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.
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)
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)
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.
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
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.
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.
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"
Lots of comments on how 45 degrees has a rational tangent. That's irrelevant, because it doesn't matter if we pass rational tangents along the way. Since it was shown that if tan 1 is rational so is tan2, when you get to tan 44 just use tan2 to skip right over tan45. This shows that even if there is a rational tangent along the way, it doesn't matter, because all you need is one contradiction to prove it wrong. Although, a quicker and neater way would get you to tan(15) where we have 2 - sqrt(3) and that cannot be rational. Done. And 45 couldn't even an issue here.
@@soundsoflife9549 You don't need to do this; whatever irrational example you use, there could be rational examples lower down (e.g. you could use tan 240 deg), but the rationality of tan 1 deg requires ALL integer angles to have rational tangents.
@@soundsoflife9549 @soundsoflife9549 I'm not attempting a proof in the comments, but it can be shown that if you divide a circle into any number of equal angle units (360, 400, 9384747, 7) - and let's call angles that are multiples of an even division of a circle "rational angles" - the only times the trig functions for a rational angle are also rational are (degrees): * sin(0), cos(0), tan(0) * sin(30) * tan(45) * cos(60) * sin(90), cos(90) as well as all the related angles elsewhere in the unit circle. Aside from these classes, all rational angles have irrational trig functions, and all rational trig functions correspond to irrational angles. (Note that radians are not equal divisions of a circle. Radians in terms of pi are, though. So 3/4 radians is not rational, but 3pi/4 radians is rational.)
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.
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.
@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.
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.
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....
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.
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.
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
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.
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.
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 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? 😂
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.
Aristotle seems to doubt that there was either a Pythagoras, or that the sect using his name were actually followers. He says "so-called Pythagoreans" a lot. I am not convinced that any other place besides the Greek milieu invented mathematical *proof*. By contrast, the pattern of triples, etc. was definitely known in many places, including India and China. As for the topic today, it looks like a great place for a teacher to set a simple question for testing understanding the logical aspects of mathematical induction.
I have to wonder if adding the story about Pythagoras sentencing someone to death on that entrance exam's answer would get you anything or just a few weird looks 🤣
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?
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.
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
There is no problem that tan(45°) is rational. A false assumption can lead to correct statements along the way and that is not a problem. As long as it leads eventually to a contradiction, the assumption is false. At the step where tan(45°) is rational, the next step is probably wrong because you are adding tan(1°) or tan(30°) which are not rational. If you stop at say 45°, you don’t have a contradiction so you have proved nothing one way or the other at that point.
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.
Q: Is tan 1 a rational number? (Justify your answer with a proof) => A: "I don't know", Proof: Lie detector. Did I pass the entrance exam for the logic course?
I rather suspect that this proof, or something rather like it is taught in Japanese schools, as if this was presented to somebody completely fresh in a time-limited entrance exam, it would surely prove to be a very demanding task indeed.
Nope. Let's apply the argument to tan(45°). Suppose that tan(45°) is rational. Using the tangent addition formula, this implies that tan(45° + 45°) = tan(90°) is rational. In the same way, it implies that tan(135°), tan(180°), etc. are all rational. There is no contradition here, so we have not proven that tan(45°) is irrational.
I would love an explenation to why this doesn't apply, by the same logic, to tan(45). I have seen some comments about this but I don't understand to be honest. I need the animations😂
short answer: the sum of two rationals is rational. but if the sum is rational, the summands need not be. so tan(45 degrees) doesnt prove anything and tan(30 degrees) does.
The notation in the video is misleading. It suggests "tan(45°) rational => tan(46°) rational", when what is actually used is "tan(45°) rational AND tan(1°) rational => tan(46°) rational".
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.
