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"!_
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
@@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.
@@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
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.
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.
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.
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)
*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)
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.
@@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
@@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
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.
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.
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.
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.
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"
I could deduce this, Im surprised about myself! I have a doubt... down the line, tan45° gives you a rational number... how is this justified? (Pls dont judge... I am not that good at math, but i love math)
There is no need to "justify" that tan(45) is rational. The assumption that tan(1) is rational may very well lead to millions of true statements, like that tan(45) is rational. But the moment that the assumption that tan(1) is rational leads to something that is untrue, you have proven that tan(1) cannot be rational.
I think the problem lies with the formulation. If tan 1 WAS rational, then everything down the line would be rational. Since tan 60 is irational (because we defined it to be that way), that means tan 1 can't be rational. However, this does not imply that every value HAS to be irrational. If everything with property A has to have property B as well, does not mean that something with property B MUST have property A as well.
Cause you cant add the 1 degree cause its irrational. So from tan(45) rational you cant say tan(46) rational since tan(1) isnt rational). We do know now that tan(x) is irrational if x divides 60
From your statement we can conclude that tan(1°) is irrational doesn't imply that tan(k°) is also irrational. But if it were rational it would imply that tan(k°) is rational.
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.
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.
@@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.
@@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
@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.
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!)
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.
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.
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)
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 .
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.
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.
@@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
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.
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
I went for a completely different route and expanded out tan as sin/cos and then sin and cos in their full exponential form, then after some algebra showed that the result is a complex fraction. I *think* that works? Do I get some points at least?
The proof should perhaps also address that the denominator does not become 0 at any step, i.e. tan(alpha)*tan(alpha +1) != 1 for any natural number alpha. Otherwise the fraction would not be a rational number anymore.
@@Anonymous-df8it 🤦♂️my calculator was in radians. I knew that property for sine and cosine, it seems obvious to me now that it would work for sin/cos as well.
@@Pasclesrm Why are you using the facepalm emoji in response to me? Did I make a fool out of myself? Also, I've noticed that you've deleted your comment...
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.
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.
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....
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.
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.
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
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
If tan(1°) was rational, you could use the ratio to construct a regular 360-gon. 360 is not of the form 2^a × (2^b + 1) for (2^b + 1) prime, but only such regular polygons are constructible as shownw by Gauß. Hence tan(1°) is not rational.
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
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.
@@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.
OK. Here's an extended question to which I don't know the answer. Is tan (a/b °) where a and b are integers ever rational except for the trivial cases of 0 and multiples of 45? I'd love to see someone answer that. EDIT: The Wikipedia entry on Niven's Theorem says that in fact tan(a/b °) is rational only for multiples of 45 degrees.
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.
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.
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.
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.
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.
@@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
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.
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? 😂
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!
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.
For someone interested: You cannote use this solution to prove tan 45 is irrational, cause (irrational*irrational) and irrational + irrational can be rational numbers!
Would you be able to say tan 1° is irrational because tan 1° = sin 1° / cos 1° and since neither of those are rational, then following the formula that a/b must be integers, tan 1° is irrational?
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.
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.
I'm thinking of using half and 1/3 angle formulas and get an algebraic expression for tan(1 deg) and then using a proof similar to proving sqrt(2) is irrational.
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
This reasoning is somewhat flaky IMO. Before reaching 60° there is 45°, whose tangent is equal to one (1). And then one can break 45° down as many times as one wishes until 1° is reached: 45° = 22° + 23°, 22° = 11° + 11°, 11° = 10° + 1°, etc. Thus, any integer arc (in degrees) between 1° (inclusive) and 45° (ditto) would have a rational tangent. Forgive me if I'm mistaken. I never excelled at this math level.
Thats not a problem. If you do it with 45°, all that you get is "if tan 1 is rational, then tan 1 is rational". Basicly, you went nowhere. In a proof by contradiction, you need to go somewhere that disproves the initial statement, but if, halfway trough, you end up in a step that "agrees" with the original statement, thats not a problem. It simply proves nothing.
@@alex2005z I see your point. I was trying to "reverse engineer" what was done in the video. But, regardless of whatever I did, I still think his argument is flawed or not too well presented. It is probably I, though, who am wrong.
