You Need To See This At Least Once

[@BriTheMathGuy]

🎓Become a Math Master With My Intro To Proofs Course! (FREE ON KZbin) kzbin.info/www/bejne/aZTdmJl-irGNedU

[@unbelievable961]

Sir , please can you recommend a book for whole geometry with there proofs...∞

[@SunHail8]

just a false logic: algebraic equation doesn't mean you can plug in any X you want to satisfy it.

[@imincent1767]

bro really dded pythagorean theorum

[@ManojkantSamal]

According to the mathematicians " zero means nothing " is not right....Zero is something, The value of zero is in the dale of experiment..... So, 1-1=1-1 1-1=(1^2)-(1^2) 1-1=(1+1)(1-1) Let 1-1=R So, R=2×R R/R=2 1=2

[@BangkokBubonaglia]

Over the years I have seen many tricks to make 1=2. They all involved either: a) division by zero. Or b) jumping a branch cut. For example, if you want to make -1 = 1, just square both sides, and then take the square root of both sides. (Hide it in symbols of course). Once you have -1 = 1, you can do anything else you want through scaling and translation.

[@methatis3013]

So either division by zero or applying reverse functions to non-bijective functions Eg sin(2π) = sin (0) => 2π = 0

[@poubellestrange7515]

There is also one more subtle way that I know of, that is making use of algebraic properties in reals that do not hold with complex numbers, like sqrt(ab) = sqrt(a)sqrt(b) and (a^b)^c = a^(bc). For example, sqrt((-1)(-1)) = sqrt(-1)sqrt(-1) and (e^(2ipi + 1))^(2ipi + 1) = e^(2ipi + 1)^2 are both false statements.

[@SpeedyMemes]

@@poubellestrange7515 I have seen the e^(2ipi +1) trick on mathologer's second channel - is that where you found it?

[@poubellestrange7515]

@@SpeedyMemes Yes, however, that (a^b)^c = a^(bc) doesn’t hold is actually a “well” known fact that complex analysis books usually cover.

[@funtamilanallrounder2700]

Then we can add 2 to both side and make 1=3 , +1 to (2=4) then divide both side by 2 , and get 1=2 , also this prove that 1=2=3

[@derechoplano]

About 35 years ago, my Math high school teacher did this 2=1 trick on the blackboard. It took me some time to figure out the problem. My teacher congratulated me in front of the class. Fond memories. Thanks. You made me smile.

[@BriTheMathGuy]

That is awesome!

[@jacklee8385]

This luckily isn't too hard to find what's wrong

[@Mono_Autophobic]

35yr ago 😶 How old r u now

[@YoursTrulyAkr]

@@Mono_Autophobicaround 45-50

[@comradelovespain5714]

@@YoursTrulyAkrman was in high school at the age of 10?

[@eltolufe2549]

A spanish divulgator did things like this some months ago. it's a really good feeling when you figure out what went wrong.

[@BriTheMathGuy]

Right?!

[@co2kp639]

Quantum fracture reference

[@diamante8864]

@@co2kp639 yesssssssss

[@el_saltamontes]

@@co2kp639 Que grande Guantum Fracture

[@jcano7526]

Crespo xd

[@solveeasyacademy186]

A mathematician once said anything could be done in mathematics as long as we are all ready to beat the consequences. I love this.

[@sumdumbmick]

and then Whitehead and Russell boldly proclaimed they'd succeeded, and Kurt Godel showed them they were actually just illiterate morons living in a fantasy world of make believe and abelian groups.

[@MichaelClark-uw7ex]

If you do enough math tricks you can make any equation equal a ham sandwich.

[@Masonova1]

This is the single best argument I have ever heard against referring to dividing both sides of an equation by a common term as "canceling"

[@mcr9822]

I had a math teacher that was a hardass about that. He insisted that we use “divide out” or whatever, to remind us that we aren’t just ignoring those elements like they never existed. We’re doing division.

[@PC_Simo]

2:55 That’s kind of like, how I first approached the problem (before knowing the formula), upon hearing the anecdote about Gauss solving the 1+2+3+…+99+100, in a matter of seconds: Realize that you can add the terms, in pairs: (1+100)+(2+99)+(3+98)+…+(49+52)+(50+51), and that there are 50 such pairs that all add up to 101, each; thus: 1+2+3+…+99+100 = 101*50 = *5050* 💡🙂.

[@PC_Simo]

7:02 From these expressions, it’s automatically clear, why the derivative of sin(x) is cos(x). Sum rule: (f+g)’ = f’+g’ (x^n)’ = n*x^(n-1) -> (x)’ = (x^1)’ = 1*x^0 = 1*1 = 1 f(x) = sin(x) = x-(x³/3!)+(x^5/5!)-(x^7/7!)+… f’(x) = 1-(3x²/2!*3)+(5x^4/4!*5)-(7x^6/6!*7)+… = 1-(x²/2!)+(x^4/4!)-(x^6/6!)+… = *cos(x)*

[@Inspirator_AG112]

*AT TIMESTAMP [**00:48**]:* If the variable's coefficient on both sides of the equation are different, then that variable is zero.

[@OptimusPhillip]

The funny thing is that the first problem could still work even after the dividing by 0, if you just handled the math a little differently. a+b=b, subtract b from both sides, and you get a=0. 2b=b can also be a true statement if and only if b=0.

