The Most Beautiful Proof sqrt(2) is Irrational

[@WrathofMath]

I keep dreaming that if enough people go watch me rap 17,000 digits of pi, KZbin will promote it like it deserves: kzbin.info/www/bejne/h6WXoZeOYtxlgKM

[@TAPABG-ANIM]

"ten hours ago"

[@wyattstevens8574]

You may not realize it, but the "classic proof" that you talk about at 2:00 actually uses the fundamental theorem of arithmetic! Have you seen the Michael Penn Math proof that doesn't, and works for every non-square n?

[@gabor6259]

3:00 The t in Tom Apostol's surname is silent.

[@Верба100]

Dude, that's creazy😮

[@oreocookiedough]

Every triangle is a love triangle when you love triangles.

[@mayankthakur5341]

❤❤❤🎉🎉🎉

[@NonEi-hs1uk]

I Love Triangles /\ / \ / \ / \ / ❤ \ /___________\

[@gallium-gonzollium]

Maybe that's why Matt Parker named his book about trig "Love Triangle".

[@Alittlebitofeverythingoffical]

That’s the first time I’ve seen the word “triangle” used in a sentence 3 times.

[@Makememesandmore]

And the next video I click on has a book called Love Triangle in the back. It's the author of Love Triangle's channel

[@mathisfun774]

As a minor mathematician and former teacher (high school, college), I am impressed by your ability to bring together all the details coherently without unnecessary repetition. (Any so-called repetition after some minimal time is not a repetition, but a timely reminder.). Just subscribed, hope to see more along this line.

[@Nikolas_Davis]

I was told by my math teacher in high school that Hippasus was _not_ killed for proving sqrt(2) to be irrational; his "crime" was revealing this secret truth to outsiders, i.e. non-members of the Pythagorean sect.

[@Makememesandmore]

lol

[@vohongphucthemultitalented]

After all it’s just a theory…

[@clemensvorbauer1183]

and i think it was not sqrt(2), it was sqrt(5)

[@Cen_t1369]

You could have simplified x+x+90=180 to get x=45 but I can see how that would lead some people to think trigonometrically

[@reubenmanzo2054]

What's the problem with that?

[@Cen_t1369]

@@reubenmanzo2054 he is trying to show a easy/beautiful proof. It's very easy to proof trigonometrically

[@Nik-ny9ue]

never change, reddit

[@w花b]

🤓 🧠💪

[@rogerphelps9939]

The angle sum of a triangle is itself dependent on the parallel lines postulate which is an axiom of geometry.

[@sugarfrosted2005]

I like how this is the idea behind the infinite descent of natural numbers.

[@kaiserquasar3178]

The algebraic idea behind this is if we start from the proposition √2=a/b for some integral a and b, we get a²=2b². Adding a²-4ab+2b² to both side of the equation leads to 2(a-b)²=(2b-a)², which is another, lesser integral solution, which completes the proof by infinite descent as it implies there is no least integral solution to the equation.

[@Makememesandmore]

thank you sir

[@viotall]

40 seconds in. Has to pause the video to say that was the most beautiful constructed pi symbol I’ve seen in my life.

[@WrathofMath]

Thank you! I practice! kzbin.info0Zal-qmcPqo

[@freedom-exe]

I used to Hate Geometry but you make me fell in love in Geometry.

[@welafobie]

I used to love geometry and this video made me hate it!

[@DrMikeE100]

As a Ph.D. mathematician / math professor, I've long favored the proof by contradiction that ultimately relies on the uniqueness of factorization. (That is, when you get down to a^2 = 2 b^2, you argue that the number of factors of 2 on the left is even but the number of 2s on the right must be odd, a contradiction.) But now that I've seen this geometric proof by Prof. Tom Apostol? It is fantastic - and I don't know how I missed it a quarter-century ago. Thanks for sharing this beautiful proof, which is indeed geometric, not algebraic or number-theoretic.

[@ertason]

are we just gonna miss the fact that this dude put the "aquatic ambience" from donkey kong country as the BG music 😭😭

[@meofamily4]

Tom Apostol taught me first-year calculus in college, at Caltech in the late 1960s. A genuinely gentle, cordial, engaging man.

[@antoniorose2461]

Thank you, Sir! You introduced us to a beautiful, concise, and powerful proof! Subscribed!

[@jay_13875]

I think a more intuitive argument is that you could repeat the process of constructing a smaller isosceles right triangle with integer sides indefinitely. The lengths of e.g. the hypotenuses of the triangles would then form a strictly decreasing infinite sequence of positive integers, which cannot exist.

