@@blackpenredpen bro you can't just write ln(0) because it doesn't exist. And you can't just come and try to break this logic. You made some serious mistakes.
I have a weird proof... so we know that anything * 0 = 0 right? well we can replace the anything with x. now we know that in the equation 0x = 0, x is all real numbers right? now we can divide the coefficient of x from the equation, and get x = 0/0 and we said anything * 0 = 0? 0/0 = anything 😮 😮 😮
@@blackpenredpen shouldn't be "contradiction"? As the TH-cam closed caption says "contradiction" and u write "contradition". But still best provement 😀
@@mondherbouazizi4433 I know how that stuff works, thanks for explaining though. Just as a reminder for you, not everyone is trying to be correct in their comments on random youtube videos. In this case I just tried to explain that ln(0) reminded me of negative infinity as that is where the limit goes and this gave me the idea that this is probably where funky stuff is happening. I did not word it correctly, as again, cba.
@@mondherbouazizi4433 Verbally it's sometimes unnecessary to say "the limit as x approaches infinity bla bla" , especially when it's clear from context that's the intended meaning.
@@mondherbouazizi4433 infinity, infinity, 1, and the last one is neither, they are the same. However these can be different in context of an equation, as these are just numbers.
As soon as he pulled out 0^0, I got hooked. Because I know there's no way to prove 0^0 is not equal to 1, so I immediately knew that was the false one, even before seeing the proof
@@zerotwo9607No, you can't because it isn't really "true", it's just a matter of definition. I pretty much agree 0^0 =1, but for some reason people confuse it with a case 0^m = 0 even though that's only true for positive values of 'm'. In fact, its value is often taken as many fields of maths, we just leave it undefined in standard maths is because our school teachers told us so, the reason for that is that they didn't themselves understand the whole thing very clearly, or simply couldn't bother to explain the technical details to the students
It’s not a matter of definition, it’s a matter of context. It’s indeterminate so depending on where you got your zeroes, it could be equal to 0 or 1 or 69 or whatever. This is why you can’t prove it’s not equal to 1 without saying what function the zeroes came from
@@anshumanagrawal346 hey so I don't actually know what I'm talking about sorry 😅, never been taught any of this, but is it that x^0 is x/x? And isn't 0/0 undefined? Because x^4/x^1 is x^3, and x^2 is just x^3/x^1, same with just x^1 it's x^2/x^1 so x^0 must be x^1/x^1, which is 0/0,
@@user-en5vj6vr2uThere's a very clear distinction between 0^0 (exact form), and some function whose base and exponent both go to 0, and according to you the greatest integer function of 0, should also be undefined as it's also an indeterminate form
It would be neat if you would prove Heron's theorem for the area of a triangle, because a lot of people have never seen it. It turns out to be fairly straightforward given the definition of sin and cos, the law of cosines, and factoring the difference of squares three times. The law of cosines is itself pretty easy to prove from the Pythagorean Theorem, definition of sin and cos, and a bit of algebra, with the lemma that sin²θ + cos²θ = 1. If you also prove the Pythagorean Theorem to start, this would be totally awesome.
Bob's number is 5. Here's why: In the first statement, Alice says she doesn't know who's number is bigger. This means she doesn't have 1 or 9. Bob also doesn't know meaning he doesn't have 1, 9, 2 or 8. Alice still doesn't know, meaning she doesn't have 1, 9, 2, 8, 3 or 7. Bob still doesn't know, meaning he doesn't have 1, 9, 2, 8, 3, 7, 4 or 6. Therefore he must have 5. ( And Alice must have 4 or 6 ).
The interesting thing about problem 2 is that in most scenarios, if you define as an axiom that 0^0 = 1, then the answers will be consistent. I guessed immediately even before I saw it from this that problem 2’s proof would probably be the faulty one (which it was). Of course setting 0^0 = 1 can lead to some issues which is why it’s normally left undefined, but there are many cases where that definition leads to consistent results.
Thanks for showing the right proof that 2^x ≠ 0. Some people accidentally provide the wrong proof: [1] Assume 2^x = 0 [2] Then, 1/(2^-x) = 0 [3] Multiply both sides by 2^-x: 1 = 0 However, this proof already relies on the fact that 1/(2^-x) is defined in [2], which tells us that 2^-x ≠ 0, telling us that we already know 2^x ≠ 0. Therefore the proof is wrong.
I believe it is the second proof because at one step, you subtract (ln 0) when (ln 0) isn't defined. If it was defined to be negative infinity, we have -infinity-(-infinity), which is infinity-infinity, which is undefined.
Well, it really depends on what you mean by "division." If you mean that we cannot multiply by the multiplicative inverse of 0, then BPRP is correct. Even in wheel theory, 0 has no multiplicative inverse. Rather, / is defined as a unary involution that specifically for 0, gives a different quantity, not the multiplicative inverse. It is an extension of division, but it can be argued whether this extension deserves to be called "division" or not. So again, it depends.
@@angelmendez-rivera351 Let's call it, "Division PRO"! "Have you been having trouble trying to divide by zero? Are you tired of having to deal with inventions from centuries in the past? Then we have just the thing for you, Division PRO™! With Division PRO™, you will be able to divide by zero and so much more! Just call ((002)-001-0001 +1)/0 to order your own Division PRO™ for only $159.99 today!"
Taking the log base 2 in proof 3 of 2^* is also doing ln0 from our definition so the proof is invalid in the same way as proof 2. Instead from that step it's much better to multiply by 2^n and take a limit as n goes to infinity!
Imagining finding a proof method that a proof method, which works, isn't correct. And then proving that the first method is incorrect, because it proves that the second method, which the first method would prove incorrect, always produces correct proofs. I hope I wrote it correctly.
I figured that proof 2 was incorrect because Ln(0) is undefined, I wondered if the proof would work as a limit, but then you would still have ln(0) after the multiplication inside of the Ln.
I was watching your video with no sound (for reasons beyond the scope of the most difficult integral...) and I thought by the title that the first two had to be valid and the third proof to be invalid. (2 legits and 1 false). And I didn't quite understand how you "worked" with ln0 is just a quantity and continued to use rules of exponents. And then the third one I couldn't find a flaw. And then came 7:00 and it all became clear! I do like the third proof a lot!
Actually, there are cases where 0^0 must equal to 1. One of those is the power series of e^x - the sum of (x^n)/n! from n = 0 to inf. If we let x = 0, we get (0^0)/0! or (0^0)/1 or just 0^0. But we know that e^0 = 1 and that the first term of exponential function's power series is 1. So...
That's why it's often rewritten as 1 + sum (n >= 1) [x^n/n!] to avoid this special case I think it's better to leave 0^0 undefined and always stay away from it '_'
Fun fact, the division sign ÷ is mainly used in computer writing, in written mathematics (at least here in Czechia) we use : The same applies to / and straight division line, where / is used in computer writing and the straight when writing by hand.
There's learning from your mistakes and then there's spinning your mistakes into a format for entertaining content. That's more impressive than proof 4.
I have a question with proof 3. Since 2^* is equal to 0, and so is 2^(*+1). Isn't the step where we go from 2^(*+1)=2^(*) to *+1=* an illegal move since you are taking the log of 0 in both cases?