Yes, it is. But the point is: IF tan 1° is rational, then tan 2°, tan 3°, tan 4°... they ALL have to be rational. However, tan 60° is not rational, we already know that. That gives us the contradiction. The fact that ONE of the values (tan 45°) in that sequence is rational, does not eliminate the contradiction. They ALL have to be rational; and, demonstrably, they are not ALL rational. QED.
*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).
The notation in the video is misleading. For example, it suggests "tan(45°) rational => tan(46°) rational", when what is actually used is "tan(45°) rational AND tan(1°) rational => tan(46°) rational".
He didn't say that all tangents from 1° to 60° are irrational. He said that IF tan(1°) is rational, then tan(60°) is rational, which is a contradiction, so tan(1°) must be irrational.
My uneducated mind: Tan = sin/cos. So any tan is rational so long as sin and cos are rational. Except sqrt(2) isn't rational and is a possible value from sin and cos. So maybe this isn't so simple.
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.
Incidentally, since 60 is highly composite, you can use the same logic to prove that the tan of 2, 3, 4, 5, 6, 10, 12, 15, 20, and 30 degrees are also all irrational. Fixed :P
Can't you use the angle sum identity & the fact that the product or sum of an irrational number & rational number is irrational? Sin(45-44) = sin (1) = root(2)*[cos(44) - sin(44)] T.f sin(1) is irrational. Do the same for cos(1), divide them & you get tan(1)
You ignore the fact that your method would have passed by Tan(45°) = 1, which IS rational. Why not prove it by working out Sin(1°)/Cos(1°), or Sin(1°)/Sin(89°), which can be written as a specific formulas?
There's nothing wrong with that. If tan1 is rational instead, then tan45 must be rational. However, if tan1 is irrational, it does not guarantee tan45 is irrational. tan45 could still be rational either way. This property doesn't work both ways.
Back at school in the 70s we learned Tan=Opposite/Adjacent. Tan 1⁰ is not like 1/1 (Tan 45⁰) It would be irrational to think that 1/45th of the angle would be rational, therefore it must be irrational. QED
I don't really understand this proof. I get that Tan 60 degrees is irrational but why did we stop there? If we had gone to only Tan 45 degrees, we get a result of 1, which is a rational number and therefore we can conclude all the others are rational too, right?
we need a contradiction and it's inevitable Basically here we used prior knowledge that tan 60 is irrational. But our initial assumption (tan 1 is rational) made it rational, and it's a contradiction => tan 1 is irrational
@@trumpetbob15 Basically, for a proof by contradiction you have to find (wait for it!) a contradiction, just one. In this case, we show that the assumption tan(1) is rational leads to the result that tan(60) is rational, which it isn't, so the assumption is wrong The fact that we pass through some numbers that ARE rational doesn't matter, they are just not the contradiction I am looking for. For example, I have a theory that everyone's birthday is Jan 17. If the first 5 people I ask do in fact have that birthday, my theory might appear to be good, but as soon as I meet someone whose birthday is say Mar 03, that's a contradiction and my theory is false. All "positive" results don't matter.
@@silver6054 My birthday is not Jan 17 so I further confirm that the theory is disproven 😂 Now we just need to work out the odds of the first 5 people having Jan 17 as the birthday as I reckon the numbers will get pretty big very quickly 🤣
Why? Non-perfect squares are numbers whose square root is not an integer. So, by this definition alone, you couldn't conclude that their square root is not a rational.
@@ilmionomenonloso No non perfect squares are not numbers who square roots are not an integer, they are numbers who's square roots are irrational. They are defined that way, nothing to prove.
@@Panqueroso12 Yes I know that, I wasn't referring to the square root of prime numbers being non perfect squares. I was talking about the square root of a non perfect square.
I don't this this works. tan(1) and tan(29) gets tan(30) which is 1/sqrt(3)--irrational. Apparently q.e.d. but continue, tan(44) and tan(1) gets us tan(45) which is 1 and rational.