[@tristanridley1601]

That little addition would make it more obvious why this is impossible, for people who are unfamiliar with this famous form of proof. Technically already covered by "there is a smallest integer" earlier, but that was much earlier and quick.

[@Aurochs330]

I think that was the point. With that new smaller triangle you can repeat the process infinitely thus making it irrational.

[@atzuras]

Which is the same as the classical proof of the infinite descent but with pictures.. it makes a lot of sense to me.

[@jaja47_coolness]

I think you could go more into how there not being a smallest isosceles triangle with integer sides, concludes that there is irrational numbers, because it's very easy for a person to say "well that just proves you can't imply that" rather than saying "oh wow, that proves that either the hypotenuse or legs have to be irrational" because I can't fully say that with the mindset of a Pythagorean Cult Member, I would be fully convinced in your proof from Tom Apostol. Thank you for listening to my Tom Talk :)

[@amarug]

Tired and tense, was gonna check out the hub but instead ended up binge watching math videos on this channel. strange times 😅

[@WrathofMath]

tmi but thanks for watching 😂

[@amarug]

@ 🤣

[@FScott-m1n]

Fun fact: When you algebra the lengths of the smaller integer sided triangle, then use similar triangles with the initial intiger sided triangle and simplify, you get the same expression as the classic proof by contradiction a^2=2b^2. The variables a and b (at least on the surface) seem to represent different things depending on which proof youre looking at -- the length of the hypotenuse and short sides in the geometry proof, and the numerator and denominator in the classic proof -- but it's interesting that the same equation appears in both proofs.

[@fredg.sanford634]

Thanks!

[@WrathofMath]

Thanks for watching and for the dono!!

[@fredg.sanford634]

@@WrathofMath -- Thanks for being such a great instructor, and not skipping any steps!

[@Michael_Kaiser108]

the visualisation of it is quite interesting.

[@thomassynths]

You only prove that sqrt2 is irrational or sqrt is an invalid mathematical object. Unfortunately the reals dont have valid constructions so one has to take as an axiom the existence of irrationals in order to complete the proof.

[@BrianHartman]

Thanks for the explanation. :) It really becomes easy when you get to the point that you show that there has to be a smallest isosceles triangle. How could you ever have a "smallest" triangle?

[@johnnye87]

You can certainly have a smallest triangle -that satisfies specific additional constraints-. There's one in this video: the smallest isosceles right-angled triangle with two integer sides, 1/1/sqrt2. The smallest isosceles triangle with all integer sides is the equilateral triangle 1/1/1. The smallest right-angled triangle with all integer sides is 3/4/5. It's trying to satisfy all three constraints, (1) right-angled, (2) isosceles and (3) all three sides of integer length, that is apparently impossible.

[@mrcatfacecat]

take a shot every time he says integer

[@Bodyknock]

My favorite proof of the irrationality of the square root of two uses the Rational Root Theorem because it’s not only very quick but it also immediately proves a large class of roots are irrational in one fell swoop. 🙂 For reference for anybody who doesn’t remember it, the Rational Root Theorem (RRT) says that if you have a polynomial of the form axⁿ + … + b where a and b are integers, then if it has a rational root p/q in reduced form it must be that p is a factor of b and q is a factor of a. Given that theorem, a special case is the polynomial xⁿ - c. If x is a rational root p/q of that, then it follows from the RRT that q is a factor of 1, i.e. that x is an integer! Therefore the only rational roots of integers are other integers, there are no rational roots of integers which contain fractional part. So this means the square root of 2 is irrational because it isn’t an integer since 2 isn’t a square of any integer. And more broadly all roots of integers are either integers or irrational numbers, and you only get integer roots when the original is a perfect powers of other of an integer (e.g. 8 = 2³ , but there are no integer and hence no rational cube roots of 9). The Rational Root Theorem is also nice here because not only does it cover a really wide category of roots of integers, it’s not even hard to prove. All you need to prove it are some basic underlying facts about prime factors and a small amount of algebra. Proving RRT takes just about as long as, say, the proof in this video. That’s not to say the proof in this video isn’t fun, it’s always neat seeing arguments that use the impossibility of “infinite descent” in the positive integers. 🙂

[@Ramkabharosa]

Suppose √2 = a/b, where a & b are pos. integers with b as small as possible. Then 2b>a>b because 2>√2>1. Now (2b-a)²/(a-b)² = (2- a/b)²/(a/b -1)² = {(2- √2)(√2- 1)}² = {2√2 - √2 - √2 + √2}² = {√2}² = 2. So, √2 = (2b-a)/(a-b) contradicting b was smallest possible pos. denominator. ∴ √2 is irrational.