[@funtamilanallrounder2700]

-1 = 1 because squaring both side give 1 Therefore , Then we can add 2 to both side and make 1=3 , and then +1 to (2=4) then divide both side by 2 , and get 1=2 , also this prove that 1=2=3

[@kromydas5063]

@Bion simply, the converse of a statement is not necessarily true. an easy example is: if person A is thirsty, person A drinks water. If person A drinks water, person A is thirsty. The second statement is not necessarily true.

[@goldenwarrior1186]

@@kromydas5063 if person A is thirsty but there’s no water nearby and no way to make water, they won’t be able to drink water. First statement disproved

[@kromydas5063]

@@goldenwarrior1186 its just an example. Assume there is water smhhhhhh

[@funnysecs4299]

I didn't knew you could prove Pythagoras theorem like that. Its very easy

[@BriTheMathGuy]

Thanks for watching!

[@urisinger3412]

what

[@rcb3921]

You should check out this Mathologer vid: kzbin.info/www/bejne/pl6ThIKNl9-Ij6s -- some really nice proofs of Pythagoras. My personal favorite is at 6:16

[@dmace14]

If you write all the natural numbers backwards and have infinite zeros to the right of them: 1 = 100000… 2 = 200000… 3 = 300000… … 9 = 900000… 10= 010000… 11 = 110000… … You can do the same thing that you did with all the real numbers between 0 and 1 to find that there is a number you don’t have on the list

[@annabashline9882]

.0, .1, .2, .3, .4, .5, .6, .7, .8, .9, .01, .11, .21, .31, .41, .51, .61, .71, .81, .91, .02, .12 … Just flip the numbers of the real series around the decimal place (1.0->0.1 24.0->0.42 345.0->0.543 (completely random examples (assuming their are no patterns with these numbers that I don't know about))), Showing that all numbers between 0 and 1 that can be represented as decimals (including repeating numbers if numbers such as 1+10+100+1000... are real numbers) could be paired with a real number showing that they are equally as large as the real numbers (I am not a mathematician so I could be very wrong).

[@MiteshJethawa]

The Thing What You Just Demonstrated Is Proof That All Rational Numbers is Equivalent to Set of All Natural Numbers . But Real Numbers Include Numbers Like π , e , π + e , e/2 , e/3 , π/12 etc supposedly Infinite of them and It's these Numbers Which Make The Difference You cannot Write π/4 (which is in between zero and one) as a Number Like This Because It is Non-Repeated Infinite The Decimal Numbers will never end And If You Decided Let It be First "n" digits of the number then what about the original number which was in it's Place . Hence We can Conclude That Real No > Rationals = Natural Numbers

[@boomerzilean]

9:14 well hold on, if you knew what cos(π) and sin(π) all along, then you could have just computed e^iπ in the first place... knowing what cos(π) and sin(π) is, is by definition equivalent to knowing what e^iπ is.

[@spoookley]

3:54 a counter point to this infinity being larger than every whole number: mirror the number over the decimal point. So 0.1 becomes 1 & 0.095 becomes 590. Since you can do this with every number between 0 & 1 that disproves that it’s a larger infinity, in turn making them equal.

[@dw6561_]

Your argument assumes that every number between 0 and 1 has a finite decimal expansion, which is false. 1/3 is a simple counterexample.

[@idrisShiningTimes]

@@dw6561_ This ^

[@PhilosophicalNonsense-wy9gy]

Both the natural number set of all numbers and the real number set of numbers between 0 and 1, can have an infinite expansion. The natural number set has infinite expansion on the left side of the decimal point, the real number set of no.s between 0 and 1 has infinite expansion on the right side of the decimal point. Thus, both these infinities are equal as long as you keep the real number set as the numbers between 0 and 1. To categorize the scales of different kinds of infinity, lets consider in how many different ways they are infinite. In natural no. set of all positive integers, two criterion are applicable. 1. The value of the number can keep increasing endlessly 2. The set has infinite expansion towards the left side Now lets compare to the second set: 1. The value of the numbers can be of infinitely different values between 0 and 1 2. It has infinite expansion towards the right side But when we consider the real no. set of all positive numbers, there comes 3rd criterion: 3. Just like the 2 criterion are applicable between 0 and 1, so are they applicable for between any other numbers, thus infinity of those numbers follow those two criteria Thus, the N set has 2 criteria while the R set has 3 criteria, thus R set of all positive numbers is a bigger infinity.

[@PhilosophicalNonsense-wy9gy]

​@@dw6561_the natural no. set also has infinite expansion but to a direction opposite to that of the R set of no.s between 0 and 1

[@godofmath1039]

​@@dw6561_ You're right. Unfortunately, KZbin commenters are stupid

[@PC_Simo]

1:00 Between the 4th and 5th iterations of the equation, we divide by (a-b), which equals 0; if, indeed, a = b. So: (a+b)(a-b) = b(a-b) |:(a-b) (a+b)(a-b) = b(a-b) |:0 -> 💥💀

[@PC_Simo]

2:00 They are equal: (a+b)² = a²+2ab+b² 4*1/2*a*b = 4/2*ab = 2ab Then, Pythagoras tells us that: a²+b² = c². So: (a+b)² = a²+2ab+b² = a²+b²+2ab = c²+2ab = c²+4/2*ab = c²+4*1/2*a*b -> (a+b)² = c²+4*1/2*a*b

[@Dsubminiature]

dividing by zero is such a barrier that prevents counter intuitive math like 2=1. it pops up everywhere like a vertical line, where all values y have the same x. the slope is undefined or 0/0. in fact, math really enforces the "zero barrier" that even if we try to define it, we get a 0 counted numbering system. in the end though all im trying to say is any problem where there could be multiple answers, there is a division by zero something. like the 0th root of x

[@martinjoster3282]

Another way I thought of to do that last one is; pk is prime pk must divide N N/pk must be a whole number N/pk = (p1*p2*p3...*pk*...*pn+1)/pk Which is also... N/pk = (p1*p2*p3...*pk*...*pn)/pk + 1/pk We're dividing and multiplying by pk, so we cancel it out... N/pk = (p1*p2*p3...*...*pn) + 1/pk p1*p2...*pn is whole, as it's a multiplication of whole numbers However, this still leaves us with 1/pk, which can only be whole if pk = 1. 1 is not prime. There is no prime "pk" that divides N.