Read up on proof by contradiction. The assumption is assumed to be true, and true is allowed to imply true. But true is not allowed to imply false. If you derive a false statement, you can conclude that the assumption was false. If you derive a true statement (like tan(45 deg) is rational) this doesn't tell you anything. You really have to think carefully about this.
but tan 45 is rational, which means somewhere in the reiteration process, the interaction between irrational numbers will create a rational number. So I dont think its a good proof. Is there a more convincing proof?
It's possible for irrational numbers to create rational numbers. If tan44 and tan1 are both rational, it means that tan45 must be rational (like in the video). However, if tan44 and tan1 are both irrational, it doesn't guarantee that tan45 must be irrational.
You need to think more carefully about how proof by contradiction works. If tan(1°) is rational, then tan(60°) MUST be rational. This is a contradiction, so it's complete and undeniable proof that tan(1°) is irrational. The part where you maybe get confused is that a false statement is allowed to imply a true statement. For example tan(1°) is rational implies that tan(45°) is rational, but this is not a contradiction. It's only a contradiction when you derive a false statement.
That's not a problem. The assumption in a proof by contradiction can lead to any number of true statements that you want, (this doesn't say anything about the veracity of the assumption) but if it leads to a single false statement (like "tan(60°) is rational") then it is a false assumption. That's how proof by contradiction works.
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.
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.
Allegedly, Pythagoras hated 2 things: irrational numbers and beans. The former he killed for, the latter he died for.
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.
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.
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.
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…
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
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.
Very smart proof by contradiction.
5:43 I think you meant to say: "I'm not sure, I will leave it to the _historians_ to decide."
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.
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! 😂
It's fun to see how Presh is slowly sliding into a bit more personal and humoristic approach in his videos.
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"
The way you pulled off that “weave” is legendary!
Lots of comments on how 45 degrees has a rational tangent. That's irrelevant, because it doesn't matter if we pass rational tangents along the way. Since it was shown that if tan 1 is rational so is tan2, when you get to tan 44 just use tan2 to skip right over tan45. This shows that even if there is a rational tangent along the way, it doesn't matter, because all you need is one contradiction to prove it wrong.
Although, a quicker and neater way would get you to tan(15) where we have 2 - sqrt(3) and that cannot be rational. Done. And 45 couldn't even an issue here.
If irrationals are between rationals, how can we prove the lower ones are irrational?
This clarification is very helpful, thanks!
@@soundsoflife9549 You don't need to do this; whatever irrational example you use, there could be rational examples lower down (e.g. you could use tan 240 deg), but the rationality of tan 1 deg requires ALL integer angles to have rational tangents.
@@soundsoflife9549 @soundsoflife9549 I'm not attempting a proof in the comments, but it can be shown that if you divide a circle into any number of equal angle units (360, 400, 9384747, 7) - and let's call angles that are multiples of an even division of a circle "rational angles" - the only times the trig functions for a rational angle are also rational are (degrees):
* sin(0), cos(0), tan(0)
* sin(30)
* tan(45)
* cos(60)
* sin(90), cos(90)
as well as all the related angles elsewhere in the unit circle.
Aside from these classes, all rational angles have irrational trig functions, and all rational trig functions correspond to irrational angles.
(Note that radians are not equal divisions of a circle. Radians in terms of pi are, though. So 3/4 radians is not rational, but 3pi/4 radians is rational.)
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)🤣
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.
@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.
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.
It's enough to go up to 30* (tan 30* = 1/√3 is irrational).
I somehow totally misread the question assuming it was asking if tan(1°) could be expressed with radicals and ended up proving it.
I instinctively thought the answer was no but don't ask me why I thought that. This was really interesting, thanks :)
Thanks for the pun at 5:50. Made me smile 😊
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....
huh
3:45, “…must be *irrational”. I believe you misspoke there. Good video!
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 ;)
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
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
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.
"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...
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.
I'd probably right "Nope" and move onto the next question that I most likely wouldn't solve WITH a calculator...
In the same way, you can demonstrate the tan(n°) is irrational when n is any factor of 60.