[@joseph-fernando-piano]

By presuming that sqrt(2) = a/b, a rational number, and then squaring both sides and multiplying by b^2, we would get 2b^2 = a^2, which would mean that for sqrt(2) to be rational, there would have to exist two perfect squares which differ by a factor of 2... I don't what the proof for that would be, but it seems like one of those things that you can just tell isn't possible just by "inspection"...

[@Alucs.E]

You're thinking along exactly the right lines! You can actually use this to come up with a proof along the same lines as the video (using the so-called "well-ordering" of the naturals - i.e., that there is a smallest natural) The proof goes like this. Suppose √2 is rational. Let a,b be the smallest denominator and numerator such that a/b=√2. Since a/b=√2, a²=2b². If a² is even, then a must be even, which means we can write a=2c. Substituting this back in, we get 4c²=2b². Simplifying we get 2c²=b². But this means that b/c=√2. We have a contradiction. We supposed a and b were the smallest denominator and numerator, but b

[@w花b]

​@@Alucs.E Very simple and easy to understand. Nice one.

[@rogerphelps9939]

Factors of squares always come in pairs so a^2 must be a multiple of 4. Therefore a is a multiple of 2 and so is b. Therefore a and b are not the smallest integers whose ratio is root 2. This is a contradiction.

[@hhhhhh0175]

if you let v_2(n) denote how many times you can divide n by 2 and still have an integer, we can prove v_2(a * b) = v_2(a) + v_2(b), because an odd number times an odd number is always an odd number. then you'd have 2b^2 = a^2 => v_2(2b^2) = v_2(a^2) => v_2(2) + 2 * v_2(b) = 2 * v_2(a) => 1 + 2 * something = 2 * something, which is odd = even and so impossible. this proof only works when v_p(a * b) = v_p(a) + v_p(b), which is only always true when p is prime. but a modification will work when p has any prime factor that occurs an odd number of times (which is a classification of the non-squares)

[@johnnye87]

This looks closely related to the impossibility of an isosceles right triangle of integer sides: 2a^2=b^2 is Pythagoras applied to an isosceles triangle.

[@Kwshal]

0:53 √2 = 1.414213.. 577/408 = 1.414215.. Differs at 6th decimal point

[@GetMeThere1]

It also demonstrates that any number of the form sqrt[2*(x^2)] where x is an integer is also irrational.

[@HollywoodF1]

It’s IR rational, not EE rational. Ir- rhymes with ear.

[@indovash]

When practicing Euclidean Geometry, the first object we come in contact with is usually the square root of 3. We have to find our way to the square root of 2 after this. In cartesian geometry, the square root of 2 is generally defined by the distance from point [0, 0] to point [1, 1]. So, yes, in "cartesian" geometry, we would likely discover the square root of 2 first. But in Euclidean Geometry, we will generally discover the square root of 3 first. It takes a small amount of work, but work nonetheless to find and properly prove the square root of 2, and you need the discovering of the square root of 3 to do it.

[@orisphera]

In other words: If √2 = a/b, then (a-b)√2 = 2b-a and √2 = (2b-a)/(a-b), therefore, if there aren't any integer solutions up to n, there aren't up to n+1, either, and therefore there aren't any at all

[@m.h.6470]

The quickest proof that √2 is irrational is this: Let's assume that it IS rational, so √2 = a/b with a and b as integers and a/b FULLY REDUCED √2 = a/b |*b b√2 = a |² 2b² = a² So you need a perfect square that is exactly twice as big as another perfect square. Both squares are perfect, because a and b are integers. But in order to have a perfect square twice as big as another, that other one would have to have 2 as a factor. So a would have to be even. As such we can write a as 2c, with c being integer. 2b² = (2c)² 2b² = 4c² |:2 b² = 2c² This means in turn, that b as well needs to be even. But now we have a contradiction: a/b is defined as FULLY REDUCED, but if both a and b are even, than both have a factor of 2, which could be cancelled, so it ISN'T fully reduced. Therefore the assumption, that √2 = a/b CANNOT be correct.

[@richardhole8429]

I never had head "irrational" pronounced with a long e.

[@WrathofMath]

i pronounce it this way because 'irrational' sounds way too much like 'a rational'

[@bobengelhardt856]

"E-rational"?? Is that a joke?

