Bijection Proof (a taste of math proof)

First part: N and Z have the same number of elements • Do We REALLY Have More...
And in this video, we will show that f is a bijection by using the definitions.
Note: "one-to-one" is called "injective" and "onto" is also called "surjective"
#bijectionproof #mathforfun
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[@blackpenredpen]

Box or +C?

[@command0_]

Box!

[@SimchaWaldman]

f(n) = [1 - (-1)ⁿ(2n + 1)] / 4

[@kwirny]

Box!

[@mrbobtehbuildah]

Box!

[@erikkonstas]

There is this feeling after a long run... Your mind has traveled many miles... It's now tired and exhausted... It has proven something big... So now it needs to relax... And what feeling is more relaxing... Than the flow of a current... Echoing its soothing sound... Passing through like a river... Engulfing you with calmness... And sinking you like a waterfall... Into its luxurious realm... This is the feeling of the pen... Releasing its ink onto the paper... Drawing a perfectly square box... And filling it with straight, gentle strokes...

[@Debg91]

The girlfriend of one of my mates at uni was studying mathematics and she did a very interesting work about the maths behind origami and that stuff, supervised by the chief professor of the algebra department, known for being as brilliant as eccentric. When she gave him a draft, he said 'it's perfect'. Then she asked if it didn't need a conclusions section and then he replied: 'we're mathematicians. Do you see that box at the end of the proof? That's the conclusion'.

[@kais.1684]

your enthusiasm for the subject makes mathematics more enjoyable for me. Thank you!

[@kwirny]

That's really cool, a box is even more satisfying than a q.e.d in my opinion

[@BrainGainzOfficial]

Loving this kind of content lately! Keep it up! 🧠💪🏼

[@ahzong3544]

That box at the conclusion means that everything you write for the proof is outside the box.

[@nikolastevic2278]

Thanks, this really helped me understand functions better as I'm just learning them in school. Looking forward to video about inverse functions :)

[@Rekko82]

Have a nice life!

[@paulhaso]

Your silly jokes and gestures just make my day

[@andrewm4731]

I watched this before am exam and met a question in that exam where I applied the knowledge. I could recall him explaining. It felt amazing. Keep up

[@Red-ot8lv]

Dont typically comment on videos but man i love your enthusiasm!

[@blackpenredpen]

I appreciate it!

[@thedeg123]

Killer video! Well explained and Helped me a lot :)

[@Ayesha_01257]

Well Explained! Thanks a lot 💕

[@leeming1317]

You remind me of a really dear student I had first year college calculus, He was so nice and loved mathematics, He had limited patience but he helped me a lot, even though I was slow. He would solve problems very interesting out of box ways, He sometimes had to explain to teacher! haha

[@justfahid5254]

y'all be making it more hard than it needs to be

[@Fidder492]

difficulty is relative. Ignorance is bliss.

[@andrewm4731]

You are a good teacher indeed

[@jayapandey2541]

The coolness in Real Analysis diverges.

[@blackpenredpen]

Are you one-to-one or onto?

[@xaxuser5033]

i m a bijection

[@Rekko82]

I dunno. I like women. Women - not men although the word "men" is included in "women". I just like them because they don't have beard (at least much).

[@mohamedhadjalla]

Good luck my friend ❤👊

[@handymandycraftsfashion7472]

I learned more from a 11 minute video rather than one and half an hour of my class😇

[@anabang1251]

First week physics/math students be like: Wow that's difficult, better not learn it and make fun of people using those terms. *exmatriculates furiously*

[@japotillor]

I enjoyed analysis, being a Latin geek, I use QED, quod erat demonstratum (though the box is cool too). I taught my HS students this...they hated me :p (our analysis teacher made us promise if we taught algebra that we would cover injection and surjection)

[@thiantromp6607]

It's quod erat demonstrandum though, not demontratum.

[@Albkiller22]

Discrete math flashback coming out... lol

[@angelmendez-rivera351]

f(n) can be expressed as [(-1)^n][n + (1 - (-1)^n)/2]/2 = [(-1)^n][(2n + 1)/4] - 1/4. The inverse function, call it g(n), can be expressed as sgn(n)[2n + (1 - sgn(n))/2] = 2|n| + [sgn(n) - sgn(n)^2]/2. Can you find an explicit formula for a bijective function between the natural numbers and the rational numbers?

