). As 90-45=45, cos(45)=sin(45), therefore tan(45)=1.

[@ercfis]

You cannote use this solution to prove tan 45 is irrational, cause (irrational*irrational) and irrational + irrational can be rational numbers!

[@MrConverse]

3:45, “…must be *irrational”. I believe you misspoke there. Good video!

[@Bodyknock]

Regarding the middle part of the video talking about proving that roots of primes are irrational, you can use the Rational Root Theorem to make a very broad, useful statement in that regard. As a reminder, the Rational Root Theorem says that if you have a polynomial equation with integer coefficients of the form aₙxⁿ + ... + a₀ = 0 , and the rational number in reduced form p/q is a solution, then p is a factor of a₀ and q is a factor of aₙ . Now take a look at the special case of the Rational Root Theorem where you want to solve an equation of the form xⁿ - n = 0 for some integer n. From the RRT we know that if x is a reduced rational solution p/q of that equation then q must be a factor of 1, which means if x is rational then it is an integer as well. In other words, all real solutions to the equation xⁿ - n = 0 must be either integers or irrational numbers. There are no purely fractional rational solutions to it! That in turn implies if you want the n-th real root of some integer c, the root must be either an integer or an irrational number. So if c isn't the n-th power of some integer than the n-th real roots of c are all irrational.

[@MichaelRothwell1]

Yes, this is my favourite proof as it immediately gives the general result about nth roots of all integers being integer or irrational without any extra work. It is very instructive to prove this without the rational root theorem (basically, prove the rational root theorem in the particular case of xⁿ=m) and observe what results from elementary number theory are needed. Interestingly, the FTA (existence and "uniqueness" of prime factorisation) is _not_ needed.

[@dijitle]

The way you pulled off that “weave” is legendary!

[@ulyssesfewl1059]

For the 30, 60, 90 triangle, surely you only need to go as far as tan (30 deg), which is 1/(root 3), which is also irrational.

[@whitepoole]

Thanks for the pun at 5:50. Made me smile 😊

[@abdulreyhan9760]

After watching ur every video my heart says ," I love you ". ❤❤❤

[@viktor-kolyadenko]

tan(30) is irrational. But tan(45) is rational.

[@GenUrobutcher]

You only need one irregular to disprove something. But not even millions of positive reinforcement is enough to prove something.

[@radiationpony8449]

The point is, that this proof for the irrationality of 1deg should also prove that 45deg is irrational. As tan(45deg) is known to be rational, this shows that the proof is at best incomplete as it only shows that not all tangent values of integer degrees are rational but does not yet state that tan(1deg) is irrational. To finish the proof you should show that in order for tan(45deg) to be rational and tan(60deg) to be irrational that tan(1deg) must be irrational I.E. the sum of irrational numbers can be rational but the sum of rational numbers cannot be irrational.

[@viktor-kolyadenko]

@@radiationpony8449, We get an equation of degree 45 of the type P(x)/Q(x) = 1. The chances of getting whole roots from it are not very high.

[@vertechua]

​@@GenUrobutcher That's something well said man!

[@lameyeast7085]

​@@radiationpony8449 i think it only goes one way Like u can use tan 1° to assume tan 2° is rational(in the proof) cuz we get - (r+r)/(1- r*r) where r is rational, so the whole equation is rational However, u can't use that tan 1° irrational to prove that tan 2° is irrational cuz then u get - (ir+ir)/(1-ir*ir) where ir is irrational, in which case the equation CAN be rational (say (π+π)/(1-√(1+π)*√(1+π)) which is rational), unlike the first case where the eq can only be rational So the proof is correct, tho correct me if im wrong

[@davidh.4649]

I wasn't able to answer the question but I do know one thing. You proved that tan 1° is irrational! Pythagoras is coming for you Presh! 😂

[@genius11433]

Can someone please explain the contradiction at 3:42? I don't get the logic.

[@wadoichimonji8522]

We are assuming that if p is a prime number then √ p is rational and therefore √ p can be expressed as some a/b where and a and b are integers. If b²p = a² and both a² and b² will have an even number of prime factors, the only way b²p will be equal to a² is if they have an equal number of prime factors so p must have an even number of prime factors as well but p is a prime number so it's can't have an even number of prime factors which is the contradiction.