[@qovro]

e is the most rational number

[@awareflag001]

Says the traditional method to prove sqrt2 irrational is by assuming its rational , refuses to elaborate. Proceeds to start his own proof the same way. (No hate , just thought it was amusing)

[@worldnotworld]

Great, but PLEASE avoid the annoying non-musical "music" in the background in future videos!

[@m.h.6470]

Why call it angle x, when it HAS to be 45°? There is NO other angle possible in an isosceles right angle triangle. They are ALWAYS 45°-45°-90° triangles.

[@farpurple]

Is there smallest 3 4 5 though?

[@jonahansen]

Very nice!

[@imnotkentiy]

amazing song choice

[@End_Oe]

The intro part reminded me of that shortade by circletoons about how tutorials never get straight into the point 😂 It's fine though if you put chapters into the video!

[@markdaniel8740]

E rational.

[@saf-shaangfox]

Six and 28 are both perfect numbers

[@derekcouzens9483]

...or as I was taught to end proofs by my maths teacher. w^5. "which was what was wanted"

[@tonyennis1787]

So, I could follow that and I am barely a trained semi-mathematician, and what I know is what remains from what I didn't really learn 40 years ago. It is odd that the ancient greeks didn't figure this out. Or no one else did until 2000 ad.

[@rdaysky]

I have an even more beautiful one for ³√2. Suppose ³√2 = m/n, or m³/n³=2, or m³=2n³, or m³=n³+n³, but Fermat’s last theorem prohibits this, QED.

[@jamesknapp64]

Well Euler proved this case well before the general case of FLT was proven in general case.

[@pranavikanyaboina2237]

8:12 x= 45

[@adamlea6339]

It is analagous to the classic proof by contradiction I've seen where you assume if sqrt 2 is rational then it can be expressed as a/b where a and b have no common factors, but then with a bit of logic you can establish a and b must be both even. I find this one more elegant.

[@lyrimetacurl0]

(just a little side maths) BD = sqrt(2) * DO AB = sqrt(2) * BO BO = BD + DO = (1+sqrt(2)) * DO therefore AB = sqrt(2) * sqrt(2) * DO + sqrt(2) * DO = (2+sqrt(2)) *DO therefore (1+sqrt(2)) * sqrt(2) = (1+sqrt(2)) + 1

[@wayneyadams]

Pi being irrational is so counterintuitive. It is the ratio of the circumference to the diameter of a circle, yet it is irrational. it hurts my brain! LOL

[@Tletna]

It's funny, I just did the simple algebra version of the proof by contradiction and it uses the same argument that a certain ratio shouldn't be simplified but it is, so sqrt(2) cannot be rational.

[@StarDotJPG]

√2 = |1+i|, which is an integer, albeit a Gaussian one. Checkmate, mathematicians.

[@Jason9637]

It's the absolute value of a gaussian integer, not a gaussian integer itself

[@AverageMathsensei]

@@Jason9637Exactly.

[@tristanridley1601]

This seems like one of the best examples of the value of imaginary numbers, and why they are the right way to interpret things.

[@Ringcaat]

This is a clever proof. However, it was too slow for my taste. I did a lot of fast-forwarding.

[@txikitofandango]

You can get a hardcover version of Apostol's calculus book for $150-170, which is comparable to other calculus books. Not sure why the paperback is $495.00...

[@noober52614]

sqrt(n) if n ≠ perfect square is always irrational

[@Namegoeshere-op9hg]

I don’t believe the assumption “there must be a smallest isosceles triangle” is true. This neglects units.

[@awareflag001]

The smallest possible integer side for the triangle is 1 , so if you can make a triangle with isosceles sides 1 smaller then that means the 3rd side wasn't integer to begin with

[@tangerian319]

So, why does this proof not disprouve the 3-4-5 triangle? Genuine question

[@tkfocht2997]

The 3-4-5 triangle isn't isosceles, so this isn't talking about that.

[@tangerian319]

@tkfocht2997 i would think similar triangles would go into play as well. But I'm assuming that you can prouv that the subdivision is also an integer?

[@stardf29]

@@tangerian319 The triangle being isoceles is key to the proof. It's what allows the smaller triangle BCD to also be isoceles, and thus CD would have to be an integer. You could do a similar construction with the 3-4-5 triangle, but since the resulting triangle would not be isoceles, there's nothing stopping CD from being a non-integer value.