I think you could just say that tan 1 is equal to sin 1 over cos 1 and for tan 1 to be rational, sin 1 and cos 1 must be integers however sin and cos are only integers when they are multiples of 90 and 1 is not a multiple of 90 therefore tan 1 is not rational.
Sentence "if k is not a perfect square, then square root of k is irrational" is false. If k=a^2/b^2, then square root of k=a/b. Or am I missing something?
*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.
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).
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.
tan^2(1)=1/(cos^2(1))-1=2/(cos(2)+1)-1 means if tan^2(1) was rational cos(2) would be too yet cos(15*2)=sqrt(3)/2 can be expressed as a polynomial with integer coefficients in cos(2) (aka chebyshev polynomial T_15), so cos(2) is irrational, so tan^2(1) is also irrational
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
Very standard proof by contradiction. Suppose that tan theta = rational. Then tan(n*theta) is rational for any integer n. But n=30 or even 60 yields a contradiction.
But tan(45°)=1, which is rational, and we can express it as tan(44° + 1°) and so on to tan(1°), can't we? Doesn't it proves that tan(1°) is rational? Please explain
If you add, subtract, multiply, or divide two known rational numbers, the result will be rational. The converse is not true. Just because the result of an operation is rational does not mean that the inputs were rational. For example, sqrt(2)sqrt(2) =2. Also, pi + (1-pi) = 1. We know that tan(45°) is rational, so if we could show that either tan(44°) or tan(1°) was rational, we could prove the other was also. In this case, they are both irrational.
No it does not. If a,b are rational, then c = (a+b)/(1-ab) is rational too. But if c is rational, that does not mean that a and b are rational. Just like b being rational means that b² is rational, but b² being rational does not mean that b is rational.
The contradiction is caused by "proving" something that is wrong (in this case that tan(60°) is rational. That tan(45°) is rational even though the shown "proof" isn't is not relevant to the argument. If assuming that tan(45°) is rational would lead to a similar contradiction there would be a real paradox.
Tan(45) is rational though. The fact is, two irrationals can combine to form a rational. (10 - pi) + (10 + pi) is rational. 2 × sqrt(2) / sqrt(2) is rational.
the logical implication (with the assumption that tan 1 is rational) that this proof is showing is (tan 1 is rational) => (tan 1 is rational AND tan 2 is rational AND ... AND tan 45 is rational AND ... AND tan 60 is rational). The second predicate is false when any of the tangents are not rational, for example (tan 60 is not rational), which means (tan 1 is rational) implies a falsehood, i.e. a contradiction.
Not hard if you know your trig formula and you know how to reason. That'd be decent entrance exam after high school. Let's suppose tan 1 is rat. Half tan formula: tan 2x = 2t/(1-t^2) where t = tan x. So tan 2 is rat. So tan 4 is rat. ... so tan 32 is rat. Besides: tan 32 = tan (30+2) Tan of a sum formula: tan (a+b) = (ta+tb)/(1-ta.tb) Therefore t32 = (t30+t2)/(1-t30.t2) t30=1/sqrt(3): irrational. Multiplying by the conjugate: t32 = (t30+t2)(1+t30.t2)/(1-t30^2.t2^2) The denominator is rational so let's check the numerator: t30+t2+t30^2.t2+t30.t2^2 = t30.(1+t2^2) + rational number is irrational because (1+t2^2) is rat. and t30 is not. therefore t32 is irrational, which contradicts our hypothesis, therefore t1 is irrational.
Isn't it enough to show the following? (Seems simpler, more straightforward):: since tan(x)=sin(x)/cos(x)... we know that sin(1 deg) is not an integer and cos(1 deg) is not an integer, thus tan(1 deg) cannot be rational. We know that Sine(x) and cosine(x) only have integer values for x=0,90,180,270,360... (x in degrees) by the unit circle definition of sine and cosine.
@@faustobarbuto Oh yeah, you're right. In fact, pi / pi = 1 is rational as well.... I must've been half-asleep when I posted this, haha. Thanks for pointing out the logical error on my part!
@@jamesdelapena5648 Actually, I made a mistake. I forgot to add the word "rational" to the end of my reply. I took for granted it was implicit, sorry. Now it is correct.