[@angrycat2789]

This was very helpful to me as I’m starting to self study analysis. Thanks!

[@Patapom3]

Great!

[@DiogoSantos-dw4ld]

From an English guy injective, surjective >> one-to-one, onto Great video though! I would've loved to see a lot more university content maths before I started my degree, love your and Dr Peyam's videos

[@BigDBrian]

"onto" ⊂ "one-to-one" because it has no symbols that aren't present in "one-to-one"

[@TheNachoesuncapo]

YAY!

[@xaxuser5033]

u looks great on the miniature ! !!!!!

[@Kaczankuku]

First I thought you proove that every function needs TWO parameters to give result of that function. For example n! apparently needs only one parameter, but not so many people see that it's multiplied every natural number instead of every tenth one if it's needed. That way 1 is the second parameter of standard factorial. Another way to proove the main statement is to express n! as gamma function which involves Euler's number in the integral definition. Again, Euler's number is the second parameter beside x (if x replaces n as a variable).

[@angelmendez-rivera351]

Kaczankuku Numbers are not parameters. The standard factorial function is only a function of one variable: by definition, of course it is every natural number. If it were every tenth number, then it would be a different function altogether, the 10-fold factorial function. The multi-factorial function does have two parameters, but as the factorial is a special case, it only has one. There is no second parameter. My guess is that you still have not learned what parameters are or how they work.

[@Kaczankuku]

@@angelmendez-rivera351 It was easier to explain using word "parameters". No matter how you call it, function needs two numbers to be defined. It is just what I intended to comment but I see you haven't agreed at all due to terms. PS. I wish you understand this case some day.

[@angelmendez-rivera351]

Kaczankuku *It was easier to explain using word "parameters".* No, it is not. "Variable works just fine, since that is the terminology everyone already uses. My point still stands. The Gamma function has one variable and one variable only. e is a constant, thus it is not a variable. *No matter how you call it, functions need two numbers to be defined.* I already proved you don't with the factorial example. *I wish you understand this case some day.* I already understand it. You're the one who does not. Come back to me when you have passed algebra 1.

[@rajkecy2223]

I have a request. Sir, will you please make a series on maths quiz problems for grade 10.

[@jgc9199]

You might need to be more specific lol.

[@FlavioPortelaVF]

A parte mais difícil não é essa é mostrar que a função tá bem definida. Faz um vídeo ensinando como mostrar que uma função é bem definida. Obrigado.

[@cks7719]

I'm having a difficult time with understanding proofs. I'm working on a problem that states n is equal to 10n. How would I explain that?

[@emperorpingusmathchannel5365]

Well it is pretty self evident why *this* function is a bijection. Define odd n= 2k-1 where k belongs in the set of the naturals Define even n=2k Note: all integers below 0 equal to -n (ie: -abs(z)=-n) Note: all integers above 0 equal to n (ie: abs(z)=n) g(k)=2k-1 and h(k)=2k There exists no integer in the set g(k) such that g(k)=h(k) since 2k-1=2k is untrue because it requires -1=0 to hold true. If you plug 2k-1and 2k in to f(n) you'll get f(2k-1)= -n f(2k)=n Since the domains are mutally exclusive and the ranges are mutaully exclusive, you have sufficiently proven that it is a bijection.

[@connorhorman]

That seems to have proven that the functions are injections, not necessarily bijections. You also have to prove that the union of the function's ranges are the codomain of the function

[@yaleng4597]

Well, I prefer QED(Quite Easily Done)

[@yaleng4597]

Good, by the way, I have just started predicate logic.

[@connory02]

so cool

[@MathZoneKH]

it hard to understanding, but I am trying to get it!

[@abirr1780]

Ok so we proved that the function is bijective. Why is this important?

[@nyx211]

Matt Elwer Bijective functions have inverse functions that undo the mapping of the original function. So, in a sense, the function and its inverse "preserve information" when applied to its inputs. This is a useful property to have if you want to reversibly encrypt data without destroying it. You can also show whether the function's domain and codomain are the same size or not. For example, N and Z are the same size, but R is larger than Z even though they're both infinite sets.

[@povilasdapsys7765]

Do a video on 8/2(2+2)=?

