How to Prove a Number is Irrational

[@hoodiegaltf]

Gonna start calling one "Legendre's Constant" to mess with people

[@publiconions6313]

Lol

[@rmdodsonbills]

Or just randomly throw in "and multiply by Legendre's Constant" when you're talking or writing about some formula. :)

[@filipeoliveira7001]

It isn’t even one though😭

[@hoodiegaltf]

@@filipeoliveira7001 did you watch the video?

[@Fire_Axus]

@@filipeoliveira7001 i want to help you but KZbin does not allow me to

[@tomkerruish2982]

Legendre's Constant is 1? All I can say is... Wau!

[@carlosarcia5714]

Love the vi hart reference

[@mantacid1221]

Even though I’m a mathematician, I appreciate when you explain the basic concepts like what an integer is and what it means for a fraction to be simplified. It makes these videos accessible to anyone I might want to show them to. So thanks for that.

[@mb-3faze]

Am I right in assuming that all rational numbers can be represented by a fraction and that at least one of the numbers making up the fraction (numerator or denominator) must be odd.?

[@Santiag2008]

@@mb-3faze correct

[@michaelremington5902]

@@mb-3faze Yes, if it is a fraction in its simplest form, otherwise, no.

[@belovedone151]

You don't have to assume a/b is in its 'simplest form.' When you have 2b^2 = a^2, and consider the number of prime factors each side has, you'll notice a^2 has an EVEN number of prime factors and 2b^2 has an ODD number of prime factors. This violates the Fundamental Theorem of Arithmetic and so we have our contradiction.

[@frimi8593]

Great point!

[@Fire_Axus]

how does that violate the fundamental theorem of arithmetic?

[@frimi8593]

@@Fire_Axus every natural number has a unique prime factorization. Therefor you will never find a pair of natural numbers a,b such that a^2=2b^2 because “a^2” must have a different prime factorization than “2b^2” on account of the fact that “a^2” has an even amount of prime factors and “2b^2” has an odd amount of prime factors edit: interestingly, this doesn't stop at 2. you will never find a pair of natural numbers a,b such that a^n=pb^n where p is any prime and n is any natural number bigger than 1

[@josephlichtblau9216]

Thank you, I never knew this rule

[@michaelremington5902]

@@frimi8593 Hello - could you please write out a proof of that, in slow steps for non-mathematicians? :)

[@soninhodev7851]

Legendres constant is... ONE??!! thats funny, and i love it! speaking of love, guess who gets their day brightened up whenever you upload?

[@TymexComputing]

Me? || You? || They?

[@soninhodev7851]

@@TymexComputing why not both!

[@thaisirina]

É aqui a reunião dos BRs? Espero não ter chegado atrasada

[@soninhodev7851]

@@thaisirina kkkkk, um dia não é atrasada, contante de legendre em mêses é atrasada!

[@TheZerwanos]

Fun addition to the conlusion of the video: Although it is not known whether pi + e is transcendental. It is easy to prove that at least one of the two numbers pi+e and pi*e is transcendental :)

[@SedgeHermit]

I know of people who, during college exams, are asked to give an example of an irrational number, and then put something to the effect of pi + e and get marked down for it.

[@Ninja20704]

Thats is quite suprising. In fact in general if a and b are transcendental numbers, then either a+b or ab are transcendental, or both.

[@jessehammer123]

Yeah, and this can be shown by considering (x-pi)(x-e)=x^2-(pi+e)x+pi*e. If the coefficients on the right were all algebraic, then pi and e would have to be algebraic. But neither is, so at least one of the coefficients pi+e and pi*e is non-algebraic or transcendental.