[@tomass420]

watching this video I remembered one interesting fact that pi can not be written as a/b (fraction), but pi is defined as circumference over diameter (C/d)

[@Grizzly01]

Meaning that at least one of C or d will not be an integer.

[@waynemartins9166]

@@Grizzly01 perhaps you meant "at least one of C or d will not be a rational number"

[@vinayegaddu2317]

Pi/1 lol

[@logan9093]

@@waynemartins9166 well he's also right, you can't have both be an integer

[@waynemartins9166]

@@logan9093 Yes he is right. What I did is to give a stronger condition (that also includes his). This is because integers are also rational numbers. Though the condition I provided can be made better, perhaps.

[@alflyover4413]

I wrote "if a = b then 2 = 1" on the whiteboard in an engineering lab once along with the equations. I came back a day or two later, and found the board covered, literally covered, with the equations. The characters were written with oddball strokes and weird fonts, apparently to keep them straight, but for every one the conclusion was inexorable: If A = B then 2 = 1. That is a solid example of what "division by zero is undefined" means. I have a notion that doing anything else with the equation once it has been reduced to 0 = 0 is suspect, maybe more than a little bogus, but I don't have a good feel for how to attack the proof.

[@DeJay7]

There is a nicer way to prove the existence of infinite primes. It's very similar, but it says that all the prime numbers (we assumed they are finite) must divide their product, so because N is their product + 1 they are 1 more than any one of their multiples, so no prime divides N. Since N is NOT PRIME (we assumed some prime was the largest prime), it is composite, but due to the fundamental theorem of arithmetic, every composite number can be written as a product of primes, but we just said no primes divide N. That's a contradiction, so our assumption that primes are finite is wrong => there are infinite primes

[@waynemartins9166]

Yeah, what he showed is that N has now become a new prime and so the logic can keep on going forever and never ever hitting the true last prime.

[@DeJay7]

@@waynemartins9166 No, N did not necessarily become a new prime. It could be a composite number divisible by primes greater that what we assumed to be the largest.

[@waynemartins9166]

@@DeJay7 ok I get you, I wasn't clear enough, what I meant is that N is prime relative to "all primes listed" and so must either itself be a prime or imply N is a multiple of another higher prime(s) (not listed) where in both cases contradicts the first argument that all primes where listed

[@DeJay7]

@@waynemartins9166 Okay I agree now And for your information, we call numbers that are 'prime relative to each other' as 'co-prime'

[@waynemartins9166]

@@DeJay7 I can sense the rigour in you, so thorough, nothing left unchallenged, keep it up

[@TerribleTonyShow]

3:55 so essentially lim x → ∞ then have a list that maps one to one with all the integers 1*x^(1/x), 2*x^(1/x), 3*x^(1/x)... everything will be 0 converging to 1

[@pastaplatoon6184]

A quick question regarding the uncountable infinity between 1 and 0, I've seen this multiple times and understood it pretty well so I never really questioned it, although I just realized; wouldn't the "add x to the xth row and repeat" trick also work on an unordered set of infinitely long and random integers as well? Therefore making a number between 0 and infinity that doesn't appear on the list of integers? Or does it have something to do with the "0." that comes before the listed numbers between 0 and 1.

[@jaredellison326]

The key is that they have to be a list of infinite strings of digits in order for the uncountability trick to work. Only then can you can ensure a difference between your generated number and each one in the list. Every natural number "between" zero and infinity terminates in a finite number of digits. While even with two characters in infinite strings the diagonal trick works to show it is uncountable, there is another trick to show countability for rational, integers, etc. .

[@jaredellison326]

Oh, in other words it has more to do with infinite string to the right of the decimal point for uncountability, than the 0 on the left.

[@pastaplatoon6184]

@@jaredellison326 that's what I was thinking, after posting this I did a bit of research, I suppose it has to do with this: when an integer has a string of infinite digits, it essentially diverges off to infinity and is no longer an integer at that point. so a list of them wouldn't exactly work for Cantor's diagonalization process as they would all technically be equal to infinity since they all have infinite digits and therefore would no longer be "integers between 0 and infinity". At least that's just from what I've read, I could be wrong on that but it felt like it made the most sense. 😅

[@-danR]

I was never comfortable with the idea that Cantorian diagonal method was a _proof_ . It seems more like a demonstration.

[@bobbun9630]

@@pastaplatoon6184Among other reasons, non-terminating decimals are required because the set of all terminating decimals is a subset of the rational numbers. It's all the rational numbers where the denominator contains only powers of two and five. Since the rational numbers are countable any subset of them is at most countable. This subset of the rational numbers has another interesting characteristic that could have been demonstrated in this introductory video, though... They are the full set of numbers with two distinct decimal representations: A representation that terminates (or ends in an infinite number of zeros, if you prefer) and a representation that ends in an infinite string of nines.

[@0x4849]

Please note the "proof" at 8:40 is not correct, since factoring out a number (i in this case) with infinite series is not a valid operation, it just happens to work out here.