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 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? 😂
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.
Aristotle seems to doubt that there was either a Pythagoras, or that the sect using his name were actually followers. He says "so-called Pythagoreans" a lot. I am not convinced that any other place besides the Greek milieu invented mathematical *proof*. By contrast, the pattern of triples, etc. was definitely known in many places, including India and China. As for the topic today, it looks like a great place for a teacher to set a simple question for testing understanding the logical aspects of mathematical induction.
I have to wonder if adding the story about Pythagoras sentencing someone to death on that entrance exam's answer would get you anything or just a few weird looks 🤣
?
P.T.: sqrt(69)
Me: sqrt(nice)
4:03 The square root of 69 is 8-something, right?
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?
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.
The word play was great.
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
Excellent question and excellent solution
You've also crossed tan 45 degrees which is in fact rational, keep it till 30
There is no problem that tan(45°) is rational. A false assumption can lead to correct statements along the way and that is not a problem. As long as it leads eventually to a contradiction, the assumption is false.
At the step where tan(45°) is rational, the next step is probably wrong because you are adding tan(1°) or tan(30°) which are not rational. If you stop at say 45°, you don’t have a contradiction so you have proved nothing one way or the other at that point.
@@bobh6728 Yes, this proves only than 1°, not any of the intermediate steps.
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.
what about tan(1 radian) instead of degrees? pretty sure it’s irrational but how do we go around proving it?
Q: Is tan 1 a rational number? (Justify your answer with a proof) => A: "I don't know", Proof: Lie detector.
Did I pass the entrance exam for the logic course?
@@Anti_Woke Well I don’t think you are allowed to use a lie detector during the exam
lie detectors are not reliable
Could a continuous function be rational for some inputs and irrational for others?
京大さんにょっす！w🐮✋
You don't even have to go to 60 degrees to prove that; tan(30 degrees) = (sqrt3)/3
Those Math Dewds play hardball... ☠☠☠
I rather suspect that this proof, or something rather like it is taught in Japanese schools, as if this was presented to somebody completely fresh in a time-limited entrance exam, it would surely prove to be a very demanding task indeed.
This proof is incomplete.
We need to also prove that tan(60) = sqrt(3).
But this can be easily proven by a well known theorem discovered by Pythagoras
Wouldn’t that “prove” that tan(45°) is irrational? but it is easy to show that tan(45°) is 1 (which is rational). Something seems wrong here.
Nope. Let's apply the argument to tan(45°).
Suppose that tan(45°) is rational. Using the tangent addition formula, this implies that tan(45° + 45°) = tan(90°) is rational. In the same way, it implies that tan(135°), tan(180°), etc. are all rational. There is no contradition here, so we have not proven that tan(45°) is irrational.
Couldn't you have stopped at 30°, because tangent(30°) = 1/sqrt(3) which we know isn't rational either?
Tan 45 is rational
@rohitdakhane9181 Indeed it is. It's 1.
But all you have to do to prove 1° is NOT rational is to prove that 1° + n° is NOT rational.
I would love an explenation to why this doesn't apply, by the same logic, to tan(45).
I have seen some comments about this but I don't understand to be honest. I need the animations😂
short answer: the sum of two rationals is rational. but if the sum is rational, the summands need not be. so tan(45 degrees) doesnt prove anything and tan(30 degrees) does.
@@theupson I think he's saying that tan 90 = tan (45 + 45) should be rational but it's undefined because 1/0
There is no contradiction at 45 degrees so the proof of contradiction will say nothing.
The notation in the video is misleading. It suggests "tan(45°) rational => tan(46°) rational", when what is actually used is "tan(45°) rational AND tan(1°) rational => tan(46°) rational".
@5:47 says doesnt want to go off on a tangent, then comes back to the tangent in question.
