bad math proofs explained: 1=2?, e=π? can't divide by 0?
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- čas přidán 15. 07. 2024
- You probably have seen "proofs" that 1=2 or e=pi but have you ever what the trick is behind all these bad math proofs? In fact, many of the bad math proofs could be explained by just one word...
0:00 proof that 1=2
2:14 "injective" means "one-to-one" & "one-to-one" means injective"
4:58 proof that e=pi
6:45 why can't we divide by zero
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"Injective" has an injective ー or one-to-one ー relationship with "one-to-one"
Better yet, bijective
@@blackpenredpen your responde is bijective with my "one-to-one" comment
@@Bruno_Noobador mind blown!
@@blackpenredpen mind-jective-blown!!
@@at7388 stimmt
5:42 fundamental theorem of engineering
also e = 2 and π = 3...
Therefore 2 = 3. If this isn't a QED...
I'm kidding
Also small angles of course, but small angles are just smaller than 180°
e = pi = 3 checks out
I get the joke, but I am a Chemical Engineer and never have I seen someone do that.
Is it something common in USA?
@@MrRogordo No one does that but it just an inside joke because mathematicians tend to care about the smallest details and generally approximations are just approximations for mathematicians. But for Engineers alot of times the approximation is used in practical purposes as well which is why there are so many jokes in the math community about approximations.
Another way of explaining this: when you "cancel" something on both sides of an equation, you are really just applying the inverse operation to both sides of the equation. The problem in these cases is that the inverse operation is not well-defined, because the operation in question is not bijective over the domain we're interested in.
In particular: sin(π) = sin(2π) ⇒ arcsin(sin(π)) = arcsin(sin(2π)).
The above is correct. The error comes in assuming that arcsin(sin(x)) always equals _x;_ it does not.
Likewise: 1^e = 1^π ⇒ log₁(1^e) = log₁(1^π). However, log₁(x) is undefined! That is, the function f(x) = 1^x has no inverse!
u should have just wrote that range of arcsin is [-pi/2,pi/2]
@@akshatsaini69It doesn't. arcsin is defined over the [-1;1] interval. You're referring to the interval in which sin(x) is injective and yields these values.
I divided by zero once, but then I took an arrow to the knee.
Math class in my school was rough.
Nice Skyrim reference!
That's an acceptable punishment for dividing by 0
I am so proud
@@blackpenredpen you just made me respect you 10000x more for getting that reference lol
It's like a function that causes a loss of information. You can't reverse it without losing information. Like having an equation and multiplying both sides by zero. Or taking a derivative, because when you do so, a constant becomes zero, and inversing the process necessitates the addition of an unknown constant C.
The same reasoning is why encryption works too. It's easy to find the value of y when you are given the value of x in sin(x) = y. It is impossible to find the value of x when you are given the value of y in sin(x) = y, because there are an infinite number of values for x that will give you value of y in sin(x) = y. So if you share the value of y, it tells you nothing about the value of x. Obviously, encryption uses different functions than a sin function, but the principal is the same: you can never figure out the original value of x if someone shares the value of y with you, so the value of y can be spread without fear. Only someone who knows relevant information about how y is generated can get back to x, which is the original message.
@@peterpike, you seem to be confusing many-to-one functions (which are what you described) with so-called "one-way" or "trapdoor" functions in computer science (which are what you probably meant to describe). These are functions which are easy to compute, but difficult to invert. It does not mean there are multiple inputs that give the same output, it only means that finding the one solution that does exist is computationally difficult.
The classic example is prime factorisation: given two primes _p_ and _q,_ computing their product _n = pq_ is trivial, but if you are only given _n,_ then determining _p_ and _q_ is very difficult. In particular, the only real way to solve this problem is by brute force, trying to divide _n_ by every prime between 0 and _√n_ until the solution is stumbled upon. For sufficiently large _p_ and _q,_ then, the problem becomes practically intractable.
@@JivanPal Actually, I did mean what I said :-) The class I took explained it using modular math, and used the example of how you could add 2 hours to a 12-hour clock OR you could add 14 or 26 hours, and you would have the same time back, so simply knowing a time is off by 2 hours doesn't tell you how many DAYS you are off by.