[@jotobrosmusic3928]

before seeing anything else, we cannot go from (a+b)(a-b)=b(a-b)=>a+b=b because we will have to divide by a-b=a-a=0 which is an error edit: called it

[@comedygamer8493]

4:27 how does the fact that you can create within the infinite set of numbers between 1 and 0 a completely new number just by taking the nth digit of each nth number and changing it means that this infinite set is bigger than the infinite set of natural numbers? cant you do the same thing with natural numbers?

[@patrikvajgel240]

Suppose you assign a number between 0 and 1 to a natural number. After you do the method described in the video, you're left with more numbers between 0 and 1 than you have natural numbers

[@comedygamer8493]

@@patrikvajgel240 but why? I am not quite sure how

[@mirkotorresani9615]

You want to prove that there isn't a bijection between all the real numbers, and all the natural numbers. We prove it by showing that any map from the natural numbers to the real numbers cannot be surjective. So let's take a map from the natural numbers to the real numbers, that maps the natural number i, to the real number n_i. As he shows, we can find a real numbers that isn't any of the n_i. So the map isn't surjective

[@angelmendez-rivera351]

@@comedygamer8493 What exactly are you confused about? You seem to just be asking "but why" to everything without really thinking about the replies you are receiving. What exactly are you questioning? What is it that is making you ask "but why"?

[@Anonymous4045]

The one about finding the sums of n natural numbers, another way to think about it is this: Say you had 1+2+3+4+5+6. 1+6, that is, the first and last term, add up to 7. 2+5, the second and second to last, also add up to 7. So do 3+4. So it can be reasoned that the total sum would be the sum of the first and last term times the total number of terms. But, since we use pairs (first + last, second + second to last, etc.) we need to divide by two. The formula can be written then as n(a(sub n) + a(sub1))/2. N is the number of terms, and 1 is the first term, so in this case, it is n(n+1)/2

[@pbjandahighfive]

The sum of the set of sequential positive integers from a -> b = (a + b) * [ (b - a + 1) / 2 ]

[@Anonymous4045]

@@pbjandahighfive not sure why you would write it like that. Essentially what you have is the formula, but a as n(dub 1) and b as a(sub n), and then added some redundant parenthesis

[@pbjandahighfive]

@@Anonymous4045 There is nothing redundant about the parenthesis unless you intend to completely ignore the order of operations or do you mean the brackets? Also most people will likely have no idea what (sub n) means and I was just offering an alternative that is more readable.

[@Anonymous4045]

@@pbjandahighfive yeah sorry, brackets. 5 * [(2+1)/2] and 5 * (2+1) / 2 are the same

[@levinsonman]

n(n+1)/2

[@PC_Simo]

2:23 Yes. 1+2+3+…+99+100 = (100*101)/2 = 10100/2 = *5050* 1+2+3+…+999+1000 = (1000*1001)/2 = 1001000/2 = *500500*

[@goldoum]

0:20 problem if a and b are equal 2 a+b x a-b =0 so that say 0=0 …………

[@MrDuck313]

Nice video (although a bit basic imo, I'm used to more advanced calculus problem from you). For this type of video I would suggest a clear "divider" between the different problems. Took me all the way to the conclusion of the Pythagorean theorem to realize that there was no connection to the first problem!

[@BriTheMathGuy]

Great suggestion!

[@sumdumbmick]

maybe you're just stupid.

[@Jo-qb6zt]

The Point with the numbers between 1 and 0, it isn't possible to list them all because it goes till infinity, but you can do the exact same thing with natural numbers, just leave away the "." in it. you can add 0s (nn>1)after the "." in both ways the number go untill infinity

[@tanxros]

This video was definitely worth a watch even though I've studied most of these concepts in my junior high and senior high school Feels refreshing

[@floridafan561]

High key flex? Or low key sarcasm ?

[@VincentLauria6]

Kantor’s theorem is not correct. If you list ALL of the numbers between 0 and 1, then that new number would also be listed.

[@TrimutiusToo]

Hidden? (a-b) = 0 just screamed me straight into face

[@BriTheMathGuy]

Nice catch!

[@mrocto329]

Exactly

[@aweebthatlovesmath4220]

Yeah

[@PC_Simo]

2:31 Yes. In general: Σ(n) = (n²-n)/2.

[@IAmMee1]

The tittle of the video would have been funnier if it was “You guys need to see this video at least twice”

[@greekfox21]

You can't simply divide both sides with (a-b), in order to do this tou need to tell that a-b≠0 because division with zero breaks math, so tou have a≠b wich can't be happening since we have set a=b.

[@frogcam1677]

You said that some infinities are bigger than others. You can always make a new number in the list of 0 to 1 by changing a number in every number in the list (which creates a new number). That is assuming your list of infinite numbers doesn't contain that number, therefore not infinite. In other words, by assuming that you can add a different number to a list of infinite numbers proves that your list wasn't infinite. You can do the same for the other list of numbers from 1 to infinity. Just add 1 to the last number in the list. The only difference between the 2 analogies is that the second one is more obvious that the list isn't infinite.

[@sridingenbaba]

I have 2 proofs that this may be false..... Let's take a=b 1st: (a+b)(a-b)=b(a-b) but (b+b)(b-b)=b(b-b) 2b*0=b*0 0=0 2nd: (a+b)(a-b)=b(a-b) a+b=b 2b=b then 2b-b=0 b=0 so ye.... these are my calculations that this may be a big oopsies

[@raydencreed1524]

For the proof that there are infinitely many primes, can’t we just notice that no prime on the finite list divides N, meaning that it must be a new prime number?