[@martinmonath9541]

At this point the fact that every natural number has a unique prime factorization is used, i.e., every natural number can be uniquely written as a finite product of prime numbers. So let's assume that the prime factor p appears in the prime factorization of a n times, where n>=0. Hence, p must appear 2n times in the prime factorization of a^2. By a similar argument, we can say that p appears 2m+1 times in the prime factorization of b^2*p, where m denotes the number of times p appears in the prime factorization of b. But then, by uniqueness of prime factorization, 2m+1=2n must hold since b^2*p=a^2. But 2m+1 is odd whereas 2n is even, so they can never be equal. Contradiction.

[@ronald3836]

​@@martinmonath9541 He uses a way too powerful theorem. It is enough to know that if prime p divides ab, then p divides a or p divides b. Now start by assuming sqrt(p)=a/b with a,b positive integers, where we pick a to be as small as possible. Since pb²=a², we know that p divides a², therefore p divides a.Thus a=pc for some integer c, and we have pb² = p²c², so b² = pc², and by the same reasoning as before we get b=pd. So a/b can be rewritten as c/d with c,d smaller than a,b. But we picked a to be minimal. Contradiction. Therefore sqrt(p) is irrational.

[@genius11433]

@@martinmonath9541 Thanks.

[@Ninja20704]

@@ronald3836 it is not “too powerful”, it is simply the fundemental theorem of arithmetic that is one of the most important and underlying theorems in all of number thoery and used for almost any number theory proof involving prime divisibilty and factorisation

[@JCCyC]

On the other hand, tan (and sin and cos) of every rational number of degrees (or rational multiple of pi, same thing) is algebraic. Wolfram Alpha tells me tan 1° is the root of a polynomial of degree 24. Bit doesn't tell me if it's expressible in radicals form.

[@PoppySuzumi1223]

Very smart proof by contradiction.

[@Physmathematicain]

Really nice proof😊

[@MPSmaruj]

That's how Pythagoras got away with sentencing his opposition to death -- he convinced the entire world he has never real to begin with.

[@japan.mathematics]

日本でとても有名。話題になった入試問題。

[@nesgoof3412]

It's fun to see how Presh is slowly sliding into a bit more personal and humoristic approach in his videos.

[@variousthings6470]

5:43 I think you meant to say: "I'm not sure, I will leave it to the _historians_ to decide."

[@maruthasalamoorthiviswanat153]

Excellent question and excellent solution

[@yb3604]

nice one laddie

[@nikhilsood9501]

The word play was great.

[@CristianBaeza-rh7zq]

Cool! My approach was, assume tan(1) is rational, then 1 degree is constructible, then any integer valued angle is constructible, contradiction.

[@bluerizlagirl]

Naïve answer: Yes, because you can draw a triangle with angles of 1, 89 and 90º, measure the sides opposite and adjacent to the one-degree angle with a ruler accurate to the nearest millimetre and divide one integer by the other. Second thoughts: Measuring is cheating! Pi will come out rational that way. There's no reason to expect the sides to be rational. After all, 45-45-90 triangle has sides 1, 1 and sqrt(2). Third thoughts: Oh, just watch the video.

[@davidpederson2905]

I learned today (listening to the audio repeatedly, and Google searches) what a "blackboard bold symbol" is. For me a serendipity moment. Spoiler: its the double line on a letter for emphasis.

[@Trulsbk]

For anyone interested, by Niven’s Theorem, tan(1)=tan(2pi/360) is an algebraic number of degree 24. This means that it is the solution of a polynomial of degree 24. So it is definitely not rational (rational numbers are of degree 1).

[@Hiroaki1990]

That problem and “Proof that π is greater than 3.05 (University of Tokyo in 2003)” are most well-known as math problems of Japanese university entrance exams.

[@unshuuLEMON324]

日本で有名な伝説の京大入試が海外の人に紹介されてて嬉しい

[@Pootycat8359]

3:56 Ahhh! The infamous words from my old calculus text: "The proof is left as an exercise for the student."

[@Pootycat8359]

I'm an ENGINEER! Gimme "Simpson's Rule," gimme "Newton's Method," gimme a number, accurate to within a few percent!