FWIW, you are correct about how prime factorization is used, particularly in things like RSA, and that's definitely more relevant today than the many-to-one functions that started off the process.
@@peterpike, how are many-to-one functions used in encryption in the way you described? Modular arithmetic is only used because it is finite and certain nice theorems (e.g. Fermat's Little Theorem) hold over finite fields. There is no utility in, for example, the fact that the equation g^x = k (mod n) has infinite solutions when x can be any integer. It still has only a finite number of solutions over {0, 1, ..., n-1}, which is what matters.
Bprp is so awesome. I wish I had a teacher like him.
Thanks to youtube, you sort of do :D at least a little bit
But I know what you mean and relate
I recommend him to all of my students.
In a sense, you *do* have him as a teacher!
@@zillow5735 what?
I wish too
5:43 blackpenredpen is engineer confirmed!?
e = pi .... = 3
To my mind, we can see injectivity in 4 ways:
-f(x)=y admit at most 1 solution in other words each image has 1 or 0 antecedent
-For all x,x1 in R, f(x)=f(x1) implies x=x1
-For all (x,x1) in R2, x≠x1 implies f(x)≠f(x1)
-A function which is strictly monotone
Doesn't need to be monotone if it isn't continuous
for me, 1^e = 1^pi is fine, we equate powers by taking log on both sides, so when we take log we get
e log1 = pi log1
but log1 = 0 so
e * 0 = 0 * pi
0 = 0,
thats why theyre equal, not because e=pi
Trying to cancel out 1^x is also a division by zero, since 1^x = e^(x ln(1)). You can cancel out e^x just fine since it's injective, but then you get x ln(1), or x * 0
Word of the day: Injective!
Yup!
Injectiven't actually
I learned about injective in my class 10
Here's how division by zero was explained by me.
It's the inverse of multiplication. 2/0 is saying that 0 x ? = 2. Obviously nothing can fit in that question mark.
So if nothing fits that "?" the answer is zero, right?
;)
No because 0 squared is still 0
In general if we divide x/0 we have to cases:
1- if x=0 then we can say the answer is a number t now 0t=0 but that means that there are infinitely many real answer solutions like pi or e or 2 etc
2- if x isn’t 0 the we say x/0=y
Then 0y=x but 0 times any number is 0 and that means x=0 but we assumed x isn’t equal to 0 henceforth we can’t divide by 0 since we get non sense results
Two cases*
0 × x = 2
If you divide by zero we get x = 2/0
The problem is division by zero approaches to "Infinity", so you can't say that 0 × Infinity = 2, divide by zero, it's undefined
I think of the result of division as this: if you have some pies and some people, how many pies does each person present get if all the pies are handed out one by one according to these rules: 1) everyone ends up with the same number of pies, and 2) You're not allowed to allocate pies to latecomers or absentees. Say you have six pies and three people, all the people present get two pies each and there's your answer - anyone who didn't join the free pie line in time is out of luck. But if there are no people present to give pies to, you never begin handing out the pies because you are not allowed to give pies to absent people. Therefore the number of pies each person gets in this scenario is not six or zero or any other number - it's unknown or indeterminate, assuming that you don't know the result beforehand (if you do know the result beforehand, there's no point asking the question and nobody gets any pies).
ps, I don't know whether imaginary pies are worth eating.
How have I never heard of this. You blew my mind right open, this is fantastic. Thank you.
Glad you enjoyed it!
Come on, everyone knows that e=pi=3
No, that's not true.
…
Not everyone knows that. This is secret engineer knowledge.
We have therefore proven that 3 is irrational Q.E.D.
@@Theraot And math meme reddit.
3^(3i) + 1 = 0
also equals to sqrt(10)
If I was asked about divisibility by zero, I'd ask them: "What times zero gives you a non-zero number?"
Can we just appreciate how fast and fluent he can switch between the two markers, red and black.😂
saying that "sin(pi) = sin(2pi) leads to pi = 2pi" is like saying that "(-2)^2 = (2)^2 leads to -2 = 2"
When I was just learning trigonometry some time ago, the sin(2pi) = sin(pi), etc was a thing that really bothered me. I wished I had this video back then because I just was taught to roll with it hahaha.