[@alexbennie]

​@@jessehammer123such a simple explanation for something that big. Awesome

[@ComboClass]

In this episode, I'll show you some surprisingly simple proofs that certain numbers like square-root-of-2 are “irrational”, and explain what has been proven about other numbers like pi, e, and “Legendre’s constant”. Timestamps of different chapters: 0:00 - Introduction 1:50 - Proving a Logarithm is Irrational 3:50 - Proving a Square Root is Irrational 9:26 - What About Pi and E? 11:36 - The Funny Tale of Legendre’s Constant 15:13 - Conclusion

[@Danny_Fantom]

8:40 - cat

[@JimmyMatis-h9y]

@Danny_Fantom that cat is very rational though 😆

[@alexanderbudianto7794]

Another surprising example is Conway's constant, the limiting ratio of the look-and-say sequence. It is irrational, but it is surprisingly algebraic, being a solution to a polynomial equation of degree 71.

[@ComboClass]

Yeah that's such a cool sequence/number

[@JoyRender]

i want to do stuff like this when i am older your way of teaching is amazing

[@JxH]

15:30 Sitting under an apple tree, there's a significant risk that you'll eventually discover gravity.

[@cameronbigley7483]

9:02 To be slightly more technical, if a/b were not the simplest form, you'd have a new fraction c/d, where a > c, and b > c, a, b, c, d are all natural numbers. What follows is an infinite sequence of descending natural numbers, which is not possible for finite a or b.

[@SigmaChuck]

totally enjoyed this ramble through the side yard of irationality, Loved the detritus--like the fishing pole and your fine collection of broken clocks.

[@rmdodsonbills]

Something that has always seemed kind of strange to me is that pi is originally defined as a ratio of the circumference of a circle and it's diameter, but since it's an irrational number, then in a perfect circle, at least one of those numbers has to *also* be irrational. Consequently, if you have a line segment that is exactly a rational number in length (say Legendre's Constant for example), then the circle you construct on that line segment has to have an irrational circumference. In the case of a line segment exactly equal to Legendre's Constant, that circle's circumference would be exactly pi.

[@mathphysicsnerd]

I believe one of Pi's earliest definitions was actually the ratio of the circle's semiperimeter to its' radius

[@TheMagicFellow]

Great to see you again Domotro!

[@Uejji]

It is true that there is a higher infinity of irrational numbers than there are rational numbers, and that practically all numbers are irrational. However, you can also prove that between any two irrational numbers there must exist at least one rational number. You can also prove that between any two rational numbers there must exist at least one irrational number. This creates an interesting fabric to the number space that, no matter how small a sample you take, you will find infinitely many rationals and irrationals, yet there are infinitely times as many irrationals as rationals.

[@eric23232323]

It's infinitely many turtles all the way down ....

[@gdclemo]

@@eric23232323 "almost all" of the irrational numbers can't be defined at all. That is, there are countably many rational numbers and uncountably many irrational numbers, but there are countably many formulae or written descriptions which define a number in some way. So most of the irrationals can never be described.

[@eric23232323]

@@gdclemo I was only making a light-hearted joke ( en.wikipedia.org/wiki/Turtles_all_the_way_down )

[@gdclemo]

@@eric23232323 Sorry I meant to reply to the poster above you... yes I did get the joke.

[@eric23232323]

"...literally be the number one.. it makes you wonder..." I see what you did there, nice alliterative flow.

[@Bingcenzo]

This is the objectively correct response to anything that Terrence Howard might have to say.

[@pufffincrazy5275]

“how to prove Terrence Howard is irrational”

[@ComboClass]

That guy thinks the square root of 2 = 1, so he'd probably think square roots are secretly a Legendre's constant type of situation haha

[@phiefer3]

A generalization for rational roots of integers. First we consider what happens when you rational number a/b to some integer power. Because raising an integer to an integer power does not grant it any new prime factors this means that if a/b was in simplest terms, then so is (a^n)/(b^n). This ultimately just tells us that if a/b is not an integer, then neither will (a^n)/(b^n). Applying this to the sqrt(2) we see that if the sqrt(2) is rational, then it must be an integer; because if it's not then we cannot square it to get an integer. And the same logic holds for any integer root of any integer. Or in other words, only perfect powers have rational roots.

[@GoSlash27]

9:55 My favorite irrational number is phi; the "golden ratio" because it's the *least* rational of all irrational numbers. 1/phi=phi-1, which causes the continued fraction method to rage- quit.