[@tkfocht2997]

@ The smaller triangle should have an integer-length side where it overlaps the larger triangle's hypotenuse, but because the triangle's not isosceles it's not required that the other side (the constructed perpendicular) be an integer length.

[@tangerian319]

@tkfocht2997 aaahhhh, okay. If I have time, I think I should try to apply this proof to a 3-4-5 triangle and see exactly where the proof breaks down

[@ApesAmongUs]

Saying X instead of 45 is confusing the shit out of me.

[@Grizzly01-vr4pn]

The measure of the angle does not matter to the proof.

[@ApesAmongUs]

@@Grizzly01-vr4pn Disguising it with a variable does not matter either.

[@Grizzly01-vr4pn]

@ That really doesn't make any sense.

[@ApesAmongUs]

@@Grizzly01-vr4pn Correct, using a variable when you know the angle doesn't make any sense. Why say A=A when you could say 45=45? You use variables when you are trying to make something generalized instead of specific, but this specifically only works on an isosceles right triangles - and all of those are 45/45/90. Every single fucking one. He ends up needing to do some similar triangles bullshit when he could just do basic subtraction. It's like using a 4 syllable thesaurus word instead of the common one that communicates more clearly.

[@Grizzly01-vr4pn]

@@ApesAmongUs giving the measure of the angle is completely unnecessary for the proof, and would introduce pointless detail that would detract from the basics.

[@BaranCemCesme]

Why use x when it's much more easier to prove that it's 45°

[@DrMikeE100]

I surmise that he did not mention x = 45 (degrees) because it is irrelevant to his point. (Ph.D. mathematician, PSU math professor speaking here.)

[@lyrimetacurl0]

Is the smallest right triangle with integer sizes 3,4,5? I thought before the video, better not say it has to do with pi because then you would also have to prove pi is irrational, thus lengthening the proof 😅

[@sonalita_]

That isn't an isosceles triangle though.

[@RM47319]

I love proofs based on infinite descent.

[@vishalmishra3046]

*New innovations are rare on KZbin lately* Nth root of any integer that is *NOT* an Nth power of some integer is *ALWAYS* irrational (no p/q matches M^(1/N)). Then, why do people in 2025 (2.5 millenniums later) still obsess with a specific number (sqrt(2) or M=N=2) and never think or talk about generalizing irrationality proof to any M,N > 1.

[@GhostyOcean]

I didn't believe the title, but I agree! Wow!

[@iras66]

To be fair I'm not a big fan of this proof. In general I'm very sceptical about visual proofs, there are a million examples where they are deceptive. And there is a lot if these "basic" geometrical facts that "everyone knows". I think a lot of these are way harder to actually prove rigurously than the sqrt(2) is irrational. Like most of the well-known proofs for the irrationality of sqrt(2) are super basic and literally uses nothing whereas this one is trickier, more difficult to see and uses a million things which are themselves way less trivial. It also does not generalize to any other number basically. So overall I'm not really seeing the point.

[@Grizzly01-vr4pn]

It isn't really a visual proof though. It's a geometric proof, which is not the same thing.

[@javiersaneiro6412]

It is also an hystorical proof. Concepts like this who are knowing since ancient greeks were actually demostrated by greeks using geometry with methods like this one. Remember that ancient greeks didn't knew algebra, that was invented centuries later by a muslim mathematician

[@iras66]

@@javiersaneiro6412 Have you watched the video? This is a modern proof. Although I must admit this is kind of something that the ancient Greeks could have found.

[@TheMoonlight1711]

so tldw, A/B/O do not search this up btw, it WILL lead you to omegaverse and i do not wish that upon anybody except the gays who want children

[@SunShine-xc6dh]

Thats a proof that the number sprt(2) doesn't exist as a number

[@ThatShushi17-mc7ct]

Rest In Proof

[@jasonthesnow]

So I understood, but I really did not understand…

[@circjit]

starting with 37, great!

[@toferg.8264]

Nice!

[@pedroteran5885]

The notion that this could be considered beautiful by anyone is perplexing to me. I lost count of how many geometric assumptions are needed: all triangles have an equal sum of angles, two triangles with proportional sides have equal angles, there exists a triangle whose sides are in an arbitrary proportion with those of a given triangle, whatever is needed to establish that D is actually an exterior point to the circle, DO and DC have the same length, and probably more that I've forgotten. What is the beauty?

[@Grizzly01-vr4pn]

Are you sure you understand the meaning of the word 'assumption'?