A better way of formulating proof is by explicitly working with the definition of division, with a/b = a·b^(-1), rather than keeping the symbol for division. The former makes it easier to see why the proof works and why it leads to a contradiction. 0^(-1) is the solution to 0·x = 1. With 0·1 = 0·17, you can left-multiply by 0^(-1), so 0^(-1)·(0·1) = 0^(-1)·(0·17), and by associativity, this is equivalent to [0^(-1)·0]·1 = [0^(-1)·0]·17. Since 0^(-1)·0 = 1 by definition, 1·1 = 1·17, hence 1 = 17. The second proof can be immediately understood to be incorrect, solely on the basis that ln(0) is undefined. Also, the assumption that ln(0^0) = 0·ln(0) is incorrect for the same reason that ln[(-1)^2] = 2·ln(-1) is incorrect. The equation ln(x^y) = y·ln(x) is not correct for every x and t: x > 0 is a requirement. The third proof does make some assumptions, but ultimately, it is still true that 2^x = 0 has no solutions. What it boils down to is that the codomain of every exponential function is C\{0}. Another example of a bad proof is the proof that 0^0 = 0/0. People say 0^0 = 0^(1 - 1) = 0/0. However, this proof method is clearly invalid, since 0^2 = 0^(3 - 1) = 0^3/0 = 0/0, yet we can obviously agree that 0^2 = 0·0 = 0. So the proof method above is invalid. Of course, the reason is simple: if you substitute a value into an equation knowing that it does not satisfy the equation, then obviously a contradiction will arise, especially if one the parts is undefined, but that does not immediately imply every part in the equation is undefined.
1:45 0/0 is indeterminate. Every complex number = 0/0. It need not be 1 both on the LHS and the RHS. The equality holds for some value of 0/0 on the LHS and some value of 0/0 on the RHS. In this case it is n and n/17 on the LHS and the RHS respectively where n is any arbitrary complex number. Confused? Let me show an example to wrap this concept around your head. Consider the equation √(9)+3=0. We know that √9=3 and √9=-3. In reality, √9=-3 is the solution for this equation. We just can't put any value of √9 as we like in the equation. In that case, 6=0 which is not the case. So the moral of this is that an equation involving indeterminates is true for some particular value of the indeterminate expression. If we insert any value we like, we may end up getting absurd results. Hence, this proof that division by zero is not possible is false. 6:30 Also, this is only satisfied by infinity. And ∞ is a real number, rather a special one. Same thing here also, ∞-∞ is indeterminate. The value of ∞-∞ on the LHS should be n-1 and that on the RHS should be n where n is any arbitrary complex number. So this proof is also false.
The first method is also false. If you consider that 0/0 = 0, then it would work. But you considered that 0/0 = 1. So it failed. But this depends on what you define what divisibility is for the zero case
1:40 "of course this and that, this and that will cancel,"said blackpenredpen. But in this case you can't cancel 0 with 0 because that would be saying that 0/0 is 1 but that is what we don't know so it is not a correct proof but I'll take your word for it.
Since we're doing a proof by contradiction, we're purposefully looking for that contradiction. You're correct that 0n = 0n wouldn't prove anything since there would be no contradiction. Instead, take m ≠ n and write 0m = 0n (this should still be true since both expressions are equal to 0). Then divide by zero and we get m = n, which contradicts the choice of m ≠ n. As for not knowing what 0/0 is, you're correct, we don't know what it equals. But we are assuming it exists (for contradiction). You can make this more rigorous by changing "divide by 0" to "0 has a multiplicative inverse." Call this inverse whatever you like (1/0 or 0^-1 or z if you'd like. Regardless, you can take the equation 0m = 0n and multiply both sides by the multiplicative inverse of 0. This leaves you with 1m = 1n (since multiplicative inverses multiply to 1 by definition). But this is a contradiction. Therefore, 0 has no multiplicative inverse.
My way of understanding why dividing by 0 is undefined and not infinite as many people like to think is when I consider the formula, f=ma. We often use this in mechanics to say that if there is no acceleration there is no resultant force but it doesn’t imply the mass is infinite, in fact it tells us nothing about the mass. The mass could be a tiny number up to an infinitely large one but we just have no way of knowing and therefore it is undefined.
The answer to Bob and Alice question is Bob has the number 5. Explanation: It's a one digit number (1 - 9), alice and bob has different number. When alice says she doesn't know whose is bigger, that means bob now knows her number is neither 1 or 9. When Bob says he doesnt know whose is bigger after alice, alice now knows his number is neither 2, 8, 1, or 9. When Alice says she still doesnt know whose is bigger, Bob knows her number is neither 3 or 7 plus the first four above. So her number's either 4,5 or 6. Lastly, bob says he still doesnt know, that means he has the number 5. He still doesnt know because alice's could be either 4 or 6. So i guess after that convo, alice would be the first to know Bob's number.
At 4:10 you had ln(0) = 1, which is not contradictory, becuase you can substitute into your other equation, 0 * ln(0) = 0 Doing so yields 0 * 1 = 0, which we know is true. Your mistake in the second proof was assuming that ln( 0 * ln 0 ) = ln 0 + ln( ln 0 ). The issue with that equation is that 0 is erasing information, and thus it can not be distributed here.
I agree, the more elegant way i would do is prove that for every real number a, star+a is also a solution, so we can conclude that the function 2^x is identically equal to 0, and then we have a contradiction
This is actually ok because we assume 2^* = 0. Since the logarithm is the inverse of the exponential function by definition, this means log2(0) = * is also true.
The #1 assuming that dividing by 0 doesn't change the field's division proprieties. The #2 ln(0) is not defined in the field. 2 lies 1 proof. #2 is incorrect for the reasoning that it is both equal to 1 and 0 at the same time which we cannot accept. #1 proof would need a field theory demonstrating we go against field assumptions if we divide: is not as easy as doing an example and calling it done. Also enlarging the field to wheel algebra we find it possible to do it.
At 1:40, you canceled the 0 in the numerator with the 0 in the denominator. I'm not sure you can do that... I might be wrong, but in order to do that, you assume that 0/0=1 But what if 0/0 doesn't equal 1? Then you can't cancel like that. What is 0/0 = 0, for example?
Actually, the reason 0^0≠1 is just an convention to make the function continuous and easy to work with in calculus. In other field of mathematics like set theory and combinatory theory, 0^0 is defined to 1 because it makes combinatorics proofs easier to deal with
*AT TIMESTAMP [**02:52**]:* HOLD ON! 🛑 _You can NOT even plug zero into a logarithm with a non-zero base!_ If you try the Above, the logarithm will immediately fail. This is also an exception to the Rules of Exponents.
Thanks! I never knew that 1=17 and mistake is spelled miSTEAK! Sounds so yummy. And also didn't knew that you can drop a c from CONTRADICTION>>>contradition
7:54 Alice and Bob are both given a different 1-digit positive number (excluding 0). They make the following statements, in order: Alice: "I know whose number is bigger." Bob: "I know whose number is bigger." Alice: "I know whose number is bigger." Bob: "I know whose number is bigger." What is Bob's number? (1) (3) (5) (7) (9) Me: "..." 📢SPOILER ALERT📢 (First Alice doesn't know whose is bigger so she can't have 1 nor 9. Then Bob also doesn't know and Alice doesn't have 1 nor 9 so Bob doesn't have 1,2,8,9 because otherwise he would know which one had a bigger number. Continuing like this, 3rd statement tells us that Alice doesn't have 1,2,3,7,8,9 and finally, the 4th statement tells us that Bob doesn't have 1,2,3,4,6,7,8,9 so Bob's number is 5.)