I was thinking about it as the video went on and... You have to understand how these functions work graphically or it's too hard to remember. Seeing them as coordinates on a circumference or as wave functions.
@@eraimattei Understanding the trigonometric functions can be tricky, but I think the problem the host comment is talking about is not about difficulty in understanding the trigonometric functions. Rather, the comment, to me, seems to be reflecting on how it can be very difficult grasp the idea that a function can have the same output for two different inputs. The idea of a function that is not injective is rather counterintuitive for people who are studying mathematics at the elementary level, because very first few operations we are introduced to can always be ”inverted.” For example, addition of real numbers is cancellable: for every real number, you can find the additive inverse. Subtraction is just a special case of addition, so naturally, it is also cancellable. Multiplication is also cancellable, as long as none of the factors is 0. This is why the fact that division by 0 is ”not possible” bothers people: because it is their first encounter of an operation you cannot cancel, and their first encounter of an operation which is not always defined.
This is also why people struggle so much with the idea of square roots and polynomial roots in general.
When I first learned trigonometry, I got nothing because I work with visual cues. I didn’t really get anything related to functions until I was introduced to the Unit Circle and the functions’ graphs. Then I finally got that trigonometry wasn’t just math, it was the study of every proportion known to humanity, a step closer to understanding the universe itself
Furthermore, the fact that the complex logarithm has infinitely many complex values is, I think, even more perplexing. Probably because we are first taught of ln(x) as the real valued inverse of e^, which are both bijective.
@@phlaxyr The reason for this confusion is because educators tend to teach functions as "expressions on variables," which is actually incorrect. The string of symbols "x^2" is not a function, this is just a monomial on an indeterminate. The strings of symbols "e^x" and "ln(x)" are also not representations of functions. On the other hand, if I write "x |-> e^x : R -> R," now I do have a function, because I have specified the domain and the codomain, as well as the mapping between elements of both sets. This is done, of course, keeping in mind that the symbols "e^x" are already well-defined a priori.
The notion that "ln(x)" is the inverse of "e^x" is only true if we can talk about functions. In this instance, you can say that there exists a function f : R -> R>0, x |-> e^x, and another function g : R>0 -> R, x |-> ln(x). Then, f°g : R>0 -> R>0, x |-> x, and g°f : R -> R, x |-> x. Therefore, g is the inverse function of f, and viceversa. However, we often abuse language and simply write "ln(x) is the inverse of e^x, and viceversa," because in mathematics, we have a genuinely bad habit of assuming beforehand that the "obvious" domain and codomain are the ones in use, unless otherwise stated, which is nonsensical, since, in light of the fact that we do work with multiple sets even within the same work, there is no such a thing as "obvious" domain and codomain. So not specifying them is simply incorrect.
Why does this matter? Because I can define h : C -> C\{0}, x |-> e^x, and notice that this function performs the same "action" on the inputs as f, but it is actually a different function from f: unlike f, h is not bijective.
You are literally doing the job of my dreams. I love this. Exciting maths. Also, my friend you don't age. bravo for your positive energy too
7:13 yeah seems pretty legit
Great Stuff as always
I wish I had a professor like you bprp
You have cleared my doubt , because by seeing the videos like 1=2 has always cofused me in a
question that math is right or wrong,great .
that gave me the freshest perspective on why you cant divide by zero
Please help me with a question please.
If we have two complex numbers , we can represent them as vectors and we can divide two complex numbers . If we take two vectors and write them as complex number . Why can't we divide the two complex numbers to get vector division?
Vectors have fundamentally different rules from complex numbers. Complex numbers are shown as vectors because it is easier to visualize, but they're not actually vectors.
Firstly, what do you mean by "vector division"?
@@ericy1817 They are vectors in the vector space of Complex Numbers over the field of Complex Numbers. :)
@@zapazap In that case, division of two complex numbers a/b would represent scalar multiplication of a one dimensional vector by the scalar vale 1/b (where b≠0 of course :-) ).
I tried to create a graph for y=0x, but, for some reason, it just flat-lined.