[@xaxuser5033]

Let n an integer > 1, and p a positive prime number. Show that: If (n divids p-1 and p divids n^3 -1) then : 4p-3 is a perfect squar. plzz any help

[@Skyler827]

"nobody will see this" 9:21

[@obinnanwakwue5735]

Well what do we have here!

[@oabcouqnguyen1519]

Q vs N plssss?

[@LavkeshMaths]

Good

[@mdajiruddin8490]

Sir u will proved that range =codomain For onto

[@christianalbertjahns2577]

That's what BPRP did

[@sloppajoe9765]

Hello May I ask why can not we treat d/dx as a fraction? Thank you:)

[@abirr1780]

SloppaJoe9 fractions are traditionally part of a whole number like 1/2. D/dx is just asking you to take the derivative of a function. If you meant dy/dx then you can treat is kind of like a fraction. This is usually with covered in a differential equations class and sometimes with implicit differentiation.

[@rishabh9389]

For fraction Both numerator and denominator should be Integer. Even numbers like root 2 or π is not fraction

[@angelmendez-rivera351]

d/dx is an operator, so it is not a fraction. d is not an element of anything, and dx is a differential element, but division by differential elements is undefined, so you cannot divide by dx to obtain d/dx, even if d were something mathematical. In reality, d/dx is nothing but notation to represent a linear operator.

[@marocplays8110]

I think you could have just prove that there is only one solution to the equation f(n)=k without going over two steps, I wonder if that’s possible

[@WarpRulez]

Why use such a _trivial_ bijection as the one between natural numbers and the integers? A much more interesting thing to prove is a bijection between the natural numbers and the rational numbers. And extra points if you can prove a bijection between the natural numbers and the algebraic numbers.

[@angelmendez-rivera351]

WarpRulez Why not go further and prove a bijection between the natural numbers and the set of all finite subsets of set of the natural numbers?

[@WarpRulez]

@@angelmendez-rivera351 Let's start with the algebraic numbers for starters.

[@angelmendez-rivera351]

@@WarpRulez Premise 1: N is equinumerous to N^2. This true by the theorem of Cantorian diagonalization. Premise 2: For all m > 1, N^m is equinumerous to N^(m + 1). Proof: fix the first m - 1 entries of the m-tuple, and the last entry can be mapped bijectively to a corresponding 2-tuple by premise 1. This produces an (m + 1)-tuple. and the first m - 1 entries of the m-tuple can be mapped to the first m - 1 entries of the (m + 1)-tuple using the partial identity map, which completes the bijection between N^m and N^(m + 1). Premise 3: By conjunction of premise 1 and 2, N^n is countable for all n > 0, since the conjunction is simply the conclusion of a proof by induction. Premise 4: The union of the sets N, N^2, ad infinitum, is countable. Proof: there exists a theorem which states that the union of countably many countable sets is a countable set. By premise 3, all the sets in the union are countable, and since n is a natural number, there are countably many such sets, hence satisfying the conditions of the theorem, hence the union is countable. Premise 5: There exists a bijection between all the polynomials of degree d and N^d. Proof: by the Fundamental Theorem of Algebra, the solution d-tuple of a given polynomial of degree d expressed in its simplest d-tuple of coefficients is unique, hence there exists a bijection between such d-tuple of coefficients and the solution d-tuple. Since a polynomial can be characterized by its d-tuple of coefficients, because if the d-tuple is different, the polynomial is different accordingly, this completes the proof. Premise 6: By the conjunction of premise 4 and premise 5, there exists a bijection between the set of solutions of all polynomials of finite degree and the union of the sets N^n starting at n = 1. Call this set of solutions of all polynomials the set S. Premise 7: The set of all algebraic numbers is a subset of S. This is true by the definition of complex numbers. Premise 8: By premise 6 and premise 7, the set of all algebraic numbers is equinumerous to the union set of N^n starting at n = 1. Premise 9: By the transitivity of the equinumerosity relation, and by premise 4, the set of all algebraic numbers is equinumerous to N. Conclusion: The set of all algebraic numbers is countable.

[@royaneone1209]

Do Real Analysis videos

[@GIFPES]

Well, if "F" from "N" to "Z" is injective, it says that there are no leftovers from one to another set. Thus, regarding cardinality, the amount of numbers or info in each set is just the same?????