That last proof also seems a little sketchy. In order to go from 2^(x+1) = 2^x to x+1 = x, you need that the exponential functions are 1-1. That fact might depend on the exponential not being 0, and if it does your proof is circular (Im not 100% sure that this is the case though). So if you can prove injectivity of exponential functions without using the fact that they are nonzero, then the proof is right, I just don't know a way to do so. You can prove that the exponentials are nonzero by doing so first for integer exponents, then rationals and then making a limiting argument to extend it to the reals. This would be the more secure way which doesn't depend on other properties of exponents.
By assuming it's a solution, it becomes defined on 0 (i.e. we assume it is defined on 0). The contradiction shows that if it's defined, it results in 1 = 0, so it must be undefined
Hey, I am researching on Riemann hypothesis for the last 3 years and successfully I have found a formula for 'a' in Zeta(a+ib) in terms of 'b' and some hard looking definite integrals which are in terms of 'a' and 'b'which I was not able to solve so I am challenging you to solve those integrals . I used Riemann xi function and Riemann functional equation for Zeta function to find the formula for 'a'. Please make a video on those integrals. Integrals are:- 1)integral from 1 to infinity of x^((a-2)/2)*f(x)*cos(ln(x)*b/2) + x^(-a-1)*f(x)*cos(ln(x)*b/2) 2)integral from 1 to infinity of x^((a-2)/2)*f(x)*sin(ln(x)*b/2) - x^(-a-1)*f(x)*sin(ln(x)*b/2) Where f(x) = summation from n=1 to infinity of e^(-(n^2)*π*x)
I think you are Indian(By username) Well I also work on Reimann hypothesis and Zeta function , and I found so many simple looking integral which are not so simple , and their exact values , infinite products and series as well using Zeta function I think you should take f(x) in terms of Jacobi theta function and use functional equation of Jacobi theta function (which also used to derive Zeta functional equation)
We cannot divide by zero Method of contradiction.. Let's suppose that on dividing any nonzero real no. a by zero we get another real no. b a/0=b Taking zero to the right side we get a= 0 x b a = 0 Which contradicts our assumption that a is nonzero.
@@danielyuan9862 a/0 = b does not imply a = 0·b. If b is a unit, then there exist elements defined by being solutions of the equation b·x = 1. If · is associative, then the above equation only has 1 solution, which we call b^(-1). If, furthermore, · is also commutative, then a·b^(-1) = b^(-1)·a, and division is defined a/b := a·b^(-1). An algebraic structure with these properties is called an Abelian group. In such a structure, you can have c = a·b^(-1), which implies c·b = [a·b^(-1)]·b = a·[b^(-1)·b] = a·1 = a. c·b = a also implies (c·b)·b^(-1) = c·[b·b^(-1)] = c·1 = c = a·b^(-1). So c = a/b is equivalent to c·b = a, but this is true if and only if b^(-1) exists, and · is both commutative and associative. In other words, c = a/b is equivalent to c·b = a if and only if a, b, c are elements of a set S, where (S, ·, 1) is an Abelian group. This is where the issue comes in. Abstract and universal algebra aside, in the vast majority of mathematics, theories assume a ring structure as a background. Suppose I have an Abelian group (S, +, 0). If (S, ·, 1) is a monoid that is left-distributive and right-distributive over (S, +, 0), then the structure (S, +, 0, ·, 1) is called a ring. A monoid is an Abelian group without the commutativity, and without the guarantee of existence of inverse elements everywhere in S. Monoids are particular in that, while inverse elements are not guaranteed to exist, whenever they do exist, they are guaranteed to also be unique. Why do I mention rings? Because in a ring, 0·x = 0 is necessarily always true. As such, there is no element of S such that 0·x = 1, and so 0^(-1) does not exist. So (S, ·, 1) can never be enriched into being a group, let alone an Abelian group, because 0 has no multiplicative inverse. You can adjoin an element q to the set S, so that in the structure (S[q], ·, 1), 0·q = 1, hence 0 and q are ·-inverses of one another. However, the inevitable consequence of doing this is that (S[q], ·, 1) is no longer monoid, but a loop, and in a loop, · is not necessarily associative, and in this case, it definitely is not associative. As such, a/0 = b still does not imply a = 0·b in this situation. Furthermore, you can modify distributivity so that (S[q], +, 0) is a monoid with commutativity, or a loop with commutativity, but you can also never have a group with (S[q], +, 0).
I think there’s a staggering flaw in the first proof as well. You implicitly imply that 0/0=1 when you attempt to cancel them out, which is wholly unjustified
I think the middle proof of 0^0 != 1 is the lie, as you can't subtract ln(0) on both sides (it is undefined) Will note that the frst proof could probably use a slightly more rigourous definition than "you cannot divide by 0" as this feels a little vague, for proof standards
If u ask me tho, the third proof is kinda invalid too, as 2^x is set to be equal to 0, so is 2^(1+x), you can't take log on both side as they are both zero.
If im to prove that, i would say it's the same as proving if x∈ℝ, then 2^x≠0 Proof by cases: Case 1: x>0 Because 2^x>0 ∀x>0, so 2^x≠0 P.S. 2^x=e^(xln2), Bec ln2>0, we can use power series of e^x to proof 2^x>0 Case 2: x=0, then 2^x=0 Case 3: x0 then 2^x =1/2^y, But from case 1, 2^y>0 ∀ y∈ℝ, 2^x >0 ∀ x∈ℝ Hence, 2^x≠0 ∀x∈ℝ
Can u just compare exponents for the 3rd case? Under our assumption, 2^star = 0 so doing any form of log is invalid hence we can't compare exponents. I would say star + 1 must also be a solution and by continuing the logic, star + 2 and star - 1 are also solutions. This means 2^any integer = 0. Then just saying 2^1 = 2 would cause a contradiction.
0^0 has multiple answers depending on how do you try to solve this problem for example if you assume 0^x=0, then 0^0=0, but if you assume x^0=1, then 0^0=1, but 0 != 1
Why assume either 0^x = 0 or x^0 = 1? How exponents work is that by definition x^0 = 1, and the other powers are achieved by multiplying or dividing numbers by x, so 0^0 = 1.
I'm worried that the first and third proofs aren't entirely valid either. For #1, you've only shown that you can't divide by 0 WHEN THE NUMERATOR IS ALSO 0, but that doesn't necessarily imply that you can't divide by 0 in any case. For #3, you make the assumption that the exponential function 2^x is one-to-one, so that there cannot be two distinct arguments with same output. Without this assumption, it doesn't logically follow that 1+STAR=STAR as you write at 6:23.
"I thought I came up with a wonderful proof showing 0^0 is not 1 [...]" If you ever come up with another one, my suggestion is to test it on other powers of 0. For example, test whether your "proof" would also show 0^2 is not 0. In my experience, pretty much every "proof" that 0^0 is not 1 also shows that 0^2 is not 0. Seeing this, we can conclude that the "proof" is flawed :)
I suspect his English is way better than you are in brpr's first language -- even if you aren't required to use the traditional symbols. I suspect your transliteration to Latin characters would also fall short.