Did bprp get new markers? Just listen to the sound they are making, sooo satisfying.
i worked on showing an induction “proof” was false and why last week. taught me a significant amount about induction. also ik this would probably be easy for everyone watching but im kinda dumb lol
Your explanation always easy to understand
You're so good. This was great.
Love ur teaching method😍
This was a cool video. I learned a lot about injective functions.
Quick question: if sine isn't injective, how do they define arcsine? It would be ambiguous, no?
They have to restrict the domain. I will make a video on it. Especially the problems like sin(sin^-1x) vs sin^-1(sin(x)). They are very confusing to students
Good mornin bprp
Simply awesome bro very good proof
Genius. Now I understand all these injective stuff. Thanks
I love the constant +c added after the integration of atoms, making our Favourite teacher "bprp"
loved this injective and non injective thing! great
It helps that I am a learning developer that functions in maths are pretty much same as functions in programming languages. You judge the result based on the *output* of the function.
Wow! Loved it! So interesting!
Thanks.
@@blackpenredpen No, thank *you*. I never thought to summarize all of these situations with the concept of being "injective". That's great insight.
Well understandable video😄
When a blue pen involved...
There is a serious matter..
Finally got satisfactory answer....thanks a lot SIR
Not only you have a board saying "integrals for you" , but it also ends with "you're welcome"....
7:37 I'll tell them it is goodn't
Why does you shirt say +C
Is there another shirt with z^2 on it?
Sir your way of teaching is inspiration for me 😀😀😀
Is "injective" also the same as saying that the function's inverse is itself not a function?
No, the inverse is a function (usually), but just not well defined in certain areas. Ie f(x) = x² is not injective, but it’s inverse is not defined for negative values. Btw, √x is itself injective.
Sir u make really inspiring maths videos ..Lots of respect and love from INDIA and Russia
Please also make a video regarding symmetricity of functions and also surjective functions.
Bprp: e=π
Engineers: Fundamental theorem of Engineering finally proved !!
Great video
as always The BEST
6:33 f(x) = 1^x is 1 for complex values of x too not just real numbers
I want to expand a little on the topic of multiplication and division by 0, since people in the comments are already creating discussion on the subject.
What do mathematicians *mean* when they *say* that division by 0 is not defined? Well, what is division? Division, be it of rational numbers, real numbers, complex numbers, matrices, or whatever structure you wish to analyze, is just multiplication by the multiplicative inverse of a number. The symbols a/b are a symbolic shortcut to write a·1/b, where 1/b means "multiplicative inverse of b." So when you talk about division by 0, what you *really* want, in rigorous terms, is to multiply by the multiplicative inverse of 0. So you have to ask yourself, does the multiplicative inverse of 0 exist? In the video, we saw that x |-> 0·x, when taken as a function from R -> R, is not injective, and not surjective. This is because 0·x = 0 for all x in R, right? So it never occurs that 0·x = 1, and so 0 has no multiplicative inverse. Aw :'(
But okay, what if we decide to add a new element to R and form a new set S, and that element satisfies the property that 0·x = 1. Can I do this? Can I form this set? Well, you can try, but in mathematics, when you form sets, you have to define them rigurously and specify what properties you want this set to satisfy in order to exist and be well-defined. So if you want to define S, you need to look at what properties you want this set to satisfy, and you need to look at how you want the operations to behave. For example, multiplication in S, as en extension of multiplication R, can it be associative? Can it be commutative? Can it distribute over addition? These are the questions you need to answer. And in general, the answer is no: multiplication in S is not associative, and it does not distribute over addition in S. Does this mean S cannot be defined? No, but it would be very difficult to find a consistent set of properties (axioms) that S has to satisfy to be well-defined that we can understand. And even if you could theoretically do it, this structure would be significantly less rich than a structure on real numbers R, for example, and it would be very likely be not useful, and maybe not interesting. So we decide not to try to create S and stick with sets like R and C instead. This difficulty in defining a set S (which, let me emphasize, it isn't necessarily strictly impossible, but it's difficult enough to put everyone off) is the idea that mathematicians are trying to communicate when they say division by 0 is not defined.