[@composerlab]

i got an alternative proof - similarly i did by contradiction where i suppose tan1 is rational tan(1 deg) =p/q for some p,q are integers, q is nonzero and gcd(p,q)=1 for simplicity, i converted to radians where tan(pi/180 rad) = p/q, hence tan(89pi/180) = q/p => qsin(pi/180) = psin(89pi/180) hence qsin(pi/180) - pcos(pi/180) = 0 then i used r formula afterwards to convert LHS to a single sin expression… for sin = 0, the thing inside the sine function = k*pi for some integer k, hennce p/q = (1-180k)/180 * pi. since (1-180k)/180 is rational, for the entire expression to be rational, pi has to be rational which is contradictory

[@ალექსანდრეოთხოზორია]

bro this is so beautifull i am crying😍😍 i would never solve this tho

[@lightyagami1752]

Basically, you can formalise this with induction. Proposition: Assuming tan x is rational, then tan nx is rational for all natural numbers n. Proof. Base case is given as an assumption. Induction step: Suppose the proposition is true for some natural number k (i.e. tan kx is rational). Then tan (k+1)x = (tan kx + tan x)/(1 - tan kx . tan x), and since the rationals are closed under the four operations of basic arithmetic, this is rational (you can substitute with a/b, c/d to convince yourself). Hence the proposition is proven by induction, given the base assumption holds. If we start by assuming that tan 1 degree is rational, then we must have that tan 60 degrees is rational. But tan 60 degrees is sqrt(3), which is not rational. A number cannot be both rational and irrational at the same time. Hence tan 1 degree is not rational (Reductio Ad Absurdum, QED). Now, an interesting problem is to show that 45 degrees is the lowest rational positive degree measure with a rational tangent. Or equivalently, that for tan k*pi, where the angle is measured in radians, to show that the smallest positive rational value that k can take while returning a rational value for tan (k*pi), is k = 1/4. Right now, this is a conjecture of mine, but an online paper by someone else (which I have not verified is correct) answers this question in the affirmative: (link)(www2).(oberlin.edu)/faculty/jcalcut/tanpap.pdf - remove the brackets.

[@smylesg]

4:55 No Chinese people were harmed in the proving of this theorem.

[@gamerzkhargoshy505]

I HAVE SEPARATE TRICKIER APPROACH 1DEGREE IS PIE/180 RADIANS SO TAN(1DEGREE)=TAN(PIE/180)=PIE/180 SINCE PIE/180 IS VERY SMALL WE CAN WRITE TAN(PIE/180)=PIE/180 AND AS IT IT CONTAINS PIE WHICH IS IRRATIONAL NO THERFORE IT IS IRRATIONAL NO .

[@cheesetasty1646]

Intuitively, since tan of 1 degree is essentially a random real number, the probability that it is rational is zero since the set of irrational numbers is uncountably infinite, while rational numbers are countably infinite. *For this to be more rigorous, you would have to show that tan(x) being rational is independent of x being rational

[@PhilipHaseldine]

Hmmm maybe I leaned on this in my belief that it was irrational before he showed the proof, not sure, thanks....

[@Straight_Talk]

There's a problem with the proof. It's logic "proves" that the tan of any angle is irrational. However, tan 45 = 1. More fundamentally, trigonometric ratios represent one side of a right triangle divided by another, which is a fraction. Which suggests that at least some trigonometric ratios must be rational. Presumably where neither the numerator nor denominator in the ratio (fraction) are either irrational or have infinitely many decimal places.

[@cheesetasty1646]

@@Straight_Talk I’m not saying tan(x) is always irrational, I’m saying that since tan(1) has no special value, it may as well just be some random number.

[@Straight_Talk]

@@cheesetasty1646 That doesn't relate to my point.

[@bandana_girl6507]

I would have actually gone about it using facts about geometric construction: Straight edge and compass can be used to construct any number which combines the operations +, -, *, and / with square roots (of which rational numbers are a subset). For a given angle, its tangent is the length of a line segment perpendicular to one leg at 1 unit from the apex and with the other end at the intersection with the angle. The constructable regular polygons have a number of sides that is a power of 2 and any number of distinct Fermat primes. Given that to get tan(1°) you have to construct a 1° angle, from which you could construct a 360-gon, if tan(1°) were constructible (not just rational), 360 would have to be of this form. However, its prime factorization is 2*2*2*3*3*5, in which 3 is repeated. Since tan(1°) is non-constructible, it is irrational. Note: it is origami-constructible, though, thanks to 3 and 5 both being Fermat primes and angle trisection being possible with origami

[@myb701]