@@artsmith1347 I suppose your comment was addressed to me? If so, I have no problem to admit that indeed English is not my native language since I'm German. And I'm aware that I regularly make my mistakes in foreign languages, this is completely normal. However this does not exclude the possibility that I also "am allowed" to draw someone's attention to their mistakes if I find some. At least in this case when the same mistake catched my eye several times. To me your comment sounds as if I had been "rude" having fun complaining about other's mistakes while claiming that I'm "perfect". Both is not the case. I emphasized that I didn't mean to offend bprp at all. It was just that one spelling mistake I found kinda funny while I keep on cherishing his mathematical content. If you can't read that from my comment above, I can't help you. 🤷🏻♂️
Absolute zero, like infinity, is an abstract number with no precision. There's no result trying to divide by absolute zero or raise absolute zero to zero because the number has no scale.
The problem I believe in math that has been stuck with so many contradiction, is that zero is not a number. Zero is the absence of any number. So using at a number you will have confusion...
Wait a minute... (0*1)/0 Can you even cancel the zeroes out? How does that work? I mean, usually cancelling factors works like... 6/3, you divide both numerator and denominator by 3, and then you get 2/1. But what factors do you even cancel out at (0*1)/0 ? Also, suppose the cancelling works out.... doesn't this result in 0/0 = 1? Oh, and... As far as I understood division, it is all about how often I can subtract the second number from the first number until it reaches 0, compared to multiplying, which is adding number x an amount of y times. So... Based on that, I actually thought of what divison by 0 actually could be. 1/0 = ? You need to subtract 0 an infinite amount of times from 1 to get it to 0 (or even more often than that, but that's the only uncertainty I have). So, let's define 1/0 as ∞. Now, what would 2/0 be? Well, you also need to subtract 0 from 2 an infinite amount of times. Sooo.... Both are infinity? And Infinity * 0 = 1 = 2 ? Well, no. The infinity from the second division is twice as large as the one from the first division. So, 1/0 is actually ∞[1], and 2/0 is ∞[2], and so on. At least that's how I would define it.
The assumption was that we can divide by zero, so that's why in that example zeros cancel out. In reality of course it doesn't work since you can't divide by zero
The 2nd cause ln(0) doesn't exist, there's no number n such that eⁿ=0, it will have to be a limit which will give you -∞ in which case your equasion is -∞+ln(ln(0))=-∞ -∞=-∞ And it's a true statement
I'd say the second proof is a false one for two reasons. First of all, 0^0 can be either 1 or 0 depending on your definition. Second, when taking the ln of 0^0 you couldn't use the exponent rule. The exponent rule can only be used if what is left is still a valid argument. In this case we were left with ln(0) which you can say is negative infinity (or undefined), but infinity is not a number so you can't really keep doing algebra on it. Anyways I loved the video and I learned a lot from it. I like seeing proofs for things, really makes you think hard things that seem trivial at first glance.
Hello. I wanted to say that I proved that n (n is a positive real number with n cannot be equal to 0) divided by 0 is defined. I am not sure of that statement because I am a student. I will show it to my teacher to see if my proof is right.
The issue when trying to prove such elementary facts is that many of the proof rules being taken for granted are less fundamental than the proved property itself. Are we actually proving anything, or is the validity of proof rules we are using actually a consequence of the proven fact being true? My former algebra professor would probably say all three of those proofs are invalid. The only rigorous way to prove those would be to start with an axiomatic theory of arithmetics and prove everything from the axioms only.
I was not expecting a double contradiction proof. Highly appreciated.
😆!!!
It was just insanely cool ! 😂👍
Is double contradiction proof always true?
Our you just cheating !
@@blackpenredpen bro you can't just write ln(0) because it doesn't exist.
And you can't just come and try to break this logic.
You made some serious mistakes.
@@extremegamingdz1309 he literally said the same
This should have been called "Two proofs and a lie."
Hey that’s a good one. Will change now. Thanks
@@blackpenredpen you didn’t 😾
@@blackpenredpen he lied!
@@blackpenredpen the lie
@@blackpenredpen how about 2 proofs 1 spoof
My favorite part is at 9:00. Get ready to bend your mind!
I found a way to prove why 0 to the power of 0 is not equal to 1, and it's legit, reply if you want to me to explain it :P
@@zerotwo9607 well, explain it boi
@@zerotwo9607 Yeah, let's see it
I have a weird proof...
so we know that anything * 0 = 0 right? well we can replace the anything with x.
now we know that in the equation 0x = 0, x is all real numbers right?
now we can divide the coefficient of x from the equation, and get x = 0/0
and we said anything * 0 = 0?
0/0 = anything
😮 😮 😮
@@crazystuffofficialchannel4406 fam in order to get x on it's own like that you'll have to divide both sides by 0, which we all know isn't possible
I wonder if he is setting up the next poll:
What is correct spelling of “contradition”?
A) contradiction
B) contradition
Nah I am done with that spelling poll
@@blackpenredpen so you didn’t spell “contradition” four times on purpose?
@@blackpenredpen shouldn't be "contradiction"? As the TH-cam closed caption says "contradiction" and u write "contradition". But still best provement 😀
The "con" in "contradition" is like the "con" in "conman". It stands for "confidence". A strong tradition of confidence.
@@Mothuzad you meant so say as con tradition right?
Means
Confidence tradition
Thank you that you write everything on board, it is much easier to follow.
😊
As soon as I saw ln(0) I was like thats negative infinity and you’re going to abuse the heck out of it 😂
and then you abused yourself instead.
@@mondherbouazizi4433 I know how that stuff works, thanks for explaining though. Just as a reminder for you, not everyone is trying to be correct in their comments on random youtube videos. In this case I just tried to explain that ln(0) reminded me of negative infinity as that is where the limit goes and this gave me the idea that this is probably where funky stuff is happening. I did not word it correctly, as again, cba.
@@mondherbouazizi4433 Verbally it's sometimes unnecessary to say "the limit as x approaches infinity bla bla" , especially when it's clear from context that's the intended meaning.
@@mondherbouazizi4433 infinity is a number. Give me one good reason not to assume a limit when calculating if infinity is unsigned like 0.
@@mondherbouazizi4433 infinity, infinity, 1, and the last one is neither, they are the same. However these can be different in context of an equation, as these are just numbers.
When he says 1:45 "New Math", I got 2 heart attacks at once.
lmao
But 2=0 so that means you were fine! Right?
In kidney
Hooray for new math, ne-he-hew math. It won't do you a bit of good to redo math. It's so simple, so very simple... That only a child can do it!!!