Well, integer+integer over (1-integer *integer) could be in the first case (tan1deg + tan1deg….) 2 over 1-1 if tan1deg is equal to 1. (We know tan 1deg is not equal to 1) I just want to pinpoint, that as a general rule you can not state, that this must be an integer, whatewer integers You are chosing… So I think it is just not true in general.
![Pythagoras Would Be Proud: High School Students' New Proof of the Pythagorean Theorem [TRIGONOMETRY]](http://i.ytimg.com/vi/p6j2nZKwf20/mqdefault.jpg)
He was sentenced to death? That doesn't seem like the most rational way to handle the situation.
It was radical though.
I dont think that this is a natural thing to do
It was his own fault for being transcendental towards Pythagoras!
Should have rationalized his decisions ref. his discovery
Only numbers can be rational or irrational, not actions.
Allegedly, Pythagoras hated 2 things: irrational numbers and beans. The former he killed for, the latter he died for.
An irrational number of beans
Did he bake beans into a pi?
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"!_
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.
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
@@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.
@@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
@@theupson Yeah, I'm totally confused with this one.
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
Also tan(pi/4)=1🤓😱🤯 pi/4=45 dgr < 60 . Not a valid proof!!!!!
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.
He should stopped at π/12
@@Mike-H_UK Yes, but you do go though a rational point as 45 degrees (pi/4 is rational) which may confuse the issue
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.
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.
And tan 18 degrees as well. You can find out this value with a regular pentagon.
I’d say, if anything, he shouldn’t have gone past 45 degrees…
@@dennisdeng3045stopping at 45° would prove nothing, as tan(45°) is rational
@@dennisdeng3045 you don't get a contradiction if you stop there
tan(15º) = (sqrt6-sqrt2)/4 :eyes:
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)
Hey what are you? a mathematician ? Just curious 🤔.
@@HackedPC I'm a math teacher at high school 😄
*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)
@@theupsonWell, prime numbers are relatively prime to each other. A problem can hide information
@@theupson Indeed, the only thing you need is that p and q are not a power of the same number.
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.
How did you type that "π"?
sin(π/90) = 2pq/(p²+q²)
You were off by a factor of 2, though it doesn't affect the proof.
@@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 ;)
@@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
@@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
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.
Bro was exiled to a unsurvivable Island and they call God killed him lol
5:43 I think you meant to say: "I'm not sure, I will leave it to the _historians_ to decide."
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.
3:45, “…must be *irrational”. I believe you misspoke there. Good video!
"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!
After six minutes of filler...
Very smart proof by contradiction.
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.
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.
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.
I instinctively thought the answer was no but don't ask me why I thought that. This was really interesting, thanks :)
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"
If he was not real, is he just...
imaginary?
I could deduce this, Im surprised about myself!
I have a doubt... down the line, tan45° gives you a rational number... how is this justified?
(Pls dont judge... I am not that good at math, but i love math)
There is no need to "justify" that tan(45) is rational. The assumption that tan(1) is rational may very well lead to millions of true statements, like that tan(45) is rational. But the moment that the assumption that tan(1) is rational leads to something that is untrue, you have proven that tan(1) cannot be rational.
I think the problem lies with the formulation. If tan 1 WAS rational, then everything down the line would be rational. Since tan 60 is irational (because we defined it to be that way), that means tan 1 can't be rational. However, this does not imply that every value HAS to be irrational. If everything with property A has to have property B as well, does not mean that something with property B MUST have property A as well.
Cause you cant add the 1 degree cause its irrational. So from tan(45) rational you cant say tan(46) rational since tan(1) isnt rational). We do know now that tan(x) is irrational if x divides 60
From your statement we can conclude that tan(1°) is irrational doesn't imply that tan(k°) is also irrational. But if it were rational it would imply that tan(k°) is rational.
if you try applying the same steps of the proof to tan(45), you just get that tan(90), ... will be rational, which is true.
Can someone please explain the contradiction at 3:42? I don't get the logic.
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.
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.
@@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.
@@martinmonath9541
Thanks.
@@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
@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.
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!)
Thanks for suggesting the problem!
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.