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.
tangent is defined as the ratio of sin to cos so yes
5:48 - Was anyone sentenced to death for too many math puns? 😏
But wait. On the way to Tan 60, you came by Tan 45. Tan 45 is rational. It is 1.
Yes, it is. But the point is: IF tan 1° is rational, then tan 2°, tan 3°, tan 4°... they ALL have to be rational. However, tan 60° is not rational, we already know that. That gives us the contradiction. The fact that ONE of the values (tan 45°) in that sequence is rational, does not eliminate the contradiction. They ALL have to be rational; and, demonstrably, they are not ALL rational. QED.
*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).
Hello Presh
If all Tangents from 1 degrre to 60 are irrational.
What happens with the Tangent of 45 degrees??????
The notation in the video is misleading. For example, it suggests "tan(45°) rational => tan(46°) rational", when what is actually used is "tan(45°) rational AND tan(1°) rational => tan(46°) rational".
He didn't say that all tangents from 1° to 60° are irrational. He said that IF tan(1°) is rational, then tan(60°) is rational, which is a contradiction, so tan(1°) must be irrational.
Let the mathematicians decide? Why not let the historians decide?
tan(1) = tan(37 - 36), proof done
My uneducated mind:
Tan = sin/cos. So any tan is rational so long as sin and cos are rational. Except sqrt(2) isn't rational and is a possible value from sin and cos. So maybe this isn't so simple.
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.
What about adding up to 45 degrees, tan(45 deg) = 1/1 ..
Bro what if we stoped at tan45° . than we would have tan1°as rational
Incidentally, since 60 is highly composite, you can use the same logic to prove that the tan of 2, 3, 4, 5, 6, 10, 12, 15, 20, and 30 degrees are also all irrational.
Fixed :P
@@mattkutschera4514 Come on! Just edit in the missing "4"!
It's driving me crazy.
4:55 No Chinese people were harmed in the proving of this theorem.
4:04 the square root of 69 is "8 something".
Pretty easy question. Let q = tan 1 be rational, then (1 + qi)^30 = rational + rational*i, so tan(30) must be rational, contradiction.
Can't you use the angle sum identity & the fact that the product or sum of an irrational number & rational number is irrational?
Sin(45-44) = sin (1) = root(2)*[cos(44) - sin(44)]
T.f sin(1) is irrational. Do the same for cos(1), divide them & you get tan(1)
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?
squares need 2 of the same thing multiplied together. 2 is an even number.
Now another question, is sin(1) rational? What about its cosine?
well knowing the formula for sin(a+b) and cos(a+b), i'm sure you can figure that one out in no time
I do appreciate a good education system but probably need to learn the basics first
My proof: I know I would need to look it up on a table.
You ignore the fact that your method would have passed by Tan(45°) = 1, which IS rational.
Why not prove it by working out Sin(1°)/Cos(1°), or Sin(1°)/Sin(89°), which can be written as a specific formulas?
You need to read the other comments. It's not relevant.
There's nothing wrong with that. If tan1 is rational instead, then tan45 must be rational. However, if tan1 is irrational, it does not guarantee tan45 is irrational. tan45 could still be rational either way. This property doesn't work both ways.
If tan(1 degree) is rational, then tan(45 degrees) is indeed rational. What's the problem? There's no contradiction there.
Sin 1/ Cos 1
Sin 1 / Sin 89
1/90/89/90
It is! yan is opposite divided by adjacent
3:45 irrational*
Back at school in the 70s we learned Tan=Opposite/Adjacent.
Tan 1⁰ is not like 1/1 (Tan 45⁰)
It would be irrational to think that 1/45th of the angle would be rational, therefore it must be irrational.
QED
Nice. SOHCAHTOA!