As soon as he pulled out 0^0, I got hooked. Because I know there's no way to prove 0^0 is not equal to 1, so I immediately knew that was the false one, even before seeing the proof
You can prove it
@@zerotwo9607No, you can't because it isn't really "true", it's just a matter of definition. I pretty much agree 0^0 =1, but for some reason people confuse it with a case 0^m = 0 even though that's only true for positive values of 'm'. In fact, its value is often taken as many fields of maths, we just leave it undefined in standard maths is because our school teachers told us so, the reason for that is that they didn't themselves understand the whole thing very clearly, or simply couldn't bother to explain the technical details to the students
It’s not a matter of definition, it’s a matter of context. It’s indeterminate so depending on where you got your zeroes, it could be equal to 0 or 1 or 69 or whatever. This is why you can’t prove it’s not equal to 1 without saying what function the zeroes came from
@@anshumanagrawal346 hey so I don't actually know what I'm talking about sorry 😅, never been taught any of this, but is it that x^0 is x/x? And isn't 0/0 undefined? Because x^4/x^1 is x^3, and x^2 is just x^3/x^1, same with just x^1 it's x^2/x^1 so x^0 must be x^1/x^1, which is 0/0,
@@user-en5vj6vr2uThere's a very clear distinction between 0^0 (exact form), and some function whose base and exponent both go to 0, and according to you the greatest integer function of 0, should also be undefined as it's also an indeterminate form
It would be neat if you would prove Heron's theorem for the area of a triangle, because a lot of people have never seen it. It turns out to be fairly straightforward given the definition of sin and cos, the law of cosines, and factoring the difference of squares three times. The law of cosines is itself pretty easy to prove from the Pythagorean Theorem, definition of sin and cos, and a bit of algebra, with the lemma that sin²θ + cos²θ = 1. If you also prove the Pythagorean Theorem to start, this would be totally awesome.
I have done those proofs already 😊
You can also prove the law of cosines from vector rules
@@mrpie3055 If you mean the dot product formula, that won't work as the dot product formula uses law of cosines AFAIK
Everyone: Mind Blowing
Bprp: Mind Bending
StandupMaths: Mind Boggling
MindYourDecisions: Mind Your Decisions
YGOAbridged: MIND CRUSH!
Bob's number is 5. Here's why:
In the first statement, Alice says she doesn't know who's number is bigger. This means she doesn't have 1 or 9. Bob also doesn't know meaning he doesn't have 1, 9, 2 or 8. Alice still doesn't know, meaning she doesn't have 1, 9, 2, 8, 3 or 7. Bob still doesn't know, meaning he doesn't have 1, 9, 2, 8, 3, 7, 4 or 6. Therefore he must have 5. ( And Alice must have 4 or 6 ).
did you comment on the wrong video
@@yoylecake313 in the sponsored segment, there is a puzzle for the viewers
The interesting thing about problem 2 is that in most scenarios, if you define as an axiom that 0^0 = 1, then the answers will be consistent. I guessed immediately even before I saw it from this that problem 2’s proof would probably be the faulty one (which it was). Of course setting 0^0 = 1 can lead to some issues which is why it’s normally left undefined, but there are many cases where that definition leads to consistent results.
Thanks for showing the right proof that 2^x ≠ 0. Some people accidentally provide the wrong proof:
[1] Assume 2^x = 0
[2] Then, 1/(2^-x) = 0
[3] Multiply both sides by 2^-x: 1 = 0
However, this proof already relies on the fact that 1/(2^-x) is defined in [2], which tells us that 2^-x ≠ 0, telling us that we already know 2^x ≠ 0. Therefore the proof is wrong.
Ooooooh fancy!!!
2 days ago
@@pmathewizard 🧐
Wait how??
Video uploaded 1 hour ago and comment 2 days ago
*Illuminati sounds play*
LOL
Peyam teach us your ways.
1:45 My teacher, every time he is correcting my assignment ;D
I believe it is the second proof because at one step, you subtract (ln 0) when (ln 0) isn't defined. If it was defined to be negative infinity, we have -infinity-(-infinity), which is infinity-infinity, which is undefined.
1:10
Me: Oh, My favorite number is 14....
bprp: *17*
Me: Oh, Nevermind.
I was searching for this comment 😅😅😅😅
“We cannot divide by 0.”
*laughs in wheel theory*
Well, it really depends on what you mean by "division." If you mean that we cannot multiply by the multiplicative inverse of 0, then BPRP is correct. Even in wheel theory, 0 has no multiplicative inverse. Rather, / is defined as a unary involution that specifically for 0, gives a different quantity, not the multiplicative inverse. It is an extension of division, but it can be argued whether this extension deserves to be called "division" or not. So again, it depends.
@@angelmendez-rivera351 Let's call it, "Division PRO"!
"Have you been having trouble trying to divide by zero?
Are you tired of having to deal with inventions from centuries in the past?
Then we have just the thing for you, Division PRO™! With Division PRO™, you will be able to divide by zero and so much more! Just call ((002)-001-0001 +1)/0 to order your own Division PRO™ for only $159.99 today!"
In 2nd one,
You could take
0°=2°
Then,
0=2 which is not possible.
Therefore, 0° is not equal to 1.
Congrats on 700k subs
@@HrsHJ Same here... are you on instagram?
Proving a proof just blows my mind
You just proved that the proof is wrong
Taking the log base 2 in proof 3 of 2^* is also doing ln0 from our definition so the proof is invalid in the same way as proof 2. Instead from that step it's much better to multiply by 2^n and take a limit as n goes to infinity!
Imagining finding a proof method that a proof method, which works, isn't correct. And then proving that the first method is incorrect, because it proves that the second method, which the first method would prove incorrect, always produces correct proofs.
I hope I wrote it correctly.
sounds like the halting problem
I figured that proof 2 was incorrect because Ln(0) is undefined, I wondered if the proof would work as a limit, but then you would still have ln(0) after the multiplication inside of the Ln.
I thought proof 3 was incorrect because it had a multiplication with 0
7:02 That bird in the background is golden
I was watching your video with no sound (for reasons beyond the scope of the most difficult integral...) and I thought by the title that the first two had to be valid and the third proof to be invalid. (2 legits and 1 false). And I didn't quite understand how you "worked" with ln0 is just a quantity and continued to use rules of exponents. And then the third one I couldn't find a flaw. And then came 7:00 and it all became clear! I do like the third proof a lot!
I have studied about indeterminate forms but today I have studied many new things from your video, I highly appreciate your work.
Actually, there are cases where 0^0 must equal to 1. One of those is the power series of e^x - the sum of (x^n)/n! from n = 0 to inf. If we let x = 0, we get (0^0)/0! or (0^0)/1 or just 0^0. But we know that e^0 = 1 and that the first term of exponential function's power series is 1. So...
That's why it's often rewritten as 1 + sum (n >= 1) [x^n/n!] to avoid this special case
I think it's better to leave 0^0 undefined and always stay away from it '_'
the end is quite satisfying with the proof by contradiction of proof by contradiction
Fun fact, the division sign ÷ is mainly used in computer writing, in written mathematics (at least here in Czechia) we use :
The same applies to / and straight division line, where / is used in computer writing and the straight when writing by hand.
There's learning from your mistakes and then there's spinning your mistakes into a format for entertaining content. That's more impressive than proof 4.
Could this become a regular series? I would definitely watch it.
6:58 the bird tho!
I have a question with proof 3. Since
2^* is equal to 0, and so is 2^(*+1).
Isn't the step where we go from
2^(*+1)=2^(*)
to
*+1=*
an illegal move since you are taking the log of 0 in both cases?
7:18 yummy!!!
You tackle some very important topics!
Awesome! 🤩🤩🤩
Thanks!
@@blackpenredpen You're welcome!
0/0 is undefined (not equal to 1) so you can't cancel them in the first proof, which makes the first proof also false
A better way of formulating proof is by explicitly working with the definition of division, with a/b = a·b^(-1), rather than keeping the symbol for division. The former makes it easier to see why the proof works and why it leads to a contradiction. 0^(-1) is the solution to 0·x = 1. With 0·1 = 0·17, you can left-multiply by 0^(-1), so 0^(-1)·(0·1) = 0^(-1)·(0·17), and by associativity, this is equivalent to [0^(-1)·0]·1 = [0^(-1)·0]·17. Since 0^(-1)·0 = 1 by definition, 1·1 = 1·17, hence 1 = 17.