[@mathphysicsnerd]

I like how phi is also the solution to the infinite radical sqrt(1+sqrt(1+sqrt(1+sqrt(1+sqrt(1+sqrt(1+sqrt(1+sqrt(1+sqrt(1+...)))...)

[@AT-27182]

Beautiful lesson. Thank you.

[@haineshoag3010]

Need a video like this for the uncomputable numbers, much harder to visualize than the rationals and the algebraic numbers, and yet they make up nearly all of the reals 😅

[@maaikevreugdemaker9210]

Legendre's cobstant is so awesome that squaring it will map onto itself. Truly a proto euler number.

[@ConManAU]

Fun fact: the last question of my final high school maths exam led us through the proof that e is irrational. And in a different year they did a different proof based on a different definition of e.

[@your_future_self]

lost it when you pulled the white board out of your jacket... genius

[@SulkendorIsAwesome]

This channel is awesome

[@JohnBerry-q1h]

Regarding the transcendental function known as the sine, and using the Taylor series representation of the sine function, how would someone prove that... • sin(45°) is irrational? • sin(30°) is rational? [note that the sum of an infinite count of rational algebraic expressions can fail to produce a known, rational number result.]

[@gdclemo]

You can show that most of the coordinates of most regular polygons are irrational, where the distance from the center to any point is one. By adding edge vectors you can construct a smaller regular polygon of the same shape. And with this you can construct a smaller one, and so on. This would be impossible if the coordinates were rational and therefore were aligned to some grid. Mathologer has a video on this.

[@walterbrownstone8017]

Step 1. Where did the number come from? It's it the sum of an infinitely repetitive equation that can never be completed? The number doesn't repeat forever, the equation does.

[@DrZip]

The most unsettling part of this video is how you write your 'a' to somehow look like a '9'. :P @2:40

[@monoman4083]

Quality as always..

[@bjornfeuerbacher5514]

Do you know if one can somehow prove that e is irrational directly using the definition lim_{n to infinity} (1 + 1/n)^n ? (Without using the binomial theorem.)

[@RyanLewis-Johnson-wq6xs]

Square root of numbers that end in 2 are irrational I don’t even have to look it up or think about it.

[@edwardblair4096]

The real issue here is how do you specify a specific number? The easiest way is to just write down the digits. This works for integers and non-repeating fractions The next easiest is to write a formula, which works for rational and some irrational ones. For other numbers, you have to come up with a more complicated method to describe the number, then prove that your definition actually IS a number.

[@publiconions6313]

Excellent vid!!!

[@triniasta]

cbrt(2)=a/b 2=a³/b³ 2b³=a³ b³+b³=a³ Contradiction by FLT.

[@TastySalamanders]

While slightly beyond the scope of the video, you touched on two related concepts so a little surprised you didn't mention that algebraic numbers (i.e. non-transcendental numbers) which included all rational numbers and a subset of irrationals has the same cardinality as the rationals (i.e. the infinity of rationals and the infinity of algebraic numbers is the same infinity [countable infinity]). Given the cardinality of irrationals is larger than rationals; if rationals have the same cardinality as rationals+algebraic irrationals, that means irrational numbers overall can only have a larger cardinality if the cardinality of transcendental numbers is larger. That is; the cardinality of the transcendental numbers is larger than the cardinality of the algebraics (rationals + algebraic irrationals).

[@johnmckown1267]

liked the video. But I'm staying away from irrational number, they just go on and about themselves.

[@MrPoornakumar]

Most "irrational" numbers that we use (leaving the series expansions) are square-roots of positive integers et cetera. It gives us a hope that some (at least) can be squared, cubed or raised to an integer, to get positive integers. Logarithms are irrational too, but some of them yield via Anti-logarithm (or raised power of a base or "Radix") positive real numbers. Thus, these may not constitute pure irrational numbers. These are different form pure irrationals, like "π" or "e" that come what may, may remain irrational. Irrational numbers can be seen as "points" on the Real number line.