[@nolategame6367]

Well yes, that's exactly that (though N is "largest prime" factorial + 1 - which leads to contradiction since it's thus not divisible by any primes)

[@zhabiboss]

Me seeing him casually divide by zero: 💀

[@marcusscience23]

When I first saw something like this, it proved any number = 0. I got freaked out and couldn’t sleep and whole night worried about my very existence. The next day everything made sense.

[@davidx9426]

the proof around 11:48 does not work, I tried it with me thinking 13 was the largest prime, however, when i multiplied 2x3x5x7x11x13+1, i got 30031 which is 59x509.

[@lolomolo]

Thought about it because my high school teacher taught me about it: at 0:30 how can you divide both side with (a-b) without knowing that (a-b) isn’t equal to 0 or a number? It’s a pretty simple thing but it’s so overlooked and my high school teacher just made everyone confused but also understood why this can’t exactly be done

[@snekburrito]

wow I wish I had stumbled across this video back when I was teaching algebra 2 to my high school classes. You have a great way of explaining the fundamentals.

[@APKZ04]

u know its wrong tho right

[@NoName-rd6et]

@@APKZ04 whats wrong with it

[@19Szabolcs91]

There being more numbers between 0 and 1 than positive integers has a very intuitive explanation. For every integer number "n" you have 1/n that is between 0 and 1, and they are all different from each other. So there are at least as many numbers between 0 and 1 as integers... And clearly, there are much, much more stuff, for example literally everything between 1/2 and 1; 1/3 and 1/2, etc.

[@edinaldoc1]

Oh wow... That made it so much easier! Literally everytime i saw someone trying to prove this, they use the exact same method as the video, and it never really clicked to me because the way it was done in the video seems unintuitive and feels like a piece is missing. But your explanation immediately makes it clear!

[@muriloporfirio7853]

The first one has a second division by 0: 2b=b, means b=0, but you divided both sides by b.

[@ilias-4252]

Yes for 2b=b we need b=0 but thats not a mistake...we could just assume that a and b are not 0 at the begining. Up to the point that he divides by a-b everything is correct, and after that point everything is also correct (by correct meaning that one step implies the next).

[@vgtcross]

All of the calculations before that work for a and b being equal to anything. At this point of the argument we have (incorrectly) proven that 2b = b for all numbers b. Since this works for all b, we can just suppose that b is not zero and divide by b, thus getting 2 = 1.

[@DeJay7]

That's actually not a division by 0, but b would equal to 0 if 2b was equal to b, but we got to that through a division by 0, a-b

[@pacomesalmon8086]

2b =b 2b-b = b-b b = 0 ; you can get b=0 without dividing.

[@dudethebagman]

In the first argument, there are actually 2 places where there is division by 0. The last step taking 2b=b and concluding 2=1 involves dividing by b on both sides. But if 2b=b, then b=0, and you can't divide by b. Although 2b=b is not implied by the original premise (since we fallaciously divided by zero already), 2b=b is a statement that is mathematically possible. But in the last step, we divide by zero again, and derive a statement that is mathematically impossible.

[@dudethebagman]

One time in high school I did something like this in class. We had a substitute teacher in a math class that day and I wanted to have a little fun with her. I had a feeling she knew less about the subject matter than some of her students did. So I went to the board to solve the problem, and I used valid deductions to establish that 3x=-5x. Then my final step was to conclude that 3=-5. I said, "Is that right?" She looked at it for a few seconds and said, "No." Then she called another student to the board to solve the next problem.

[@HenryZhoupokemon]

Here’s a better proof 1 = 2 We can write x = 1 + 1 + …. + 1, with x # of 1’s. Then x^2 = x * (1+1+…+1) = x + x + … + x, or x plus itself x times. For example, 3^2 = 3 + 3 + 3; 4^2 = 4 + 4 + 4 + 4, etc. So now we have x^2 = x + x + … + x, and we can take the derivative of both sides f(x) = x^2 = x + x + … + x f’(x) = 2x = 1 + 1 + … + 1 = x Since we had x number of x’s, since taking the derivative is linear, we just add each individual derivative to get 1 x times, or x Thus, we have 2x = x 2 = 1 And notice we didn’t divide by 0 since x was any number, not just 0 (for example we used 3 and 4 for examples for x)

[@Micha1996]

In the first "proof" you could also rather subtract b from both sides to get a = 0 for all a, rather than using a = b to get 2 = 1. (Though neither are accurate due to division by 0).

[@themolten8]

how to 0=1: have the true statement x(0) = y(0), where x ≠ y complexify simplify remove 0 from both sides x = y, but x ≠ y 3(2) + 4(2) = 14 remove 2 from both sides, you are essentially dividing by 2 3 + 4 = 7 by removing 0 from both sides you are dividing by 0 which is why you get a contradiction

[@wiggles7976]