I also proved this by doubling the argument!!! Except I stopped at tan(32-2) to arrive at a contradiction 😊
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.
Skip the first 6 minutes of the video because it has nothing to do with the problem in question.
I disagree, as the first 6 minutes includes the proof of why root 3 is irrational, which is needed for the final proof
So you are telling that first 6 mins is irrational ? 😂😂😂
@@universalphilosophy8081 At least the first sqrt(35) minutes, anyway
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)
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.
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 .
Cool!
My approach was, assume tan(1) is rational, then 1 degree is constructible, then any integer valued angle is constructible, contradiction.
trivial. small angle approximation. sin 1 = 0. 0/cos 1 = 0. 0 is rational. hence, tan 1 = 0 is rational. qed
Cos 1 = 0
U outplayed ur self
After watching ur every video my heart says ," I love you ". ❤❤❤
The way you pulled off that “weave” is legendary!
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.
tan(30) is irrational. But tan(45) is rational.
You only need one irregular to disprove something. But not even millions of positive reinforcement is enough to prove something.
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.
@@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.
@@GenUrobutcher That's something well said man!
@@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
Thanks for the pun at 5:50. Made me smile 😊
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.
You made a verbal error at 3:44 stating, "therefore sqrt(b) must be rational"
It's enough to go up to 30* (tan 30* = 1/√3 is irrational).
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
The question is whether there is a right-angled triangle whose legs are natural and form angles of 1° and 89°.
I somehow totally misread the question assuming it was asking if tan(1°) could be expressed with radicals and ended up proving it.
It's fun to see how Presh is slowly sliding into a bit more personal and humoristic approach in his videos.
I went for a completely different route and expanded out tan as sin/cos and then sin and cos in their full exponential form, then after some algebra showed that the result is a complex fraction. I *think* that works? Do I get some points at least?
The proof should perhaps also address that the denominator does not become 0 at any step, i.e. tan(alpha)*tan(alpha +1) != 1 for any natural number alpha. Otherwise the fraction would not be a rational number anymore.
@@Pasclesrm tan(1°)*tan(89°)?
@@Anonymous-df8it 🤦♂️my calculator was in radians. I knew that property for sine and cosine, it seems obvious to me now that it would work for sin/cos as well.
@@Pasclesrm Why are you using the facepalm emoji in response to me? Did I make a fool out of myself? Also, I've noticed that you've deleted your comment...
@@Anonymous-df8it I am facepalming my own stupidity
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.
Replace all instances of pi/180 with pi/4...
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.
this isn't a right triangle, and even if it was, the angles wouldn't be consistent for all values of x
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! 😂
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....
huh
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.
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)🤣
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.
Excellent question and excellent solution
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
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
If tan(1°) was rational, you could use the ratio to construct a regular 360-gon. 360 is not of the form 2^a × (2^b + 1) for (2^b + 1) prime, but only such regular polygons are constructible as shownw by Gauß. Hence tan(1°) is not rational.
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
Hmmm maybe I leaned on this in my belief that it was irrational before he showed the proof, not sure, thanks....
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.
@@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.
@@cheesetasty1646 That doesn't relate to my point.
What about tangent of 45 degrees? I think this logic is faulty.
OK. Here's an extended question to which I don't know the answer. Is tan (a/b °) where a and b are integers ever rational except for the trivial cases of 0 and multiples of 45? I'd love to see someone answer that.
EDIT: The Wikipedia entry on Niven's Theorem says that in fact tan(a/b °) is rational only for multiples of 45 degrees.
I think that Presh has personal hate on Pythagoras...
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.
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.
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.
I thought it was irrational but I wouldn't have been able to prove it
8:36 if we continue to tan(45)=1 that is rational
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.
You cannote use this solution to prove tan 45 is irrational, cause (irrational*irrational) and irrational + irrational can be rational numbers!
I don't think that this resolves or proves that tan 1° is irrational. Isn't tan 45° rational? The explanation is not completely proven.
yes i agree
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.
Now, time to watch your proof :D
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.
@@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
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.
Also, it is important in my proof that a and b are coprimes, sorry for not mentioning ;)
Kyoto unversity is second most difficult in Japan.The unique and interesting question is very famous in Japan.