That square root of 69 💀
Brain 70% rotted
@@Kero-zc5tc bro also tried to sneak in sqrt 420/10
8-something iirc
I don't really understand this proof. I get that Tan 60 degrees is irrational but why did we stop there? If we had gone to only Tan 45 degrees, we get a result of 1, which is a rational number and therefore we can conclude all the others are rational too, right?
we need a contradiction and it's inevitable
Basically here we used prior knowledge that tan 60 is irrational. But our initial assumption (tan 1 is rational) made it rational, and it's a contradiction => tan 1 is irrational
@@asdbanz316 I'm still not sure I understand but thank you for trying to explain it.
@@trumpetbob15 Basically, for a proof by contradiction you have to find (wait for it!) a contradiction, just one. In this case, we show that the assumption tan(1) is rational leads to the result that tan(60) is rational, which it isn't, so the assumption is wrong The fact that we pass through some numbers that ARE rational doesn't matter, they are just not the contradiction I am looking for. For example, I have a theory that everyone's birthday is Jan 17. If the first 5 people I ask do in fact have that birthday, my theory might appear to be good, but as soon as I meet someone whose birthday is say Mar 03, that's a contradiction and my theory is false. All "positive" results don't matter.
@@silver6054 Thank you. This video kimd of glossed over that part where we can skip through the results that are rational as that is what confused me.
@@silver6054 My birthday is not Jan 17 so I further confirm that the theory is disproven 😂 Now we just need to work out the odds of the first 5 people having Jan 17 as the birthday as I reckon the numbers will get pretty big very quickly 🤣
In base pi, we have pi = 10
Isnt tangent just sin/cos
and then sin/cos of one degree is a fraction of non intergers, so its irrational?
No, (1/7) / (4/31) is a fraction of two non-integers but it is rational.
how did chatgpt fair on this one?
You don't need to prove that a non perfect square integer is irrational, that's the definition of a non perfect square.
Why? Non-perfect squares are numbers whose square root is not an integer. So, by this definition alone, you couldn't conclude that their square root is not a rational.
@@ilmionomenonloso No non perfect squares are not numbers who square roots are not an integer, they are numbers who's square roots are irrational. They are defined that way, nothing to prove.
@@maxhagenauer24 The fact that the square root of a prime number is a non perfect square is not a definition, but a theorem
@@Panqueroso12 Yes I know that, I wasn't referring to the square root of prime numbers being non perfect squares. I was talking about the square root of a non perfect square.
@@maxhagenauer24 Well, if it's just a matter of definitions I guess that there's no right or wrong.
I don't this this works. tan(1) and tan(29) gets tan(30) which is 1/sqrt(3)--irrational. Apparently q.e.d. but continue, tan(44) and tan(1) gets us tan(45) which is 1 and rational.
Read up on proof by contradiction. The assumption is assumed to be true, and true is allowed to imply true. But true is not allowed to imply false. If you derive a false statement, you can conclude that the assumption was false. If you derive a true statement (like tan(45 deg) is rational) this doesn't tell you anything. You really have to think carefully about this.
but tan 45 is rational, which means somewhere in the reiteration process, the interaction between irrational numbers will create a rational number. So I dont think its a good proof. Is there a more convincing proof?
It's possible for irrational numbers to create rational numbers. If tan44 and tan1 are both rational, it means that tan45 must be rational (like in the video). However, if tan44 and tan1 are both irrational, it doesn't guarantee that tan45 must be irrational.
You need to think more carefully about how proof by contradiction works. If tan(1°) is rational, then tan(60°) MUST be rational. This is a contradiction, so it's complete and undeniable proof that tan(1°) is irrational. The part where you maybe get confused is that a false statement is allowed to imply a true statement. For example tan(1°) is rational implies that tan(45°) is rational, but this is not a contradiction. It's only a contradiction when you derive a false statement.
Well, the ancient cultures killed people for lesser reasons....
But Tan 45° = 1 --> rational
That's not a problem. The assumption in a proof by contradiction can lead to any number of true statements that you want, (this doesn't say anything about the veracity of the assumption) but if it leads to a single false statement (like "tan(60°) is rational") then it is a false assumption. That's how proof by contradiction works.