An Introduction to Mathematical Proofs 

Hate to break it to you, but 1 != 5040

I can prove that 6 = 3!

@@broorI can prove that 3!= 7

Assume that the numbers 1 and 7 indicate a summation of vectors, whose starting and ending points are same in both summations. Using the vector theorem, if you start the vectors from the end point of the nose of the number 1 until its foot, you get the summed vector that looks more or less like this - _\_ Again, use vector theorem for the number 7 taking the direction starting from its nose to its foot. We see that the sum is _\_ Both are facing the same direction. If the nose of number 1 is tilted downwards, the length may not be the same between both 1 and 7. However if its straight maybe we can then say its equal. Here we proved that 1 = 7, or _\_ = _\_ , which also means 1 || 7.

​@@broor I approve of factorial

Ah true, we did actually call this "proof by contra-positive" the same thing at uni but i see what you mean. But i guess then this negation of an implication is what i wanted to teach here anyways, just gave it a bit of a wrong name. Thanks for paying such close attention to detail!

@@0prcent No problem, I found you from your Jordan curve video, truly remarkable content

@@0prcent _"How to prove anything"?_ Quite the ambitious thumbnail, though imo out of reach for the introduction. Good 3b1b animation is not a cure-all silver bullet, after all.

​@@ultimaxkom8728 The title is obviously hyperbole.

@@ethanbottomley-mason8447 Yes, it's clickbait.

thank you so much!! (and thanks for the algorithm of course)

Yeah this would be really helpful 🙏🙏

​@@SlightSmile very would be an understatement

Really helpful for my financial status

aw thanks!

thx! so happy i got to use it in this video :)

oooh that's what it is I was digging it the entire time

the full music list is also in the description btw if youre curious about any of the other songs

all the best!

thank u

Baba is baba and not you

Baba is win?@@user-sk4kg4hr3k

@@user-sk4kg4hr3k now prove by contradiction (contra-positive)

baba

Les go you also recognised the bg music

I will bring 4 gpa.

good luck dude!

@@filipus098 thanks.

Good luck

good luck friend!

im not sure how accurate my portrayal of the proofs are but if youre intrigued i can definitely recommend it, especially with analysis topics the approach tends to give you a much deeper understanding for whats actually going on

best of luck!

Thanks for the feedback! And also yeah for implications specifically it might be helpful to review the truth table (which i probably should also have covered in this video now that i think about it) for that operator specifically, then it should make sense why we want both the result and the condition to be wrong.

Eh its like assuming that something you want to be true isnt and then showing that leads to an absurdity. Eg: Not INF many primes, ie can be expressed as {p1, p2, ... pn} What about some val k = p1*p2*p3...*pn + 1. This is clearly larger than any element in our set of primes; so it would be composite. By fundamental theorem of arithmetic, it can be written as a product of primes, since it is not prime itself, but composite. Take any one of those nums from that set of primes, and divide our k value by it; it should be fine. But it's not, because we are adding 1/(some prime >= 2). So our k should be able to be divided by a prime, but it isn't. This is absurd, so the set of primes we claimed actually isn't all of the primes. (Here Euclid made k bigger than all of the primes by multiplying them, and adding the 1 to set it up so it wouldnt divide evenly, as the smallest prime is 2). I have also seen this a lot with Well ordering principle. My god, that principle was abused the hell out of in the first half of this quarter and ik i still need to remember the proofs for the final too. Bezout's said: ax + by = min{ax' + by' E N: x', y' E Z} and we could show there is a min by making x' = a, y' = b, = a^2 + b^2 which is > 0, since when doing the gcd(a, b) [which is what youre proving in bezouts], at least one of a or b is nonzero. Then we have this ax + by and set up an inequality with gcd(a, b) one way by using the definition of gcd(a, b), and an inequality with the division algo. In the division algo part, you end up getting some "r" value from the bq+r that *would* be the new min element of the set we defined. But we already defined ax+by as the min element, so this r cannot exist in N. Hence, it = 0. [This is just a layout of it. I liked watching learnifyable for these proofs] Div algo proof itself used this idea also when assuming for sake of contradiction that r >= b, so r - b >= 0, but this would be min element. and for showing uniqueness. Oh yeah or irrationals: assume rationality, show impossible. Ie, sqrt(2) = a/b, a, b E Z, and b != 0. Then doing some ops: 2b^2 = a^2. Well since square numbers are only div by 2 if they are even, then a is even. 2b^2 = (2k)^2, k E Z 2b^2 = 4k^2 b^2 = 2k^2. Now we have the same thing, and can do that rule to b. But that's kinda crazy, is it not? we can keep expressing it in terms of 2*itself? We keep on dividing the numbers in question by 2, essentially, like how we said a = 2k, so k = a/2. This is crazy, at some point if we keep dividing by 2, we will not have an integer, then the equality could no longer hold true. This is also known as a disproof "By infinite descent". I know it because the name is hella cool. TLDR: Unlikely u will come up with these early on yourself. as you learn, it is better to see examples than to struggle through with it yourself. All skills require some degree of memory.