[@ComboClass]

The "pure irrational" numbers you're describing aren't really a clear term, but what you are referring to is similar (although not identical) to the difference between transcendental numbers vs. algebraic irrational numbers.

[@MrPoornakumar]

@@ComboClass Thank you ! Both (transcendental as well as algebraic) numbers can't be separated on Irrationality basis, but suffer the same. For instance Cos60º= Cos(π/3)= 1/2 & Tan45º= Tan(π/3)= 1 etc. "e", "π" and "φ" (magic number) are such - whatever one does with them, they remain irrational. The only rule or way to transform an irrational number is to divide it with itself, to yield "unit" (the multiplicative identity).

[@Anonymous-df8it]

@@MrPoornakumar Since (2(phi)-1)^2=5, it is indeed possible to "transform" phi by combining it with only integers to get a rational number

[@gdclemo]

Any irrational number that can be defined has some "connection" to the rationals. If you use some function such as sin, cos that expands to an infinite sum then you can define its inverse and you've got a formulaic connection back to the rationals. There are uncomputable values like Chaitin's constant which depend on which computer programs terminate, which can never be known. I think this may or may not be irrational. There are only countably many definitions though, so only a countable subset of the uncountable irrationals can be defined at all.

[@Anonymous-df8it]

@@gdclemo But surely you can order every definable number by placing their shortest definitions in alphabetical order, then apply Cantor's diagonal argument, resulting in a new definable number?

[@nicolascalandruccio]

Is it possible for a root of a polynomial of degree greater than 4 with integer coefficients to be a transcendantal number?

[@MuffinsAPlenty]

No. An algebraic number, by definition, is the root of a polynomial with rational (or equivalently integer) coefficients. So all such numbers would, by definition, be algebraic.

[@okaro6595]

Funny I had thought that has there ever been a case where they did not know if a number was irrational or rational and it was proven to be rational. This still is somewhat had example as it is an integer.

[@ComboClass]

Yeah Legendre's constant is an example of that. Integers are rational numbers (like 1 can be expressed as 1/1).

[@okaro6595]

@@ComboClass Yes, but it would be cooler if we knew that the number is for example 2.45487..... and it would ne found out to be rational. That was with numbers kike pi. We knew their approximations.

[@sterlingsimmons8309]

Love your channel. Any idea what the most known number of digits after the decimal point is for a number that is known to be rational?

[@ComboClass]

Thanks! Assuming you mean irrational there: you can construct irrational numbers with non-repeating patterns (like 0.01001000100001...) that we would "know" all of. As far as numbers that emerge more naturally, probably whatever the current record for pi is.

[@sterlingsimmons8309]

@@ComboClass Thanks for the reply. I actually mean terminating rational numbers. What is the most known number of decimal places for a terminating rational number?

[@sterlingsimmons8309]

But I suppose that since the denominator can be infinitely large then so can the number of decimals places. Hey, no dumb questions right?🥴

[@ComboClass]

Yeah any questions are great. We could create an arbitrarily large terminating decimal pattern (in base ten) for a rational number by creating 1/(2^n) for large n.

[@danielrhouck]

I donʼt know if this is a valid proof, but my favorite proof of the irrationality of a number is for the *cube* root of 2. As usual, itʼs a proof by contradiction. Suppose ∛2 is rational, and in particular ∛2 = a/b, where a and b are integers. Then, we have 2 = a³/b³ 2b³ = a³ b³ + b³ = a³. However, this is impossible, by Fermatʼs Last Theorem. Therefore, ∛2 is irrational. ∎ (I donʼt know if this is a valid proof because I donʼt know if the known proof of Fermatʼs Last Theorem relies on ∛2 being irrational, so this might be circular.)

[@TymexComputing]

Really interesting creative proofs for 2 and 3 and logs and roots - i didnt know them - was expecting some other rational induction contradiction proof. Havent heard about Legendre constant but i simply remove B from that equation - much easier to remember and not much of a meaningfull error while estimating the number of primes :), even ChatGPT says that BIG numbers (> 10^80) are so big that its almost impossible to randomly choose one, then mathematicians should focus on small numbers (10^65) and there the B is higher than 1. All good yo ya all!