Another way to prove sqrt(2) is irrational: Write sqrt(2)=m/n for integers m and n, with n != 0. Thus, 2n^2=m^2 => 2n^2 - m^2 = 0. Two numbers a and b are congruent mod k if k divides their difference, i.e. k|(a-b), or equivalently kc = a - b for some c. Since k*0=0, we have 2n^2 = m^2 mod k for all suitable k. Pick k=5, and let's find a solution (henceforth, all numbers are in modular arithmetic). First let's write the squares mod 5 for convenience: 0,1,4,4,1. Now let's go through all possible cases, n=0,1,2,3, or 4. The first case is not possible. If n=1, then m^2 = 2, but 2 has no square root since we listed all the numbers that do have square roots. Hence n=1 isn't possible. If n=2, then m^2 = 2(2)(2)=8=3, and again, this is a number with no square root so there exists no m to be paired with n=2. If n=3, then m^2 = 2(3)(3)=18=3, and again no m exists to satisfy the equation. if n=4, then m^2 = 2(4)(4)=32 = 2, and no number exists that can be squared to give 2, so no we have no solution. In all 5 cases, there is no solution. Thus, no m and n exist to satisfy 2n^2 = m^2 mod 5. If there was a solution in the integers for 2n^2 = m^2, there would be a solution to 2n^2 = m^2 mod 5, thus by contrapositive, there is no solution in the integers for 2n^2 = m^2. I like this proof because it reduces the problem of finding the square root of 2 in the rationals (an infinite set of numbers that we could never exhaustively search) to the problem of finding the square root of 3 or 2 in the integers mod 5 (a finite set that we could exhaustively search).

[@markgraham2312]

Action at 3:33 is invalid since 1) It is not a standard algebraic rule, since one cannot divide by 0 on both sides of an equation. 2) a - b = 0, if a = b.

[@MrHotBagel]

3:02 I like this derivation. Something similar was probably done to prove the sum of arithmetic sequences formula since it's essentially the same: S=n(a1+an)/2 where in this case a1 = 1 and an=n (since the subscript value/sequence number matches the actual value of the term)

[@PhilosophicalNonsense-wy9gy]

a1=1 a×1=1 a=1

[@PhilosophicalNonsense-wy9gy]

Both the natural number set of all numbers and the real number set of numbers between 0 and 1, can have an infinite expansion. The natural number set has infinite expansion on the left side of the decimal point, the real number set of no.s between 0 and 1 has infinite expansion on the right side of the decimal point. Thus, both these infinities are equal as long as you keep the real number set as the numbers between 0 and 1. To categorize the scales of different kinds of infinity, lets consider in how many different ways they are infinite. In natural no. set of all positive integers, two criterion are applicable. 1. The value of the number can keep increasing endlessly 2. The set has infinite expansion towards the left side Now lets compare to the second set: 1. The value of the numbers can be of infinitely different values between 0 and 1 2. It has infinite expansion towards the right side But when we consider the real no. set of all positive numbers, there comes 3rd criterion: 3. Just like the 2 criterion are applicable between 0 and 1, so are they applicable for between any other numbers, thus infinity of those numbers follow those two criteria Thus, the N set has 2 criteria while the R set has 3 criteria, thus R set of all positive numbers is a bigger infinity. (Note that creating new numbers for R set of no.s between 0 and 1 also does work for the Natural no. set of all positive integers)

[@Проекттвояноваядевушка]

03:54 you can list numbers like this: .0 .1 .2 … .9 .11 .12 … .19 .21 .22 And etc It will guarantee that you will never miss a single number. This is the same thing as we did in natural numbers 1,2,3,4,…

[@19Szabolcs91]

False. Right off the bat, you missed 1/100 = .01 for example. Not to mention all the irrational numbers such as 1 / sqrt(2)

[@Проекттвояноваядевушка]

@@19Szabolcs91 Nice argument! Now I finally understand why it is impossible to list this numbers. Thank you so much. You made my day.

[@IMAN_IIT_PATNA]

When you are cancelling, (a-b) from both sides, you should mention, that a≠b, but previously you mentioned, that a=b, so this is incorrect

[@VietDuckProductions]

Man, this video made me realize there are probably *REALLY* sneaky ways to divide by zero I haven't caught onto yet.

[@castonyoung7514]

3:50 I've seen this "proof" done before, but can't you just do the same thing on the other side of the decimal point (i.e. the integers)? In either case, you are assuming that 10oo = oo (because base 10).

[@hussainfawzer]

the example shown here (3:17 ) to illustrate that there are more real number between 0 & 1 than the set of integers, is very obvious and I could understand that in my own way, but if someone asked me to explain that, I don't know whether I could do a justifiable job. Here I can not understand the explanation given by BRITHEMATHGUY.

[@larsscheele9914]

Small nitpick about the Cantor argument: It is inherently assumed that if two numbers between 0 and 1 have different decimal representations (i.e. they do not agree on at least one figit) that they are not equal. This is not true in general as for example 0.09999999999... = 0.10000000... Luckily it can be shown that ending a number on all digits 9 is the only "non-unique" representation, so excluding these cases yields a valid argument.

[@SimonClarkstone]

You can also adjust differently: turn 1 into 2, and all other digits into 1.

[@angelmendez-rivera351]

Excluding those cases yields a valid argument because they form a proper subset of the set of all rational numbers is countable. This much has to be stated, though.

[@cmilkau]

division by zero, a-b=0 my calculus professor did this too, called him out, he said he was testing our attention, but didn't give a correction.

[@SBC-41]

So many of these caught me so off guard. This man is a legend at explaining math. Really intrigued. Keep up the good work!

[@keshavgoyal1325]

Where is the mistake? x² = x+x+x....(x times) Taking derivative on both sides 2x = 1+1+1....(x times) 2x = x 2 = 1

[@sharpfang]

You didn't differentiate the "(x times)", which is also a part of the equation and a function of x. The "(x times)" is a multiplication operation, so x+x+x+...(x times) can be equivalently written as x*x. Applying the derivative of multiplication, (x*x)' = x*x'+x'*x = x*1+1*x = 2x

[@Lucaazade]

mn = (1 + … + 1)n = n + … + n if, and only if, m = 1 + … + 1. There is no derivative of integer functions because they’re not continuous. (Dots don’t have gradients.) It is ambiguous whether the first or second line is the one that’s wrong - the first is wrong if x isn’t an integer and the second is wrong if x is an integer. Obviously writing them both down is always wrong. :) Edit: Sorry for all the edits.