Of course, there are other things you can try. Look at what wheel theory has done, for example. But in wheel theory, 0 has no multiplicative inverse, and division is defined as a generalization of the multiplicative inverse, rather than being strictly defined in terms of the multiplicative inverse.
It's been said that the greatest contribution Bourbaki made to mathematics was vocabulary: *injection, surjection, bijection, disk, circle, etc.*
Yes, it used to be unclear when a professional mathematician talked about 'circle', was this about the 1-D curve or the 2-D region? Today the 1-D curve is called 'circle' and the enclosed 2-D curve is called 'disk'. At least that is what I have been told. :)
But I really *really* like BPRP's application of 'injective'. Well done!
5:49 the best part so far
amazing, thanks
Wow! Never learned about injective before.
Well I have one such proof 0 factorial equals 1 factorial; canceling out factorial on both sides we get zero equal to one.
let's divide both sides by zero, that seems pretty legit
for e=pi
We can resolve this equation with ln, but we obtain 0=0
One day i will meet you sir as a student.❤️ You are just amazing
I hadn't watched this guy for a while and now he has a beard?
Want a detailed Ostrogradski method (integrating rational functions) lecture.
I like the T-Shirt You are wearing today!
haha why did i never hear of this. i need to inject "injective" into my vocabulary lmao
Have a like
Question for 4:40
What if a line is perfectly horizontal? Like if it was y=1 or even y=69
Does that count as injective or not?
No
We all know that all these proofs are wrong but still we are watching...
Because we love bprp..
Huge appreciation from India🇮🇳🇮🇳.
I love this reason👍🏻
Bprp i want your shirt ... But I ant differentiation one .... Is there any of yours ?
Nevertheless the right explznation and I never had it, not divding by 0 thus I felt it more and more as a esotoric magic dogmas during my studies such a way I also became very anxious xith pdychiatric conséquence because of it. So thank you !
I don't see anything wrong with e=π. Also, I'm an engineer.
/s
Oh god oh rip
Wait, you’re telling me they aren’t the same?
Like the engineer's proof that all of numbers are prime.
1 is traditionally ignored.
3 works
5 works
7 works
9 merely a tolerance error
11 works
Proof by enumeration. Sorted.
@@trueriver1950 lol
so injective means that the inverse function would also be a function right? but im guessing that even if it isnt injective you can write f(x)=f(x') --> x=x' if the horizontal line would cross the function at only one point.
Nice ,Injectivity is a wonderful subject .
I haven't watched your videos in a while and wow what a beard
Sir in first proof if we say sin(π)=sin(2π) so we take general solutions that is for sin(nπ)=0 n=1,2,3....
And what is difference between injective and bijective function and how to check function is onto or into?
So if I can put this in terms that an elementary school student might understand. A function is anything that has only one specific output for a certain input. In other words, if I put any value into a function and I know what that function is, I can calculate what comes out. If each output from a function is unique, and matches one and only one specific input, that function is injunctive. If I can have different inputs giving me the same output, that function is not injunctive. Do I have that right?
ceil(e) = floor(pi)
Change my mind
dont wanna
You are correct. In fact, you can define an equivalence relation ~ such that x ~ y ceil(x) = floor(y) or ceil(y) = floor(x). You can define the equivalence class [x] = {y | y ~ x}, and this implies [e] = [π].
Engineers were right all along :P
@@angelmendez-rivera351 Thankyou for this explanation! Btw engineers aren't right they're just lazy to calculate the fractions :)
Throughout the whole video my younger brother kept on wondering that which Pokemon is there inside that Pokeball
Also (not quite the point of the video with its being about injective and non-injective functions)
0*2 = 0*1
(0/0)*2 = 1 (divide both sides by 2)
(0/0) = 1/2
and
2 = (0/0)*1
2 = (0/0)
Contradiction
This is not a contradiction unless you explicitly require multiplication to be associative. In particular, if I have the equality 0·1/0 = 1, I can left-multiply by 2, leaving 2·(0·1/0) = 2·1 = 2, and this is still free of contradictions. The problem emerges when you assume that 2·(0·1/0) = (2·0)·(1/0) must be true. This precisely implication is where division goes wrong. In other words, a set with an element 1/0 cannot have an associative product that distributes over addition.