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? 😂
well( 16 = 53 -37) and 15 =45- 30. 16-15 = 1. 53 ,37 , 90 is angle of 3,4,5 triangle. you could find value of tan 1. its pretty short
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!
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.
For someone interested:
You cannote use this solution to prove tan 45 is irrational, cause (irrational*irrational) and irrational + irrational can be rational numbers!
is this basically a method of descent or ascent with proof by contradiction?
Without watching, I would guess that 1 degree is a fraction of pi, which is irrational. So the tan of it may be also irrational.
We don't have to make it to 60°. tan 45° = 1, which is a rational number.
Would you be able to say tan 1° is irrational because tan 1° = sin 1° / cos 1° and since neither of those are rational, then following the formula that a/b must be integers, tan 1° is irrational?
You have to prove that sin1 or cos1 is irrational.
4:04 the square root of 69 is "8 something".
Sure ok, but in fight who would win? "mind your decisions" or "3blue1brown"?
what about tan(1 radian) instead of degrees? pretty sure it’s irrational but how do we go around proving it?
Could a continuous function be rational for some inputs and irrational for others?
In the same way, you can demonstrate the tan(n°) is irrational when n is any factor of 60.
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.
Is there a way to proof that the ratio of 2 irrational numbers is always irrational?
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.
I'd probably right "Nope" and move onto the next question that I most likely wouldn't solve WITH a calculator...
I'm thinking of using half and 1/3 angle formulas and get an algebraic expression for tan(1 deg) and then using a proof similar to proving sqrt(2) is irrational.
That's how Pythagoras got away with sentencing his opposition to death -- he convinced the entire world he has never real to begin with.
This can also be solved with maclauren expansion of sin(1)/cos(1)
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
This reasoning is somewhat flaky IMO. Before reaching 60° there is 45°, whose tangent is equal to one (1). And then one can break 45° down as many times as one wishes until 1° is reached: 45° = 22° + 23°, 22° = 11° + 11°, 11° = 10° + 1°, etc. Thus, any integer arc (in degrees) between 1° (inclusive) and 45° (ditto) would have a rational tangent.
Forgive me if I'm mistaken. I never excelled at this math level.
Thats not a problem. If you do it with 45°, all that you get is "if tan 1 is rational, then tan 1 is rational". Basicly, you went nowhere. In a proof by contradiction, you need to go somewhere that disproves the initial statement, but if, halfway trough, you end up in a step that "agrees" with the original statement, thats not a problem. It simply proves nothing.
@@alex2005z I see your point. I was trying to "reverse engineer" what was done in the video. But, regardless of whatever I did, I still think his argument is flawed or not too well presented. It is probably I, though, who am wrong.
@@faustobarbuto its not really poorly explained, it just assumes you already know what a proof by contradiction is.
Can we use Taylor's series expansion , it give pi terms which indicates irrational
I think you could just say that tan 1 is equal to sin 1 over cos 1 and for tan 1 to be rational, sin 1 and cos 1 must be integers however sin and cos are only integers when they are multiples of 90 and 1 is not a multiple of 90 therefore tan 1 is not rational.
Smart
Actually this is kinda bad cuz you gonna prove both cosine and sine are irrational after some thought lol
Sentence "if k is not a perfect square, then square root of k is irrational" is false. If k=a^2/b^2, then square root of k=a/b. Or am I missing something?
*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.
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).
@5:47 says doesnt want to go off on a tangent, then comes back to the tangent in question.
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.
bro this is so beautifull i am crying😍😍 i would never solve this tho
3:56 Ahhh! The infamous words from my old calculus text: "The proof is left as an exercise for the student."
I'm an ENGINEER! Gimme "Simpson's Rule," gimme "Newton's Method," gimme a number, accurate to within a few percent!
For a different challenge, show that the square of tan 1° is irrational.
It isn’t
tan^2(1)=1/(cos^2(1))-1=2/(cos(2)+1)-1 means if tan^2(1) was rational cos(2) would be too
yet cos(15*2)=sqrt(3)/2 can be expressed as a polynomial with integer coefficients in cos(2) (aka chebyshev polynomial T_15), so cos(2) is irrational, so tan^2(1) is also irrational
2=1+1, 4=2+2, 8=4+4, 16=8+8, 32=16+16, 30=32+(-2), but tan(30) is irrational is faster.