[@wqltr1822]

Just a guess, maybe the rule which states: 'the derivative of the sum of functions = sum of the derivative of each function' might only work when the number of functions is a constant, whereas x is not a constant

[@nikhilnagaria2672]

You forgot to use chain rule: While taking derivative of x+...+x (x times), it's going to be 1+...+1(x times) + x+...+x(1 times) = x+x=2x as it should be :)

[@angelmendez-rivera351]

If x is a real number, then it is not the case that x^2 = x + x + ••• (x times). This is only true if x is a natural number, because (x times) is incoherent otherwise. Here, sqrt(2)^2 = sqrt(2) + sqrt(2) + ••• (sqrt(2) times). Explain: what the hell does it mean to do anything "sqrt(2) times"? Nothing, is what it means. It is nonsensical. If x is a natural number, then x^2 = x + x + ••• (x times) is indeed true, differentiation of functions N -> N works differently than for functions R -> R, since the natural numbers are well-ordered, and thus, not densely ordered. This means that the derivative must be evaluated as (x + 1)^2 - x^2 = 2·x + 1. This is the discrete derivative. To differentiate the other side, one must make use of the fact that f(x + 1)·g(x + 1) - f(x)·g(x) = f(x + 1)·[g(x + 1) - g(x)] + [f(x + 1) - f(x)]·g(x) = [f(x + 1) - f(x)]·[g(x + 1) - g(x)] + f(x)·[g(x + 1) - g(x)] + [f(x + 1) - f(x)]·g(x), so that [1 + 1 + ••• (1 time)] + [1 + 1 + ••• (x times)] + [x + x + ••• (1 time)] = 1 + x + x = 2·x + 1, meaning that 2·x + 1, so there is no contradiction.

[@Nightcrawler333]

2:37 can you please explain how you got (n+1) + (n+1)+.....+(n+1) from the previous step. You say "add it to itself" but my tiny brain can't figure out how this is done..

[@professionaldumbass1953]

im not very good at math but i think n = 3 n+1 = 4 S = 3+2+1 2S= 3 2 1 +3 +2 +1 now each number has an addend that can be added to and produce a sum of n+1 and then the rest of the addend split among the numbers less than it so like this 2S= 3 2 1 +3-2 +2+1 +1+1 ^ split among the rest of the addends 2S= 3 2 1 +1 +2 +1 +1 +1 2S= 4 2 1 +3 +2 2S= 4 2+2 1 +3-2 +2 ^now take 2 and add it to the number above to get 4 2S= 4 4 1 +1 +2 2S= 4 4 1 +1-1 +2+1 ^give the rest of the addend (1) to the the last number 2S= 4 4 1 +3 2S= 4 4 4 n = 3 n+1 = 4 2S= (n+1)+(n+1)+(n+1)

[@danielwilkowski5899]

0:32 - cancel (a-b), is essentialy dividing by zero, since a=b

[@danielwilkowski5899]

Additionally, 2b=1b does have a solution, and it's b=0, since 2*0 is 1*0.

[@v1ru5official]

i dont understand about the statement Pk must divide 1 is impossible, anyone can explain?

[@user1ejej]

wait if a = b then a-b should be zero right because we could write a-b as b-b because a equals b so we can’t cancel them from both the sides.

[@Nikioko]

2:31: This genius way was found by Carl Friedrich Gauß as a schoolboy, when the teacher asked him to sum up the numbers from 1 to 100, and he finished after just a few seconds.

[@schrodingcheshirecat]

in a funny way, if 1/0 = infinity and 0/0 = 1 it sort of works out for 1 = 2. let 1 = 2 ,0/0 = 1, and 1/0 = infinity 0 + 1 = 1 + 1 now divide both sides by zero 0/0 + 1/0 = 1/0 + 1/0 1 + infinity = infinity + infinity 1 + infinity = infinity infinity + infinity = infinity and so, these 2 values are equal but with a consequence: if 1/0 = infinity, and 0/0 = 1, resulting in 1 = 2, then all other mathematical operations go haywire 1 = 1+1 = 1+1 + 1+1 = 1+1 + 1+1 + 1+1 + 1+1 etc.

[@tst_09]

Congratulations Bri, you’ve successfully made Math addictive

[@parthpandey2030]

3:09 I plugged in the Ramanjuan Summation into this formula and got infinity when n = infinity instead of -1/12

[@Nebulisuzer]

a=b a²=a×b a²-b²=a×b-b² because a=b, when he does a-b it's equal to zero a+b×0=b×0 remove zero from both sides (fault) a+b=b 2b=b divide by b 2=1

[@argusvoidstar4177]

"But 2 doesn't divide 1" Me who doesn't know this stuff: *Laughs in decimals*

[@Arxareon]

1:19 usually.. An example when you can? :D

[@THE_HONOURED_ONE_LOL]

1/0 = infinity, done

[@Ameliahorn6582]

He made an enitre video about assigning it a value :D

[@yousef1342]

Calculus students be like 😂

[@blakeks8520]

"THAT type of infinite is uncountably large" yeah no shit

[@owentang546]

Very interesting. I saw all of these before, except the Euler's Formula proof.