aw thank you! glad that helped with not making it seem too overwhelming

blender for the 3d animations, manim for the math/latex animations and then edited everything in davinci resolve

i agree the whole thing about the names of different parts of an implivation even got myself confused while i was making the video, should have made that clearer. i dont think the different usages of if only if would be that confusing to a beginner since it wouldnt really matter to them in the first place and i was thinking about also including negations somewhere but it honestly felt a bit too intuitive for me to specifically talk about them, maybe thats not the case but lets see what other people think as well thanks a lot for taking the time to write down mistakes/things to improve like these, ill take everything into consideration and i really appreciate it!

manim for everything 2d/math stuff, blender for general 3d animations and editing in davinci resolve thanks!

@@0prcent impressive work :) waiting for more

@@0prcent Thanks man

There are some things i didnt cover here but as a next step it would probably be best to take a look at some actual proofs (my recommendation would be proof-based book tackling a beginner area of mathematics, i started with analysis back then). I could do another video about advanced proof techniques/tricks but my next video will most likely be about something very different and after getting the basics of proofs down, usually the hardest part is just having an idea

i havent really heard of any solvers being covered at uni. not sure what kinda course you would have to attend for that but it would certainly be an interesting topic

@@0prcent Looks like there are only independent efforts by some professors ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-FDx0nXFQloE.html and there is also the Xena project. I think it might be useful to include it in education because it looks like there is future potential for it. Imagine being a first year student playing that game adam.math.hhu.de/#/g/leanprover-community/nng4

oh just looked it up and i didnt know that "whole numbers" was an ambiguous term. integers is of course what i meant but might be a bit of a translation mistake then, thx for making me aware of this

bros a hater

@@filipus098 what that mean!

thx!

proof: it was revealed to me in a dream

@NibberPancake || @0prcent The statement 2ⁿ > n² ⇒ n ≥ 0 means that if 2ⁿ > n² then n must be positive, not that if n is positive then 2ⁿ > n². A mistake was still made though. He said that 2ⁿ > n² ⇒ n ≥ 0 which is false. This is because 2ⁿ > n² is true at n ∈ (-0.76666469..., 2) ∪ (4,∞), this includes all negative numbers within (-0.76666469...,0). His mistake was assuming that n < 0 ⇒ n² > 1 instead of n < 0 ⇒ n² > 0. You can visually see this in Desmos by draphing n² and 2ⁿ and seing where the curve of 2ⁿ is above that of n², typing 2ⁿ > n² will show you directly.

Hey, if you're interested in math and science please check out my channel! I am new but soon I will make more content!

i like what languages like coq and agda are going.

i know exactly what you mean, i also come from a programming background and the biggest thing im missing in maths is descriptive variable names. in lots of longer proofs its impossible to track all of the A,B,C,D,x,y,z and 10 different greek letters all being used in conjunction.

remember, n is always an integer

Hab ich auch überlegt

(and it seems this one got too) edit: this one only partially