[@reetasingh1679]

Infinity can be famously unintuitive. Here it intuitively does feel wrong, but a rigorous maths proof just cannot be wrong

[@GIFPES]

@@reetasingh1679 Sure that I am just making a naif inference, I am not stating for that for I have not made this proof myself, yet, for a first-degree function, odd kind of, such as f(x)=y=Kx; K belonging to the integers, so it would be symmetric to the point (0,0), I think, intuitively that there will be no leftovers inside the counter-domain for all domain values will achieve the counter-domain one to each one, so we would get an injective function. PS: I do not know if the english word I used is correct to call the "X" set for domain and "Y" set as counter-domain, because in portuguese we say "domínio" and "contra domínio", in an F(x)=y=function formula.

[@reetasingh1679]

@@GIFPES There is no right or wrong in your choice of naming... Whatever you end up calling your domain and counter domain by, will be what is correct for you. But your names are indeed conventionally correct.

[@alexdemoura9972]

Injection is any relation between two sets let's say Domain set and Counter-domain set (or Image set, if you wish). The Bijection is an one-to-one Injection, no leftovers as you said. And that is used to prove the same Cardinality (size) of those two sets, as BPRP made in a previous video between N-set and Z-set. Injection can be understood as an airplane with passengers, if all passengers are seated and all seats are taken then that is a Bijection relation between passengers and seats. In this video now, BPRP is proving that the function he used to prove the same Cardinality of N-set and Z-set in previous video, is really a Bijection function. What he did, ending with an END OF PROOF tombstone character. Otherwise if there are any leftovers in any of the sets the Injection is a Surjection ('Sobrejeição' in Portuguese???) or a Non-surjection, meaning there are vacant seats on the airplane (a good flight), or there are passengers standing in the aisles and kids on the lap of their mothers with all seats taken (a terrible flight).

[@alexdemoura9972]

By the way, functions of Surjection or Non-surjection cannot make a satisfactory proof of Cardinality of sets, even if the sets have same number of elements. Even if the airplane have a number of passengers as the same number of the seats, some passengers can be standing on the aisles or kids can be on mothers' laps, making some seats vacant - those passengers and seats are becoming elements not related to a correspondent element in other set.

[@mathsquare1]

when f(x1)>0 but f(x2)

[@MrLulinous]

i have an exam today

[@omargaber3122]

Happy Eid. For Muslims, non-Muslims and the whole world💓❤💚💖💗 Peace prevails in the world😍

[@MaksymCzech]

There's a huge bijection in my pants right now!

[@sadanandmukkanawar640]

How to integrate [ (x+2)+ x.logx ]/x

[@angelmendez-rivera351]

Sadanand Mukkanawar [(x + 2) + x·log(x)]/x = (x + 2)/x + log(x). The antiderivative of log(x) is trivial, and (x + 2)/x = 1 + 2/x, whose antiderivative is also trivial.

[@supernovaw39]

eeeniway

[@ermacster8009]

I want to be as good as you , that's why I study even in summer by myself

[@rishabh9389]

Before coming to this channel I use to think only Indian, chinese and some other Asians are good at maths This is the number of non Asians here 👇👇

[@GIFPES]

As a brazilian I must state in the name of my people: when we are not making a barbecue or playing samba, we are within math in our minds!!!...lol....

[@rajeshdas6539]

Sup rishav pant Tired of playing cricket all day long?

[@Peter_Schluss-Mit-Lustig]

@@GIFPES same for me here in Germany if we are not drinking beer or working we are trying to prove the continuum hypothesis and riemanns conjecture

[@GIFPES]

@@Peter_Schluss-Mit-Lustig eheheheeheh....

[@bilmiyorum2247]

He

[@TheScienceBiome]

Loved this video! Also, I created a video on how Calculus helped fight HIV, and it would mean a lot to me if you could check it out.

[@XDjUanZInHO]

Fourth :D

[@Lastrevio]

we do these in 10th grade

[@rajeshjangid7224]

r u chinies sir

[@Rekko82]

I am not gonna waste 11+ minutes of my life for this video. Decades ago math used to be useful and about numbers. Now it is just ks and f(x)s and shit and never useful (unless you become mathematician and teach it to students who never use it).