[@kravovchik]

What ist that music that keeps playing in all of your videos?

[@ComboClass]

They are soundtrack beats I made myself

[@kravovchik]

@@ComboClassthey sound cool

[@elcolicous]

I love your cat and your channel 😻😻

[@vegenzohrabian5340]

Great video!!!

[@Porcuponic]

Is it possible that some numbers might turn out to unprovably rational or irrational? would it need some new classification of number? It seems like the halting problem would imply that we could construct a definition of number that can't be computed and therefore can't arrive at an answer on its irrationality. I guess what I'm asking is, do we know that a proof for pi + e exists?

[@ComboClass]

That could be possible. We don't necessarily know that a proof exists for pi+e (although in that example, I would suspect it probably does).

[@TymexComputing]

@@ComboClass and @Porcuponic on the other hand i postulate that in mathematical formulae we can observe only a countable number of irrational constants :D - mathematics only uses a finite language with finite sentences that include some finite number of functions like trigonometry(), log(),power(,) these sentences can only be finitely many but on the other hand the number of mathematical sentences grow exponentially with length but to reach that exponential complexity we need to write infinitely long sentences... and nobody would read that :) I recommend taking a look at Rob Pike APL calculator with some strange iota() function and ** and more and his abstract work - he works at google and has own homepage robpike io but its main page is full of sheet()*

[@alaricsnellpym]

Look at en.wikipedia.org/wiki/Chaitin%27s_constant for an example of a uncomputable number due to the halting problem!

[@TymexComputing]

My posts have been deleted :(

[@Porcuponic]

The more I think about it the more I realize that an uncomputable number is necessarily irrational. If the process for calculating it halts at a single ratio, it is rational, if the process continues on forever it’s irrational.

[@ThomasHaberkorn]

Such a great video thanks

[@TymexComputing]

i postulate that there is only countable number of irrational numbers in mathematical formulae - if you find new ones then their power will all also be countable - to reach an uncountable number of irrational numbers we would need some irrational maths which i hope this Combo Class is leading us to :) -- Rob Pike

[@TymexComputing]

Hey Folks! Domotro - thank you for patroning Mathlogger! :)

[@ComboClass]

If you mean Mathologer, yeah I am one of his Patreon supporters since I think he does great things on his channel :)

[@jackkalver4644]

I figured out the logarithm proof myself

[@vishalmishra3046]

*Irrationality of Nth Roots* nth (n>1) root of any integer that's not a perfect nth power of an integer is always irrational and the same applies to the nth root of any rational number that's not a ratio of 2 nth powers of integers. *Example* 100 is not a perfect cube so it's cube root is irrational. Similarly, the cube-root of 100/27 is also irrational even though denominator is a perfect cube (but not numerator is this simplified fraction). In this way, the rules can be incrementally *GENERALIZED* until most Algebraic numbers are covered and eventually the set of exceptions can be made arbitrarily small.

[@vishalmishra3046]

*Another Example* IF A is Irrational and B is Rational (including any Integer) or vice versa, then *ALL* of these are *IRRATIONAL* , A+B, A-B, AxB and A/B.

[@RyanLewis-Johnson-wq6xs]

rational numbers can be expressed as a fraction whole number and repeating decimal.

[@Rougarou99]

Physics: π ≈ e ≈ 3, therefore π+e=6. QED

[@AnoNymous-dh2sv]

The real Beast channel.

[@greatgoldengoat2194]

I was just wondering how we knew that Pi was irrational!

[@diribigal]

KZbinr TheGermanFox turned that proof that Euler's number is irrational into a song. I think the title is "eulers number is irrational - proof & song"

[@sugarfrosted2005]

I prefer the proof of irrationality of the square root of 2 using infinite descent because it's a more generalizable strategy

[@vez3834]

Could you elaborate?

[@RossPfeiffer]

proofs sound harder than calculous . i should really study

[@Porcuponic]

I checked your channel and you write code. Proofs and code are essentially the same thing, it's just a set of steps based on a set of initial assumptions. In code those assumptions are a set of commands set by the language you're writing in. In math it's a set of definitions and axioms. Just think of math (or fields of math) as a language and write a program that outputs a conclusion.