Like most ancient sources, the little information there's about Pythagoras makes studying him really fun! We're kindasure he did exist, even if he might have not been a "greater than life" figure that the Pythagoreans admired regardless, but since there's no writings of him left, it's all muddy, and it's hard to sift the propaganda and legend from reality. The original source did say that the Pythagorean that discovered irrational numbers later died from drowning. But everything else comes from later, very inconsistent sources, with the story going from the original, to mixing with an account of the Pythagorean that discovered the regular dodecahedron drowning at the sea, most stories don't even refer to Hippasus by name lol. The particular tale that Pythagoras himself executed Hippasus for his discovery is probably less than 50 years old. Hell, we aren't even sure that it was Hippasus who proved that Sqrt(2) is irrational, we're just as unsure of that as we are of Pythagoras proving the Pythagorean Theorem, or if one of his students/succesors did.

[@PhilipHaseldine]

Like Jesus, although he was not famous for maths (unless you count the feeding of the five thousand and dividing up a small amount of food)🤣

[@TDSONLINEMATHS]

Mathematics as a subject serves as a basics to all subject which is generally accepted at all levels of educational ladder and it plays a unique role in the development of each individual....

[@bubtb-yl8lu]

huh

[@watchmakerful]

It's enough to go up to 30* (tan 30* = 1/√3 is irrational).

[@justlikeyourfathersaid]

But extending this logic it will imply that tg(45)=1 is irrational. The only thing I can find that really proves this is Nivens theorem. And that’s wayyyyyy too complex for a school question

[@zmaj12321]

@@justlikeyourfathersaid The proof demonstrated in this video does not imply than tan(45) is irrational. The proof merely shows that tan(1) and tan(60) are irrational. It does not actually make any claim about the values in between.

[@maneeshaliyanapatabendy1481]

this video is like the Cars animated movie trilogy part 1 - math part 2 - math lore, the tragic origin of irrational numbers and the tyranny of Pythagorus part 3 - back to math

[@bobbyhillthuglife]

Historically, while the existence of Pythagoras himself is doubted, the existence of a school of philosophers with a keen interest in mathematics known as the Pythagoreans is not. They very much existed, whether or not their supposed founder was actually real. So, when we refer to something as "Pythagorean" we are not necessarily referring to Pythagoras himself, but rather the teachings of the "Pythagoreans"

[@AH-xs3hg]

If he was not real, is he just... imaginary?

[@byronwatkins2565]

The part that always confused me was that pi is defined to be the ratio circumference/diameter; this seems to make pi rational by definition. Eventually, I was forced to figure out for myself that circumference and diameter cannot simultaneously be rational.

[@JohnLeePettimoreIII]

4:04 the square root of 69 is "8 something".

[@pianoplayer281]

Consider a right angle triangle with side lenghts 1-x, \sqrt{2x-x^2} and \sqrt{1+x}. The angle is 1 degree. Then tan1=\frac{\sqrt{x}\sqrt{2-x}}{1-x} and whether x is rational or not, the expresion for tan1 will be irrational.

[@snc6344]

this isn't a right triangle, and even if it was, the angles wouldn't be consistent for all values of x

[@deept3215]

I somehow totally misread the question assuming it was asking if tan(1°) could be expressed with radicals and ended up proving it.

[@bernhartschmieder9401]

I think you gotta change the edition style or the editor because it's very confusing that what you say and what he writes are different things and one has to guess, referring to the beginning.

[@santripta]

trivial. small angle approximation. sin 1 = 0. 0/cos 1 = 0. 0 is rational. hence, tan 1 = 0 is rational. qed

[@creativelad3680]

Cos 1 = 0 U outplayed ur self

[@chrishobein3742]

This is interesting. When I first heard this question, I thought tan (1°) had to be rational: Assume we have a triangle with angles 1°, 89° and 90° and the cathetes a and b. Any other triangle with the same angles would be similar to our assumed triangle, which means for any triangle with above angles, there is a factor k, such as that its cathetes are k*a and k*b. So tan (1°) would be ka/kb (assuming a is the cathete across from the 1° angle). The k's cancel out, leaving a/b. I thought there had to be intergers or at least rational numbers a,b that would construct a right triangle with a 1° angle. But I guess, after watching your proof, it is impossible to construct a right triangle with a 1° angle and rational sides a and b.

[@PhilipHaseldine]

I thought it was irrational but I wouldn't have been able to prove it

[@MrTomyCJ]

When asked for a random number, always say it's irrational, you have a 100% chance of being right!