The second proof can be immediately understood to be incorrect, solely on the basis that ln(0) is undefined. Also, the assumption that ln(0^0) = 0·ln(0) is incorrect for the same reason that ln[(-1)^2] = 2·ln(-1) is incorrect. The equation ln(x^y) = y·ln(x) is not correct for every x and t: x > 0 is a requirement.
The third proof does make some assumptions, but ultimately, it is still true that 2^x = 0 has no solutions. What it boils down to is that the codomain of every exponential function is C\{0}.
Another example of a bad proof is the proof that 0^0 = 0/0. People say 0^0 = 0^(1 - 1) = 0/0. However, this proof method is clearly invalid, since 0^2 = 0^(3 - 1) = 0^3/0 = 0/0, yet we can obviously agree that 0^2 = 0·0 = 0. So the proof method above is invalid. Of course, the reason is simple: if you substitute a value into an equation knowing that it does not satisfy the equation, then obviously a contradiction will arise, especially if one the parts is undefined, but that does not immediately imply every part in the equation is undefined.
I was so happy to see how happy he was when he found the contradiction
The contradiction is this video has 3 legit proofs
Lol true that
1:45 0/0 is indeterminate. Every complex number = 0/0. It need not be 1 both on the LHS and the RHS. The equality holds for some value of 0/0 on the LHS and some value of 0/0 on the RHS. In this case it is n and n/17 on the LHS and the RHS respectively where n is any arbitrary complex number. Confused? Let me show an example to wrap this concept around your head. Consider the equation √(9)+3=0. We know that √9=3 and √9=-3. In reality, √9=-3 is the solution for this equation. We just can't put any value of √9 as we like in the equation. In that case, 6=0 which is not the case. So the moral of this is that an equation involving indeterminates is true for some particular value of the indeterminate expression. If we insert any value we like, we may end up getting absurd results. Hence, this proof that division by zero is not possible is false.
6:30 Also, this is only satisfied by infinity. And ∞ is a real number, rather a special one. Same thing here also, ∞-∞ is indeterminate. The value of ∞-∞ on the LHS should be n-1 and that on the RHS should be n where n is any arbitrary complex number. So this proof is also false.
Love this format! Do more of these please
Cancelling In (0) from both sides was wrong.
⭕
The first method is also false.
If you consider that 0/0 = 0, then it would work.
But you considered that 0/0 = 1. So it failed.
But this depends on what you define what divisibility is for the zero case
Or how about...
0/0 = #
Such that # is something that is not in the real numbers
It looks like it is coherent
“What’s your favorite number” me: Says 17, he then immediately proceeds to say 17
Again, he heard 17 is my favourite number.
1:40 "of course this and that, this and that will cancel,"said blackpenredpen.
But in this case you can't cancel 0 with 0 because that would be saying that 0/0 is 1 but that is what we don't know so it is not a correct proof but I'll take your word for it.
Since we're doing a proof by contradiction, we're purposefully looking for that contradiction.
You're correct that 0n = 0n wouldn't prove anything since there would be no contradiction. Instead, take m ≠ n and write 0m = 0n (this should still be true since both expressions are equal to 0). Then divide by zero and we get m = n, which contradicts the choice of m ≠ n.
As for not knowing what 0/0 is, you're correct, we don't know what it equals. But we are assuming it exists (for contradiction). You can make this more rigorous by changing "divide by 0" to "0 has a multiplicative inverse." Call this inverse whatever you like (1/0 or 0^-1 or z if you'd like. Regardless, you can take the equation 0m = 0n and multiply both sides by the multiplicative inverse of 0. This leaves you with 1m = 1n (since multiplicative inverses multiply to 1 by definition). But this is a contradiction. Therefore, 0 has no multiplicative inverse.
Dividing by zero is like the question "How long do I have to stand still to get to another place" 😂
the answer to both is "however long it takes someone to kick you across the room for your crimes."
My way of understanding why dividing by 0 is undefined and not infinite as many people like to think is when I consider the formula, f=ma. We often use this in mechanics to say that if there is no acceleration there is no resultant force but it doesn’t imply the mass is infinite, in fact it tells us nothing about the mass. The mass could be a tiny number up to an infinitely large one but we just have no way of knowing and therefore it is undefined.
The answer to Bob and Alice question is Bob has the number 5.
Explanation:
It's a one digit number (1 - 9), alice and bob has different number.
When alice says she doesn't know whose is bigger, that means bob now knows her number is neither 1 or 9.
When Bob says he doesnt know whose is bigger after alice, alice now knows his number is neither 2, 8, 1, or 9.
When Alice says she still doesnt know whose is bigger, Bob knows her number is neither 3 or 7 plus the first four above. So her number's either 4,5 or 6.
Lastly, bob says he still doesnt know, that means he has the number 5. He still doesnt know because alice's could be either 4 or 6.
So i guess after that convo, alice would be the first to know Bob's number.
Perfect explanation!
The moment Ln showed up I had a bad feeling. Anything with euler cant be trusted to act normal
At 4:10 you had ln(0) = 1, which is not contradictory, becuase you can substitute into your other equation, 0 * ln(0) = 0
Doing so yields 0 * 1 = 0, which we know is true.
Your mistake in the second proof was assuming that ln( 0 * ln 0 ) = ln 0 + ln( ln 0 ). The issue with that equation is that 0 is erasing information, and thus it can not be distributed here.
1st suspicion: "wait, isn't 0^0 actually 1"
2nd suspicion: "ln0"
2:15
Phone calculators:
and we took that personally
2 because i tried to think what e^iz would be 0 but there would be no solution, so i concluded ln0 was undefined and cant be used like a variable
expanding a bit bc to get 0, the angle would have to be 0 +2npi, but e^0 = 1 so that is impossible.
1:13 I ACTUALLY SAID 17 WTF
lol!!!
The third proof is also invalid, you cannot take log of 2 to the star on the right because 2 to the star is 0. That would be undefined again.
I agree, the more elegant way i would do is prove that for every real number a, star+a is also a solution, so we can conclude that the function 2^x is identically equal to 0, and then we have a contradiction
This is actually ok because we assume 2^* = 0. Since the logarithm is the inverse of the exponential function by definition, this means log2(0) = * is also true.
This can be rectified by instead noting that 2^x is strictly increasing on R.
I am just amazed how the heck did you read my mind 17 is my favourite number ❤️
Mine is too i almost dropped my phone when i heard him say that!
The #1 assuming that dividing by 0 doesn't change the field's division proprieties. The #2 ln(0) is not defined in the field. 2 lies 1 proof. #2 is incorrect for the reasoning that it is both equal to 1 and 0 at the same time which we cannot accept. #1 proof would need a field theory demonstrating we go against field assumptions if we divide: is not as easy as doing an example and calling it done. Also enlarging the field to wheel algebra we find it possible to do it.
At 1:40, you canceled the 0 in the numerator with the 0 in the denominator.
I'm not sure you can do that... I might be wrong, but in order to do that, you assume that 0/0=1
But what if 0/0 doesn't equal 1? Then you can't cancel like that. What is 0/0 = 0, for example?