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
Very standard proof by contradiction. Suppose that tan theta = rational. Then tan(n*theta) is rational for any integer n. But n=30 or even 60 yields a contradiction.
But tan(45°)=1, which is rational, and we can express it as tan(44° + 1°) and so on to tan(1°), can't we? Doesn't it proves that tan(1°) is rational? Please explain
No. One contradiction is enough to prove the assumption false.
If you add, subtract, multiply, or divide two known rational numbers, the result will be rational. The converse is not true. Just because the result of an operation is rational does not mean that the inputs were rational. For example, sqrt(2)sqrt(2) =2. Also, pi + (1-pi) = 1. We know that tan(45°) is rational, so if we could show that either tan(44°) or tan(1°) was rational, we could prove the other was also. In this case, they are both irrational.
No it does not. If a,b are rational, then c = (a+b)/(1-ab) is rational too. But if c is rational, that does not mean that a and b are rational.
Just like b being rational means that b² is rational, but b² being rational does not mean that b is rational.
The contradiction is caused by "proving" something that is wrong (in this case that tan(60°) is rational. That tan(45°) is rational even though the shown "proof" isn't is not relevant to the argument. If assuming that tan(45°) is rational would lead to a similar contradiction there would be a real paradox.
Read zsoltnagy5654 comment
Tan(45) is rational though.
The fact is, two irrationals can combine to form a rational.
(10 - pi) + (10 + pi) is rational.
2 × sqrt(2) / sqrt(2) is rational.
the logical implication (with the assumption that tan 1 is rational) that this proof is showing is
(tan 1 is rational) => (tan 1 is rational AND tan 2 is rational AND ... AND tan 45 is rational AND ... AND tan 60 is rational).
The second predicate is false when any of the tangents are not rational, for example (tan 60 is not rational), which means (tan 1 is rational) implies a falsehood, i.e. a contradiction.
Not hard if you know your trig formula and you know how to reason.
That'd be decent entrance exam after high school.
Let's suppose tan 1 is rat.
Half tan formula: tan 2x = 2t/(1-t^2) where t = tan x.
So tan 2 is rat.
So tan 4 is rat.
...
so tan 32 is rat.
Besides: tan 32 = tan (30+2)
Tan of a sum formula: tan (a+b) = (ta+tb)/(1-ta.tb)
Therefore
t32 = (t30+t2)/(1-t30.t2)
t30=1/sqrt(3): irrational.
Multiplying by the conjugate:
t32 = (t30+t2)(1+t30.t2)/(1-t30^2.t2^2)
The denominator is rational so let's check the numerator:
t30+t2+t30^2.t2+t30.t2^2
= t30.(1+t2^2) + rational number
is irrational because (1+t2^2) is rat. and t30 is not.
therefore t32 is irrational, which contradicts our hypothesis,
therefore t1 is irrational.
Isn't it enough to show the following? (Seems simpler, more straightforward)::
since tan(x)=sin(x)/cos(x)... we know that sin(1 deg) is not an integer and cos(1 deg) is not an integer, thus tan(1 deg) cannot be rational. We know that Sine(x) and cosine(x) only have integer values for x=0,90,180,270,360... (x in degrees) by the unit circle definition of sine and cosine.
Not sure if I really got your point, but 3/8 and 7/8 are not integers, but (3/8)/(7/8) = 3/7 is rational.
@@faustobarbuto Oh yeah, you're right. In fact, pi / pi = 1 is rational as well.... I must've been half-asleep when I posted this, haha. Thanks for pointing out the logical error on my part!
@@jamesdelapena5648 Actually, I made a mistake. I forgot to add the word "rational" to the end of my reply. I took for granted it was implicit, sorry. Now it is correct.
Well, integer+integer over (1-integer *integer) could be in the first case (tan1deg + tan1deg….) 2 over 1-1 if tan1deg is equal to 1. (We know tan 1deg is not equal to 1) I just want to pinpoint, that as a general rule you can not state, that this must be an integer, whatewer integers You are chosing… So I think it is just not true in general.
Great proof - lovely distraction and under 10 minutes!