[@jean-michelgilbert8136]

I would have converted to radians then used tan x = sin x / cos x. By the small angle approximation, which is totally valid for one degree, sin x = x, so we have a multiple of pi which is irrational divided by cos x which we don't need to care whether it's rational or irrational anymore because irrational divided by anything (except zero or infinity) is irrational.

[@ralphwang1434]

I had a different proof: If tan(1 deg) was rational, we could (with compass and straightedge) construct an angle of 1 deg, which would allow us to construct an angle of 20 deg, which is previously known to be impossible. Therefore tan(1 deg) is irrational. Kinda the same concept as what was presented but a different base case

[@heresjohny996]

You made a verbal error at 3:44 stating, "therefore sqrt(b) must be rational"

[@wannabeactuary01]

Great proof - lovely distraction and under 10 minutes!

[@SergeyKorytnik]

Now the question is whether tan(1 degree) algebraic or transcendental?

[@yinjohn23]

It is algebraic. The same reasoning works. tan(60x) is a rational function r(x) (with rational coefficients) in terms of tan(x). Now tan(1 degree) is the solution to r(x) = sqrt(3). Thus, x is algebraic.

[@siddu6003]

Applying Limits would be a great way to solve this

[@MylesCubes]

I don't fully understand limits but how could they help here?

[@Nikioko]

The question is whether there is a right-angled triangle whose legs are natural and form angles of 1° and 89°.

[@Panqueroso12]

Theorem: tan(x) is an increasing in the first quadrant. Proof of thm: the derivative of tan(x) is 1/(cos(x))^2, which is always positive, so tan(x) is increasing. Now, we know that 0 < tan(1) because tan(x) is in the first quadrant. Also, from the previous theorem, tan(1) < tan(45), and so 0 < tan(1) < 1. So if tan(1) = a/b, then a = b tan(1). But a is some integer, and so tan(1) has to be some integer, as b is an integer as well. This is not possible because there is no integer strictly between 0 and 1. Edit: above, I assumed a/b to be a simplified fraction.

[@Panqueroso12]

Now, time to watch your proof :D

[@rohit71090]

Ummm hypothetically ... say a=2 and b=4 .... Why does tan(1) have to be an integer then ? tan(1) can be simply 0.5 which can be written as 2/4 or 1/2.

[@Panqueroso12]

@@rohit71090 You missed the point of a proof by contradiction: IF tan(1) is rational, THEN tan(1) has to be an integer, according to my argument ;) I am not claiming that tan(1) is an integer as a result

[@amazuri3069]

As someone already said, your proof is flawed. Specifically, "So if tan(1) = a/b, then a = b tan(1). But a is some integer, and so tan(1) has to be some integer, as b is an integer as well." To see why this isn't true, we can just use a simple example. Let's say: tan(1) = 0.5 a = 1 b = 2 Now we can rewrite your equations with these values: 0.5 = 1/2 1 = 2*0.5 As you can see, just because our product of the multiplication is an integer, doesn't mean every part of the equation is an integer.

[@Panqueroso12]

Also, it is important in my proof that a and b are coprimes, sorry for not mentioning ;)

[@Halega101]

I'd probably right "Nope" and move onto the next question that I most likely wouldn't solve WITH a calculator...

[@gustavoabate6242]

P.T.: sqrt(69) Me: sqrt(nice)

[@kdno421]

@5:47 says doesnt want to go off on a tangent, then comes back to the tangent in question.

[@조연우-m1m]

That "Q.E.D." made me laugh 🤣

[@vishalmishra3046]

*More General Theorem* All rational numbers are Algebraic numbers and No transcendental number is rational number. All 6 trigonometric functions (sin, cos, tan and their inverse) are exact-value computable to an algebraic number if the angle is an integer multiple of 3 degrees. For all other integer multiples of 1 degree, the result is transcendental and not exact value computable (no real radical exact-value expression). So, tan 1 degree is not exact-value computable to a real-radical expression like tan 3 or 6 or 9 degrees. So, tan 1 degree is transcendental and therefore cannot be rational.

[@vishalmishra3046]

This can be deeply generalized - e.g. tan π/7 and tan π/9 are both transcendental but for any integer n, tan of n π/D is algebraic and exact-value computable if D = 2^32 - 1 = 4,294,967,295, which I think is pretty awesome if you ever tried to calculate its exact value expression for sin, cos or tan 2π/D (n=2).

[@DiamondSaberYT]

Very cool one!