[@johnnye87]

"Beauty is truth, truth beauty -- That is all / Ye know on earth, and all ye need to know."

[@thomastang2587]

why don’t you just say x=45 degrees?

[@Grizzly01-vr4pn]

Because the measure of the angle is irrelevant to the proof.

[@johnwiltshire8763]

There is no such thing as an "irrational number". Here's a proof: Let: S = 14142135624....(n digits of sqrt(2))/10^(n-1) By definition, S is a rational number because it is the ratio of two integers and it remains rational as n increases. Let: e = sqrt(2) - S We can make e as small as we like by making n sufficiently large. In the limit, e =0, and, if the difference between two quantities is zero, then they must be equal. Since S is rational, it follows that sqrt(2) must also be rational.

[@DominiqueB1]

Extremely verbose, but a nice proof :-)

[@Hyperfictionalgoogologist]

i feel irrational

[@pranavikanyaboina2237]

2/sqrt(2) !!!!

[@WilliamWizer]

I fail to see why this is the most beautiful proof of anything. since all ratios can be expressed in lowest terms (numerator and denominator are coprimes): assume √2 = a/b with a/b being in lowest terms. square both sides: 2 = a²/b² multiply both sides by b²: 2*b²= a² a must be even. (let it be a=2c): 2*b²= (2c)² split the square: 2b²= 2²*c² divide both sides by 2: b²= 2*c² b must be even. we just proved that a and b are both divisible by 2 but, they must be coprimes (no common divisors) that's a contradiction. for those that got confused. here is an easier way to think about it: let N, X and Y be natural numbers: if N² is divisible by X then, either X=Y² or N² is divisible by X² if Y exists, √X is a natural number. if Y doesn't exist, √X is irrational.

[@erkintek]

Too much talk, nice papers to write,

[@MagnificentCreature]

Does this proof method of proving the smallest really work? How are you so sure that just because the smallest doesn’t exist means the scaled up version won’t exist? Even with your reasoning of the scaling, I am not quite convinced.

[@brian8507]

Recursion my friend.... start with a right Isosceles triangle of ANY size... and do what he did above over and over and over again... iterating over and over again on the newly made triangle that's smaller... and u can keep doing this till infinitely small... still carrying the logic that all sides on each Recursion are integers Sorry ... mathematician here.. I think the video maker forgot to highlight that point... or it was subtle

[@jolkert]

think of it this way - the "scaling" thing is to say that if √2 is rational, you must be able to construct an isosceles right triangle where all sides are integers - using the method shown, you can could construct a smaller isosceles right triangle where all sides are integers from any other such triangle no matter the size - the previous point implies that there is no smallest such triangle, as a smaller triangle can always be produced by repeating the process - this implies that there is not a smallest positive integer - there *is* a smallest positive integer (namely: 1) - therefore an isosceles triangle where all sides are integers cannot be possible - which in turn implies that √2 cannot be rational

[@MagnificentCreature]

@@jolkert i see! This explanation wins it for me. I am convinced thank you!

[@EdMatthewMorales]

Smallest as in zero loll

[@nathanbingham7369]

First?

[@nathanbingham7369]

Maybe?

[@WrathofMath]

i'm always first, or maybe I can be 0th and so then you're first

[@anon_y_mousse]

Every number is rational if you use the right representation. A formula works just as good as a single symbol, it just requires more work.

[@DrMikeE100]

That "every number number is rational" statement is manifestly false. Pi is irrational no matter how you represent it, for instance. You probably should defer commenting on math topics if you don't fully understand the definitions. (Ph.D. mathematician, PSU math professor speaking here.)

[@anon_y_mousse]

@@DrMikeE100 It's a single symbol to represent it, π, seems pretty rational to me. Now, speaking as someone who is not a robot, you might wish to have your controllers install a humor chip.

[@javiersaneiro6412]

@@anon_y_mousse Rational from ratio, not rational from reason. It is not the same thing. π is not rational because it can't be expresed on a ratio a/b with a and b integers

[@anon_y_mousse]

@javiersaneiro6412 I think I've found a new form of sniping. I wouldn't call it nerd sniping because it belies a lack of intelligence on your part and the other dingus who responded, but it's a bit unwieldy calling it humorless robot sniping. Maybe just robot sniping.

[@filipsperl]

Ugly and needlessly complicated. No wonder it took 15minutes to explain the most basic proof ever.

[@MarkAhlquist]

You should do better

[@Grizzly01-vr4pn]

Confused you, did it?

[@sorenriis1162]

Nice!