Actually, the reason 0^0≠1 is just an convention to make the function continuous and easy to work with in calculus.
In other field of mathematics like set theory and combinatory theory, 0^0 is defined to 1 because it makes combinatorics proofs easier to deal with
*AT TIMESTAMP [**02:52**]:*
HOLD ON! 🛑 _You can NOT even plug zero into a logarithm with a non-zero base!_ If you try the Above, the logarithm will immediately fail. This is also an exception to the Rules of Exponents.
One of the best explainer of mathematics on TH-cam...
@1:41
2 why is the first proof correct?
By asuming, that we can divide by 0, we can not conclude, that 0/0 = 1.
Or do i miss something?
When in the first proof your favorite number is 1.
Thanks! I never knew that 1=17 and mistake is spelled miSTEAK! Sounds so yummy.
And also didn't knew that you can drop a c from CONTRADICTION>>>contradition
and 0=e?!!!
haha 1=0
7:54
Alice and Bob are both given a different 1-digit positive number (excluding 0). They make the following statements, in order:
Alice: "I know whose number is bigger."
Bob: "I know whose number is bigger."
Alice: "I know whose number is bigger."
Bob: "I know whose number is bigger."
What is Bob's number?
(1) (3) (5) (7) (9)
Me: "..."
📢SPOILER ALERT📢
(First Alice doesn't know whose is bigger so she can't have 1 nor 9. Then Bob also doesn't know and Alice doesn't have 1 nor 9 so Bob doesn't have 1,2,8,9 because otherwise he would know which one had a bigger number. Continuing like this, 3rd statement tells us that Alice doesn't have 1,2,3,7,8,9 and finally, the 4th statement tells us that Bob doesn't have 1,2,3,4,6,7,8,9 so Bob's number is 5.)
That last proof also seems a little sketchy. In order to go from 2^(x+1) = 2^x to x+1 = x, you need that the exponential functions are 1-1. That fact might depend on the exponential not being 0, and if it does your proof is circular (Im not 100% sure that this is the case though). So if you can prove injectivity of exponential functions without using the fact that they are nonzero, then the proof is right, I just don't know a way to do so.
You can prove that the exponentials are nonzero by doing so first for integer exponents, then rationals and then making a limiting argument to extend it to the reals. This would be the more secure way which doesn't depend on other properties of exponents.
6:17 you’re applying ln on 2^(*+1) = 2^*, but assuming 2^* = 0, it’s undefined too. Is there 2 demonstration false ?
By assuming it's a solution, it becomes defined on 0 (i.e. we assume it is defined on 0). The contradiction shows that if it's defined, it results in 1 = 0, so it must be undefined
Hey, I am researching on Riemann hypothesis for the last 3 years and successfully I have found a formula for 'a' in Zeta(a+ib) in terms of 'b' and some hard looking definite integrals which are in terms of 'a' and 'b'which I was not able to solve so I am challenging you to solve those integrals . I used Riemann xi function and Riemann functional equation for Zeta function to find the formula for 'a'.
Please make a video on those integrals.
Integrals are:-
1)integral from 1 to infinity of x^((a-2)/2)*f(x)*cos(ln(x)*b/2) + x^(-a-1)*f(x)*cos(ln(x)*b/2)
2)integral from 1 to infinity of x^((a-2)/2)*f(x)*sin(ln(x)*b/2) - x^(-a-1)*f(x)*sin(ln(x)*b/2)
Where f(x) = summation from n=1 to infinity of e^(-(n^2)*π*x)
I think you are Indian(By username)
Well I also work on Reimann hypothesis and Zeta function , and I found so many simple looking integral which are not so simple , and their exact values , infinite products and series as well using Zeta function
I think you should take f(x) in terms of Jacobi theta function and use functional equation of Jacobi theta function (which also used to derive Zeta functional equation)
I am in 11 grade , and I leave maths 2months ago due to pressure to gain marks
I think I shouldn't look here
@@pardeepgarg2640 okay I will try
@@pardeepgarg2640 okay, I know, indian educational system is not very good.
We cannot divide by zero
Method of contradiction..
Let's suppose that on dividing any nonzero real no. a by zero we get another real no. b
a/0=b
Taking zero to the right side we get
a= 0 x b
a = 0
Which contradicts our assumption that a is nonzero.
Your proof is invalid. There is no such a thing in mathematics as "taking the 0 to the other side." There is no mathematical operation for this.
it's more like by definition, division is the inverse of multiplication, so
a/0 = b
implies
b*0 = a
and the rest is the same
@@danielyuan9862 a/0 = b does not imply a = 0·b. If b is a unit, then there exist elements defined by being solutions of the equation b·x = 1. If · is associative, then the above equation only has 1 solution, which we call b^(-1). If, furthermore, · is also commutative, then a·b^(-1) = b^(-1)·a, and division is defined a/b := a·b^(-1). An algebraic structure with these properties is called an Abelian group. In such a structure, you can have c = a·b^(-1), which implies c·b = [a·b^(-1)]·b = a·[b^(-1)·b] = a·1 = a. c·b = a also implies (c·b)·b^(-1) = c·[b·b^(-1)] = c·1 = c = a·b^(-1). So c = a/b is equivalent to c·b = a, but this is true if and only if b^(-1) exists, and · is both commutative and associative. In other words, c = a/b is equivalent to c·b = a if and only if a, b, c are elements of a set S, where (S, ·, 1) is an Abelian group.
This is where the issue comes in. Abstract and universal algebra aside, in the vast majority of mathematics, theories assume a ring structure as a background. Suppose I have an Abelian group (S, +, 0). If (S, ·, 1) is a monoid that is left-distributive and right-distributive over (S, +, 0), then the structure (S, +, 0, ·, 1) is called a ring. A monoid is an Abelian group without the commutativity, and without the guarantee of existence of inverse elements everywhere in S. Monoids are particular in that, while inverse elements are not guaranteed to exist, whenever they do exist, they are guaranteed to also be unique. Why do I mention rings? Because in a ring, 0·x = 0 is necessarily always true. As such, there is no element of S such that 0·x = 1, and so 0^(-1) does not exist. So (S, ·, 1) can never be enriched into being a group, let alone an Abelian group, because 0 has no multiplicative inverse.
You can adjoin an element q to the set S, so that in the structure (S[q], ·, 1), 0·q = 1, hence 0 and q are ·-inverses of one another. However, the inevitable consequence of doing this is that (S[q], ·, 1) is no longer monoid, but a loop, and in a loop, · is not necessarily associative, and in this case, it definitely is not associative. As such, a/0 = b still does not imply a = 0·b in this situation. Furthermore, you can modify distributivity so that (S[q], +, 0) is a monoid with commutativity, or a loop with commutativity, but you can also never have a group with (S[q], +, 0).
1:25 my favorite number is actually 16 but in arabic we write it just like bprp's 17: (١٦)
the bonus part is amazing.
I think there’s a staggering flaw in the first proof as well. You implicitly imply that 0/0=1 when you attempt to cancel them out, which is wholly unjustified
On the firts demostration, if you assume that we can divide Real set becomes to IR={0} a real boring set!