[@Iomhar]

Skip the first 6 minutes of the video because it has nothing to do with the problem in question.

[@Stephen_The_Waxing_Lyricist]

I disagree, as the first 6 minutes includes the proof of why root 3 is irrational, which is needed for the final proof

[@universalphilosophy8081]

So you are telling that first 6 mins is irrational ? 😂😂😂

[@zzzaphod8507]

@@universalphilosophy8081 At least the first sqrt(35) minutes, anyway

[@thomaskolar90]

It isn't, it provides a result tyat is used later on (and also provides a simpler example for a proof of irrationality by contradiction, which is why it's good that it's first)

[@DavidDavid-yp3ln]

I think the first 6 minutes provide a great context for the problem and the solution. It's not only about Maths but about logic and reasoning.

[@madi112233]

Proof by induction and contradiction. Brilliant

[@DaneBrooke]

Sine and cosine of 1 degree are algebraic, but contain surds; Possibly the surds cancel when dividing sine by cosine. Watch this gripping video to find out!

[@lapaget1]

Fortunately, we know that tan(45°) equals 1 which is an integer and a rational. It then shows that tan(44°)=tan(45°-1°)=(tan(45°)-tan(1°)/(1+tan(45°).tan(1°)) is also a rational, etc., down to tan(1°) is a rational, which is the initial assumption. But as tan(15°)=2-sqrt(3) which is an irrational, it contradicts tan(15°) being a rational. So tan(14°)=tan(15°-1°) is irrational, down to tan(1°) is irrational after all.

[@lemontree-n7i]

0:00 guys video starts at 5:55

[@patentpendulum]

He was sentenced to Math.

[@nilusingh377]

SOLUTION 1 We can use induction here. Let prove that tanx is always rational for all x. The basic step is to prove that tan1 is rational. Here, let us suppose that tan1 is rational. Let tanx be rational. We prove that tan(x+1) is rational. Also, tan(x+1)= tanx+tan1/1-tanxtan1 which is in rational/rational form. This gives that tan(x+1) is rational. Hence from principal of mathematical induction, tanx is always rational. Now, consider tan30=1/√3, which is indeed an irrational Number. Hence a contradiction arises. This is because of our supposition that tan1 is rational. Hence we conclude that tan1 is irrational. SOLUTION 2 We know that tanx=x for small angles(in radians). 1degree=π/180 Tan1=tanπ/180=π/180. Since π is irrational, hence π/180 which is tan1 is also irrational.

[@llchan]

@MindYourDecisions This is Larry Chan who emailed you this problem. Thanks for taking it up. By the way, looking at the comments, some people seem to be confused because tan 45 is rational and they misunderstood your argument and thought you've also proved that tan 45 is irrational. In the original Japanese video (link in your description), the math teacher doubles the angle instead of adding 1 degree at a time, resulting in the following chain: tan 1 rational => tan 2 rational => tan 4 rational => tan 8 rational => tan 16 rational => tan 32 rational => tan 64 rational => tan (64-4)=tan 60 rational. This is a contradiction because tan 60=sqrt(3). That may clear up some of the confusions that people have.

[@7636kei]

Oof, looks like the solution given in the original video (as you described, anyway) was similar to what I came up with: 1/ notice that if tan(a) and tan(b) are rational, due to how the tangent of sum/difference of angle formula works, tan(a+b) _and_ tan (a-b) are bound to be rational too 2/ assuming tan(1°) _is_ rational, the chain of consequences would be: if tan(1°) is rational -> tan(1° + 1°) would be rational -> tan(2° + 2°) would be rational -> tan(4° + 4°) would be rational -> tan(8° + 8°) would be rational -> tan(16° + 16°) would be rational -> tan(32° - 2°) would be rational -> sqrt(3)/3 would be rational (oof!)

[@davidhowe6905]

Thanks for suggesting the problem!

[@cannot-handle-handles]

Great comment; it should be pinned! I also thought it would be more elegant to either double the angle or argue that tan(5°) = tan(2° + 3°), tan(10°) = tan(5° + 5°), and so on.

[@randomdude9135]

I also proved this by doubling the argument!!! Except I stopped at tan(32-2) to arrive at a contradiction 😊

[@ercfis]

For someone interested: You cannote use this solution to prove tan 45 is irrational, cause (irrational*irrational) and irrational + irrational can be rational numbers!