[@vez3834]

​@@Porcuponic I don't write code and struggled with proofs in uni.

[@lolbajset]

isn't it a given that the sum of two irrational numbers is also irrational? if a number has infinitely many decimals, doesn't that remain true regardless of what number you add to it?

[@ComboClass]

Not necessarily. As a simple example, (square root of 2) and (3 minus the square root of 2) are both irrational but they add up to 3.

[@quentind1924]

If A is irrationnal and B rationnal, B-A will always be irrationnal. But A (irrationnal) + (B-A) (irrationnal) = B (rationnal)

[@markstyles1246]

You could, in fact, say the number 1 is Legen... ... ...wait for it... ... dre's Constant.

[@prosimion]

thank you

[@不只是谁所无名找小二]

3:13 what?

[@michaelcutler6118]

Cool cat that walked by. And I know that this is random but it is cool that you feed that squirrel in one of your videos. I like both cats and squirrels

[@lemonsys]

Root of 2 must be a fraction that can be infinitely simplified - kinda awesome

[@publiconions6313]

Question about the end: Are there any two irrational numbers that can sum to a rational?

[@ComboClass]

Yes, that's possible. Maybe I'll cover some examples in another video sometime.

[@jakobr_]

Yes. pi + (-pi) = 0. You can shift this around a bit to use any irrational number, and sum to any rational number, but the core idea is always the same

[@Kyle-nm1kh]

Have you ever seen the equation: e^(i*pi)+1=0?

[@BenAlternate-zf9nr]

Trivially so. For any rational Q and irrational R, define X = Q - R. Then irrational X, R sum to rational Q.

[@quentind1924]

A rationnal number plus or minus an irrationnal number will always be an irrationnal number. Therefore, take any irrationnal number (let’s call it A) and any rationnal number (let’s call it B). B-A will be irrationnal, A is an irrationnal number, but A+(B-A)=B, a rationnal number So for example, square root of 2 + (2-square root of 2)=2

[@RyanLewis-Johnson-wq6xs]

Log[2,3]=Log[2,1.5]+1 which is an irrational number

[@Luper1billion]

Would be so cool if pi plus e equaled a rational number

[@thomaskember3412]

I think a step has been left out in the proof for the square root of 2. If a perfect square is even, it may seem obvious that its square root is also even. But this proposition still has to be proved.

[@ComboClass]

We have to choose which things we take for granted when writing a proof, because you don't want to have to re-prove every basic thing (like that 1+1=2, or a+b = b+a) within every proof. It's true that in this episode I somewhat took the "if a square is even so is the square root" as a pre-existing assumption, although I did briefly explain the logic that if any (x) was odd then (x^2) would also have been odd. If you wanted to prove that part within the proof instead of using it as an assumption, you could use the definitions of even and odd number as 2k or 2k+1 for different integers k and do some arithmetic. But any proof will still have "missing steps" unless it includes dozens of other proofs that people consider already proven/obvious.

[@phyarth8082]

proof by contradiction or in Latin "reductio ad absurdum" = Simplification to absurd :)

[@hkayakh]

6:08 cat spotted in the background

[@stvp68]

I’m fascinated that in geometry, irrational lengths are really simple and very much not infinite. Makes me think that numbers are the real problem. 😁

[@ZapOKill]

no fire?

[@ComboClass]

I don't have fire in every video. With that said, there will be some in the next episode ;)

[@LordMarcus]

Y'all got a BIG cat.

[@ComboClass]

Dandelion is pretty big but it's mostly fluff

[@tpros6289]

I'm wrong in my head about math. I need someone to correct this for me, I don't care if you make me feel dumb. If sqrt2 =a/b then 2=a^2/b^2. Then a^2= 2b^2 . That tells my brain a is simply twice b. Logically I know that's wrong, partially because you have to still find sqrt2. In my mind I see two rectangles, one is defined as being twice the other, so I see a perfect square and a rectangle with one side having the same length as the square, and twice that on the other dimension. That tells me a = 2b which I'm sure is the first part I'm wrong about. Next, does that mean the fraction translates to 1/2 as a=2b or does that translate to 2/1? I've worn my brain out doing this, I must be tired.