Very topical
- i/(2*pi) = 1/0
Got from here.
exp^(i*2*pi)=1
i*2*pi=ln(1)
1/0 = -i/(2*pi)
Any explanation?
Thanks b(blue , black) pen , red pen for proper explanation. DrRahul Rohtak Haryana India
@@at7388 No sir .I am PhD in physics .My name is Rahul only and Gupta is my family name .
ii think i got the idea of “injective” but the reason why we cant divide by zero is because it gives the expression that goes to infinite ♾ , though am not sure what that means maybe u could explain it ?
"When you can snatch the pebble from my hand, then you will be ready to leave."
5:44- e=pi=3. said every engineer ever.
e=pi? Engeneers: normal stuff
Long time no watch. what a beard! more muscles too?
My favorite one of these:
Theorem: -1 = 1
Proof:
1=1 -1=-1 ( -1)/1=1/(-1) sqrt((-1)/1) = sqrt(1/(-1)) sqrt(-1)/sqrt(1) = sqrt(1)/sqrt(-1)
cross multiply ->
sqrt(-1) * sqrt(-1) = sqrt(1)*sqrt(1) -> -1=1 qed.
Assume that we have two variables a and b, and that: a = b
Multiply both sides by a to get: a2 = ab
Subtract b2 from both sides to get: a2 - b2 = ab - b2
This is the tricky part: Factor the left side (using FOIL from algebra) to get (a + b)(a - b) and factor out b from the right side to get b(a - b). If you're not sure how FOIL or factoring works, don't worry-you can check that this all works by multiplying everything out to see that it matches. The end result is that our equation has become: (a + b)(a - b) = b(a - b)
Since (a - b) appears on both sides, we can cancel it to get: a + b = b
Since a = b (that's the assumption we started with), we can substitute b in for a to get: b + b = b
Combining the two terms on the left gives us: 2b = b
Since b appears on both sides, we can divide through by b to get: 2 = 1
The problem is that a - b = 0. You divided by 0. In other words, your cancellation assumed that 0/0 = 1, while also using the associative properties of multiplication, which is not at all compatible with dividing by 0.
@@angelmendez-rivera351 Well explained.
An interesting corollary to this explanation of the flaw in the reasoning is that in the usual method for solving partial fractions we do exactly that.
It works, so it's an easy way to find A and B in practice, but proving why it works is actually quite hard, and finding A and B without dividing by zero is harder to learn...
@@trueriver1950 Yes, you are right. Partial fraction decomposition is something that, when done rigorously, is rather tricky and needs precautions.
One way you can try to justify the mamipulations is telling students these cancellations produce the correct constants because the terms that are being cancelled only have a removable discontinuity. Of course, this assumes that they have already learned about the concept of removable discontinuity correctly, which is often not the case.
Ultimately, this is what makes partial fraction decomposition difficult to teach appropriately: students will need prerequisite knowledge that, in theory, was taught to them already, but knowing how the system of education is, they likely did not learn it.
another explanation that doesn't use injective function: a/b is defined as the unique number that multiplied by b gets you a. 5/0 isn't defined because no number like that exist, 0/0 doesn't exist because there isn't a unique number that multiplied by 0 gets you 0.
When I am covering why dividing by zero does not work, I tend to relate it to a multiplication problem instead. For example, 12 / 4 means the same as: what number when multiplied by 4 gives you 12? The answer to the division and the multiplication is 3. Let's try it with 0. 0/0 =? means: What number, when multiplied by 0, yields 0? Well that works with any number. That is no good. We want a single answer for division. If the student has not yet taken calculus, I would mention that in their Calculus class, we will cover limits, which will usually let us figure out which 0/0 we are actually dealing with. Then I ask the student about 3/0, which means: What number, when multiplied by 0, gives 3? There are no real answers. We have gone from feast to famine.
dividing by 0: inversion of nothing is everything.
Iove that shirt
My teacher just finished the relation ans function chapter but never taught about this .... That is interesting........
So do all injective functions then have to monotonically increase or decrease?
Next time someone asks that, divide him by 2.
is this one of these so-called lockdown beards?
Is 1^x the least injective function?
Ohh , now you teach about functions 👍