[@wallace3199]

a similar algebraic equation was shown to me by my classmates which they had probably seen in a reel. I got to prove it was wrong and found, as this video points out, like in the fourth line in the starting of this video, there is a multiplication by 0, as a=b and a-b=0, and it is not possible to cancel out zero. hence 2=1 or anything like that is theoretically incorrect.

[@Yadobler]

6:10 that's how A1,A2,A3,A4... ISO sized paper works! the ratio of length to width is sqrt(2) if you fold A4 in half lengthwise, you get A5 size, and likewise if you put 2 A4s lengths against each other, you get A3! because if A4 is L * W, then the ratio L/W is equal to W/2L (for A5) and 2W/L (for A3) L/W = 2W/L (L/W) ^ 2 = 2

[@Fytrzaczek21]

3:30 normal guy: "wtf" guy after set theory class: "yeah what's up with that"

[@rossholst5315]

My problem with assuming that the natural numbers can be listed is that there is no unique way to list them sure you can list them 1,2,3,4 but you can have any unit of interval. Meaning maybe I am counting the number of millions in infinity or the numbers of 1/10’s in infinity. Basically we can always add zeros behind your number 1. And we can add any finite amount of zeros without ever getting your final number any closer to an infinity. For there to be more decimal numbers than natural numbers there would need to be more zero positions behind the decimal point than there are zeros in the infinity of the natural numbers. The reason we cannot list the decimal numbers is simply because we are trying to count back from infinity. But you have the exact same problem if you try to count back the natural numbers from infinity. With the natural numbers you can also always make a new number that is not in your list as well using the same diagonal trick.

[@thatonedude-6819]

4:04 why can’t you do the changing digits with integers?

[@Calculusgoat]

The algebraic use on 0:35 is 0=0 which you cant substitute which makes the expression wrong

[@jacobf.1691]

1:00 b = 0

[@timecubed]

I think the issue was that, though in fact a + b = b, and a does in fact equal b, it also means that a = b - b = 0, and since a = b, therefore b = 0. So saying that 2b = b is in fact correct but to cancel out the b's you'd have to do 0/0 which is indeterminate Edit: I was right in a different way, there were two divisions by zero and I just caught the second one

[@karadelik4478]

1:48 Bruh area of square is a^(2) imagine a square that has 16 square meters area and a=b 2a × 2a = 8 × 8 = 64

[@yreaz1808]

The Sum of the Arithmetical Progression is actually a mandatory subject in my country

[@dannypipewrench533]

Which country is that?

[@michallesz2]

a^2 - b^2 = ab - b^2 => 0=0 (a+b)(a-b) = b(a-b) => 0=0 2b=b => 2*0=0 => 0=0 => a=b , a=0 , b=0

[@cesarjom]

Proof by contradiction is a classic argument used in mathematics, every beginner student should master its many patterns of application.

[@GalileanInvariance]

10:45 Should say 'That is, every composite number can be written as a [unique] *product* of primes.'

[@pedrosso0]

0:32 since a=b by definition, a-b is 0 so you cannot cancel them.

[@BriTheMathGuy]

exactly!

[@namantenguriya]

Prime proof was stunning. 😲 Nyc vid 🙂

[@BriTheMathGuy]

It really is!

[@chrisg3030]

Interpret writing the infinity sign ထ as starting an endless count, like you hit the ^ in the number bar when making an online payment but the mouse jams and it never stops. You go out for a beer or a week's vacation and when you come back it's still counting. Then you write down another to the right: ထ ထ (assuming you follow the left-to-right writing convention). That second infinity will of course be smaller because you started later, so ထ > ထ.

[@videakias3000]

x=0 2x=0 x+x=0 x=-x x/x=-x/x 1=-1

[@ryanhollstein4164]

My question is if you are able to put any number as an exponent and the base is always 1 then shouldn't any x equal any number that can fit in x so for example 2 equals 3 or 9 equals 35 or whatever you can imagine

[@anon.9303]

In that first one, you don't even have to see the division by zero to tell what's wrong because you *can't* divide both sides by b. If that is to be treated as an equation, the coefficient of the variable has to go along, just making it so that b=0.

[@gabrielgomes242]

4:13 ok, so if you can write all those numbers randomly then lets write them in another way .1 .2 .3 .4 .5 .6 .7 .8 .9 then you would say: but .10 = .1 so every time you get another digit you will put a zero to the left of the numbers that are already in and continue like this: .01 .02 .03 .04 .05 .06 .07 .08 .09 .10 now do what you said at 4:12 well you can't because the infinity downwards is bigger than the infinity to the right so, technically that argument is not right

[@larsscheele9914]

Well, this way you obtain an (ordered) list of all real numbers with a finite part after the decimal dot (because all numbers in your list end after a certain time) and this subset of the numbers between 0 and 1 is indeed countable. But a number like 1/3 = 0.3333333... is not part of the list. That is why the argument given above (given by Cantor) is the usual way to do it: Suppose there is ANY ordering, whatever it looks like, then you can do this the way he decribes. By the way: In "your" ordering all numbers extend infinitely to the right with 000000... so his argument works as well, by extending this way.

[@bgmarshall]

@@larsscheele9914 I do wonder, why can't you do the same thing saying there are unlimited zeros to the left of any natural number and diagonal the other way?

[@edinaldoc1]

@@larsscheele9914 Yours was literally the best way to understand this concept i have ever heard on the internet. The fact you separated the real numbers with a finite end from the ones that go on infinitely is exactly the missing piece. The video just went straight to the numbers that go infinitely so it felt like something was missing

[@ThePiMan0903]

Nice video BriTheMathGuy!