[@BenAlternate-zf9nr]

If a² = 2(b²), that means a square with side length a has twice the area of a square with side length b. The ratio a:b is sqrt(2) ≈ 1.414.

[@Chris-op7yt]

we need a pure math that is not hindered by pesky properties of numbers such as integers, odds and evens, irrationality etc. you should be able to calculate the radius of a (real) circle whose area is four (4), without getting silly irrational number results.

[@rmdodsonbills]

So, you want math that doesn't have any numbers in it? Considering that the math we do currently is all based on the ability to count things, to do what you're asking, you'd need to sacrifice the ability to say that 1 + 1 = 2.

[@Chris-op7yt]

@@rmdodsonbills : a special branch of pure maths where things arent suddenly harder or you need approximations, because of particular values used. for example: if you say that square root of a number is some smaller number multiplied by itself, but you cant actually calculate it for many cases, then your everyday mathematics doesnt actually work, because you cant give an answer for square root of say 2.

[@rmdodsonbills]

@@Chris-op7yt Yes, and what I'm saying is that if you were somehow able to create a system like that, you wouldn't be able to use it to know how many apples I'm holding.

[@nova_supreme8390]

Ask it if it thinks the Earth is flat. It works for most irrational numbers but some are trickier. Especially pi is exceptionally rational when it comes to all things spherical so stay on your guard.

[@Chomta]

Drop inverse trigonometry

[@rogerkearns8094]

One is Legendriest number...

[@Grafyte]

4:33 -fraction- *ratio

[@EvanEscher]

41/29 is a VERY CLOSE APPROXIMATION of root2 Just like how 22/7 is an approximation of pi

[@MuffinsAPlenty]

Indeed! And it makes sense since 41/29 is a convergent of the continued fraction for sqrt(2). So yes, in some sense, it's one of the "best" rational approximations of sqrt(2), based on the size of the denominator.

[@felipedutra5276]

Is there any really important fraction? Relevant constants kind of always end up being irrational 😔

[@mr.redlazer9173]

15:02 nice

[@scientious]

The background music sucks. It's completely unnecessary and makes it harder to hear what is being said.

[@turingmachine7905]

Obviously, Zerone is the number between 0 and 1.

[@Kyle-nm1kh]

That's the sound a racecar makes as it zooms by

[@RedShiftedDollar]

Proof by contradiction is problematic in this context. The reason is that a very similar problem can be constructed in which you prove that infinity is irrational by contradiction when in fact the contradiction simply means infinity does not exist as a rational number. It also does not exist as an irrational number. This is to say that encountering the contradiction is not enough to prove irrationality. It can only prove that your assumption is not true. So if you assumed a rational solution exists, the proof by contradiction result becomes “a rational solution does not exist.” To affirmatively prove irrationality you must also prove that a solution exists and is irrational. However we already know no rational solution exists so we must assume an irrational solution exists and then prove the assumption true. Good luck. These are some reason why it is better to think of pi as a concept or an equivalence class of integer sequences rather than a number. It is also describable as an algebraic relationship which is also a concept. Like infinity which has an unbounded most significant digit, irrational numbers have unbounded least significant digits. The unbounded nature is what makes these mathematical things concepts rather than numbers. Infinity and irrational numbers have in common the fact that they contain an infinite amount of information. So in summary, proof by contradiction works, but it only works to prove the negation of the assumption. There does not exist a rational expression for sqrt2. This is not the same as claiming sqrt 2 exists and is irrational. Making this unjustified leap is one of the greatest logical blunders of modern mathematics.

[@SedgeHermit]

>mtg >mathematics Man this guy's interests intersect with mine.