I think the middle proof of 0^0 != 1 is the lie, as you can't subtract ln(0) on both sides (it is undefined)
Will note that the frst proof could probably use a slightly more rigourous definition than "you cannot divide by 0" as this feels a little vague, for proof standards
When you use a proof by contradiction to disprove your previous proof by contradiction 🤯
If u ask me tho, the third proof is kinda invalid too, as 2^x is set to be equal to 0, so is 2^(1+x), you can't take log on both side as they are both zero.
If im to prove that,
i would say it's the same as proving if x∈ℝ, then 2^x≠0
Proof by cases:
Case 1: x>0
Because 2^x>0 ∀x>0, so 2^x≠0
P.S. 2^x=e^(xln2),
Bec ln2>0, we can use power series of e^x to proof 2^x>0
Case 2: x=0, then 2^x=0
Case 3: x0
then 2^x =1/2^y,
But from case 1, 2^y>0 ∀ y∈ℝ,
2^x >0 ∀ x∈ℝ
Hence, 2^x≠0 ∀x∈ℝ
1:43 "Whoa! New math"
Can u just compare exponents for the 3rd case? Under our assumption, 2^star = 0 so doing any form of log is invalid hence we can't compare exponents. I would say star + 1 must also be a solution and by continuing the logic, star + 2 and star - 1 are also solutions. This means 2^any integer = 0. Then just saying 2^1 = 2 would cause a contradiction.
6:30 if ☆=∞, ∞-∞ is impossible so 3 is false
3:d proof is also incorrect (or incomplete). It could be multiple solutions.
The trouble is the way we think about exponents involves division. Why is 4^0=1? Because 4^2/4=4^1 and 4^1/4=4^0, and 4^1/4=1.
At 10:00, couldn’t you just device both sides by ln(0) resulting in 2 = 1?
that's with assuming ln0 ≠ 0 but it is in this case, so no.
I knew the false proof was gonna be the second one as soon as I saw 0⁰, however when I saw the bonus proof as to why, it really messed with my brain
0^0 has multiple answers depending on how do you try to solve this problem
for example if you assume 0^x=0, then 0^0=0, but if you assume x^0=1, then 0^0=1, but 0 != 1
Why assume either 0^x = 0 or x^0 = 1? How exponents work is that by definition x^0 = 1, and the other powers are achieved by multiplying or dividing numbers by x, so 0^0 = 1.
I'm worried that the first and third proofs aren't entirely valid either.
For #1, you've only shown that you can't divide by 0 WHEN THE NUMERATOR IS ALSO 0, but that doesn't necessarily imply that you can't divide by 0 in any case.
For #3, you make the assumption that the exponential function 2^x is one-to-one, so that there cannot be two distinct arguments with same output. Without this assumption, it doesn't logically follow that 1+STAR=STAR as you write at 6:23.
He is skillful in maths and thinks only about maths everytime.did not expect this proof wow great
3 because the log function has many values, so both star and star+1 could be solutions
"I thought I came up with a wonderful proof showing 0^0 is not 1 [...]"
If you ever come up with another one, my suggestion is to test it on other powers of 0. For example, test whether your "proof" would also show 0^2 is not 0. In my experience, pretty much every "proof" that 0^0 is not 1 also shows that 0^2 is not 0. Seeing this, we can conclude that the "proof" is flawed :)
Ahhhh! You actually showed this. Very good! I should have kept watching before commenting. Very nice :)
You always used to write "contradition" during the video, just as a hint. 😉 (This does not make the mathematical content worse of course. :))
He forgot the +C
Edmund Schluessel You win
@@stlemur Lol there's no better way to express this. 👍🏻😁
I suspect his English is way better than you are in brpr's first language -- even if you aren't required to use the traditional symbols. I suspect your transliteration to Latin characters would also fall short.
@@artsmith1347 I suppose your comment was addressed to me? If so, I have no problem to admit that indeed English is not my native language since I'm German. And I'm aware that I regularly make my mistakes in foreign languages, this is completely normal. However this does not exclude the possibility that I also "am allowed" to draw someone's attention to their mistakes if I find some. At least in this case when the same mistake catched my eye several times. To me your comment sounds as if I had been "rude" having fun complaining about other's mistakes while claiming that I'm "perfect". Both is not the case. I emphasized that I didn't mean to offend bprp at all. It was just that one spelling mistake I found kinda funny while I keep on cherishing his mathematical content. If you can't read that from my comment above, I can't help you. 🤷🏻♂️
Absolute zero, like infinity, is an abstract number with no precision. There's no result trying to divide by absolute zero or raise absolute zero to zero because the number has no scale.
4:15 its kinda weird to see zeros there...
The problem I believe in math that has been stuck with so many contradiction, is that zero is not a number. Zero is the absence of any number. So using at a number you will have confusion...
Wait a minute...
(0*1)/0
Can you even cancel the zeroes out? How does that work?
I mean, usually cancelling factors works like... 6/3, you divide both numerator and denominator by 3, and then you get 2/1.
But what factors do you even cancel out at (0*1)/0 ?
Also, suppose the cancelling works out.... doesn't this result in 0/0 = 1?
Oh, and...
As far as I understood division, it is all about how often I can subtract the second number from the first number until it reaches 0, compared to multiplying, which is adding number x an amount of y times.
So... Based on that, I actually thought of what divison by 0 actually could be.
1/0 = ?
You need to subtract 0 an infinite amount of times from 1 to get it to 0 (or even more often than that, but that's the only uncertainty I have). So, let's define 1/0 as ∞.
Now, what would 2/0 be?
Well, you also need to subtract 0 from 2 an infinite amount of times. Sooo.... Both are infinity? And Infinity * 0 = 1 = 2 ?
Well, no.
The infinity from the second division is twice as large as the one from the first division.
So, 1/0 is actually ∞[1], and 2/0 is ∞[2], and so on.
At least that's how I would define it.
The assumption was that we can divide by zero, so that's why in that example zeros cancel out. In reality of course it doesn't work since you can't divide by zero
Because 1/0 can result in negative infinity or positive infinity defending on whether you take it as 0- or 0+
The 2nd cause ln(0) doesn't exist, there's no number n such that eⁿ=0, it will have to be a limit which will give you -∞ in which case your equasion is
-∞+ln(ln(0))=-∞
-∞=-∞
And it's a true statement
Wouldn’t 3 be wrong since after you removed the star you got 2^1 = 2 which is true?
17 is actually my favourite number and I was really surprised when you guessed it correctly O.O
I'd say the second proof is a false one for two reasons.
First of all, 0^0 can be either 1 or 0 depending on your definition.
Second, when taking the ln of 0^0 you couldn't use the exponent rule. The exponent rule can only be used if what is left is still a valid argument. In this case we were left with ln(0) which you can say is negative infinity (or undefined), but infinity is not a number so you can't really keep doing algebra on it.
Anyways I loved the video and I learned a lot from it. I like seeing proofs for things, really makes you think hard things that seem trivial at first glance.
Hello. I wanted to say that I proved that n (n is a positive real number with n cannot be equal to 0) divided by 0 is defined. I am not sure of that statement because I am a student. I will show it to my teacher to see if my proof is right.
The issue when trying to prove such elementary facts is that many of the proof rules being taken for granted are less fundamental than the proved property itself. Are we actually proving anything, or is the validity of proof rules we are using actually a consequence of the proven fact being true? My former algebra professor would probably say all three of those proofs are invalid.
The only rigorous way to prove those would be to start with an axiomatic theory of arithmetics and prove everything from the axioms only.