[@markotrieste]

I went for the statistical proof. We know that irrational numbers are an infinity order more than rationals; degrees are an arbitrary choice; tangent is a transcendent function. All this makes one wonder, what are the odds that I picked a rational number? 😂

[@thecatofnineswords]

At 3:49, how is this true? How can you say there are an odd or even number of factors of p on either side?

[@PhilipHaseldine]

squares need 2 of the same thing multiplied together. 2 is an even number.

[@IsYitzach]

You could have stopped at tan(15 degrees). Many people are saying to stop at tan(30 degrees)=sqrt(3)/3. But tan(15 degrees)=2-sqrt(3) happens sooner. This also avoids tan(45 degrees) which is 1 and rational.

[@zmaj12321]

There is no reason why tan(45) needs to be avoided.

[@denelson83]

tan (π/180) is irrational because π/180 is not a constructible angle. Only fractions of a full turn where the prime factorization of the denominator includes only powers of two and distinct Fermat primes are constructible. And rational numbers are a proper subset of constructible numbers.

[@EvidLekan]

It is evident tangent of 1° is irrational if one understands the impossibility of certain constructions using only a straightedge and compass. Constructing a 1° angle is impossible with these tools, as certain angles are not constructible (see Gauss-Wantzel theorem). Consequently, their trigonometric ratios, such as sine, cosine, and tangent, are also non-constructible. Since all rational numbers are constructible, proving that a 1° angle is not constructible (spoiler: it is not, as it would require the trisection of any rational angle) leads to the conclusion that the tangent of 1° is irrational. A background in undergraduate-level geometry is recommended to fully understand this argument.

[@lexiu6036]

1. check for tan of 2, 4, 8, 16, and 32, they all would be rational (assuming tan(1) is rational) 2. calculate the value of tan(32-2), it also should be rational (again, according to the assumption) 3. tan(30) is 1/sqrt(3) which is known to be irrational, which makes a contradiction. qed.

[@rosiefay7283]

If tan 1° were rational, it would be possible to construct a regular 360-gon by classical means. But even the regular 9-gon is not so constructible. So tan 1° is irrational.

[@vladyslavkryvoruchko]

Great exercise

[@littlestonesadventures3939]

Proof by contradiction and induction❤

[@antoniopedrofalcaolopesmor6095]

1 = tg(45) = tg(44+1) = (tg44+tg1)/(1-tg44*tg1), then it must be that tg44+tg1 = 1-tg44*tg1 I never would have antecipated that this equality holds. That's quite surprising for me.

[@aba_dab_o]

5:48 - Was anyone sentenced to death for too many math puns? 😏

[@tobyfitzpatrick3914]

Video starts at 5:55

[@Moofinxd]

This can also be solved with maclauren expansion of sin(1)/cos(1)

[@apoorvvyas52]

Very beautiful question and very elegant solution too.

[@amankamath6076]

Could we use approximation? Say 1° = pi/180 rad And we know tan(1) = sin(1)/cos(1) Then sin(pi/180) ≈ pi/180 & cos (pi/180) ≈ 1 Then tan(pi/180) ≈ pi/180 which isn't rational.

[@professorhaystacks6606]

Are we allowed to accept that tan(60 degrees) = sqrt(3) without proving it? We'd need some justification from the class's set-up for that. Maybe we need to start with Euclid's postulates.

[@mattheww4862]

This is a nice systematic proof. A less elegant/rigorous idea, could we express tan x as a taylor series expansion, replace x with pi/180 (since 1 degree is pi/180 radians)? The resulting series has powers of pi. Given pi is transcendental then presumbly this infinite sum should be irrational.

[@Anonymous-df8it]

Replace all instances of pi/180 with pi/4...

[@satrajitghosh8162]

Let tan 1 ° be rational. Hereby tan 3 ° = (3 tan 1° - tan ^ 3 (1°)) /( 1 - 3 tan ^2 (1°)) would be rational Extension of this argument gives tan (9 °) would also be rational Hence tan (18°) would also be rational. But sin (18°) = (√ 5 - 1)/4 Hereby cos (18°) = √ ( 10 - 2 √ 5)/4 tan (18°) = ( √ 5 - 1) /√ ( 10 - 2 √ 5) this is irrational

[@mayukhintesarislam306]

friggin elegant!!

[@ああ-j1w7c]

京大さんにょっす！w🐮✋

[@malvoliosf]

In the same way, you can demonstrate the tan(n°) is irrational when n is any factor of 60.