[@Tenraiden]

I love you

[@dejaphoenix]

How do we know that these proofs by contradiction aren't just math shenanigans in disguise? For example, we can "prove" adding two positive integers results in a positive integer, and yet 1+2+3+4+... = -1/12. So, how can we say that an assumed simplest form fraction being able to reduce indicates a contradiction? Or that 2^a = 3^b doesn't make sense in some insane way?

[@ComboClass]

You can't actually prove that 1+2+3+4+... = -1/12 under the regular definitions of "+" and "=". That's a misconception.

[@ConManAU]

Mathematicians spent a lot of time in and around the 19th and 20th Century carefully constructing a mathematical system where these proofs work, and in general any proof undergoes a lot of scrutiny to check that it actually is valid. You can build a different system where 1+2+…=-1/12 is actually true, but in doing so you will give up some other useful property of arithmetic or logic.

[@tristantheoofer2]

⁠@@ComboClassyeah ive always wondered how tf 1 + 1/2 + 1/3... etc etc was equal to -1/12 because in my mind it should trend to infinity

[@Kyle-nm1kh]

​@tristantheoofer2 yeah it doesn't really. They had to change the rules to make it like that

[@tristantheoofer2]

@@Kyle-nm1kh yeah that would make sense to me, especially considering that theres no negatives in that equation at all

[@seionne85]

If we invented a counting system based on the concept of Planck units, are there any assumptions we could make based around that? Or postulates, I Don't know the lingo lol, just wondering if there may be some other form of mathematics that's being used 10 million years from now, and calculus, infinities, and irrational numbers are studied in history class lol but I probably should just watch your show sober instead

[@blobberberry]

I think what you're describing is a discrete field of numbers - which do exist, and are thought about! Some people think the existence of a smallest physically meaningful length implies that continua don't exist in reality. I remain unconvinced. But even if that were true, we wouldn't stop using the number line! It would just turn out to be an extremely convenient approximation of reality. And from the pure maths side of things, there are so many interesting aspects to the reals that will be studied basically for the rest of time, simply because it's fascinating.

[@rmdodsonbills]

So, you're supposing a counting system where fractions are impossible? Seems like that would be less than useful. I mean, we tried that with the Natural numbers and rapidly determined that most divisions were undefined.

[@producer_of_juni]

what is going on

[@SimchaWaldman]

11:23 I sent you an email.

[@CompanionCube]

A number is irrational when she let‘s her feelings control her

[@JohnBerry-q1h]

no muppets?? 🎼sun~~ -ny 🎶day!! 🎼keepin' the 🎵🎶 🎼clouds~~ a-way!~~🎵🎶

[@JohnWiltshire-yp9bj]

There is a rather obvious fallacy in all this. let xn = 141421356....(n digits)/10^(n-1) For all e>0 we can find a value of n such that (Sqrt(2) - xn) < e ALL values of xn are the ratio of two integers, and that remains true if we let e approach zero. The fallacy in the so-called proof is to consider the value of sqrt(2) to have an infinite number of digits while constraining the integers to a finite number of digits. It follows that irrational numbers do not exist and the whole concept is completely pointless (pun intended😁).

[@TheGrimStoic]

easy - you only need to ask it about the moon landing or globe earth

[@Kyle-nm1kh]

So youre a time traveler is what youre saying

[@MrLemonsChannel]

maybe pi is 1

[@ComboClass]

I don't know, I'm pretty sure it's at least 1.5

[@tristantheoofer2]

@@ComboClassnahhhh its gotta be atleast 1.618

[@Kyle-nm1kh]

Pi is one actually if you rotate the circle in the Z dimension by 90 degrees. Then the circumference would = the diameter. AKA it becomes a line

[@Gamr-bc6kp]

play it at 2x speed and he sounds like ben shapiro

[@mantacid1221]

Why would you curse us with this knowledge?

[@bracodescanner]

Hot tiktok today

[@jota5044]

You only proved that the square root of 2 cannot be rational, you didnt prove that irrational exists

[@robertveith6383]

A real number is either rational or irrational. The square root of 2 is a real number. If it was proved that it cannot be rational, then it follows it is irrational.