2 legit proofs & 1 false proof

Рет қаралды 162,973

Күн бұрын

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0:00 cannot divide by 0
2:14 0^0 is not 1
4:25 2^x=0 has no real solutions
6:56 comment your answer below
7:03 the false proof (I am not saying if 0^0=1 or not)
7:23 learn more on Brilliant
8:16 bonus (mind-bending logically writing)
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Пікірлер: 784

@andrycraft69 3 жыл бұрын

I was not expecting a double contradiction proof. Highly appreciated.

@blackpenredpen 3 жыл бұрын

😆!!!

@hugonamy7504 3 жыл бұрын

It was just insanely cool ! 😂👍

@filipve73 3 жыл бұрын

Is double contradiction proof always true? Our you just cheating !

@extremegamingdz1309 3 жыл бұрын

@@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.

@nikhilnagaria2672 2 жыл бұрын

@@extremegamingdz1309 he literally said the same

@Some_Guy77 3 жыл бұрын

This should have been called "Two proofs and a lie."

@blackpenredpen 3 жыл бұрын

Hey that’s a good one. Will change now. Thanks

@Adambenhmida0000 2 жыл бұрын

@@blackpenredpen you didn’t 😾

@lolzhunter 2 жыл бұрын

@@blackpenredpen he lied!

@asheep7797 2 жыл бұрын

@@blackpenredpen the lie

@k_wl Жыл бұрын

@@blackpenredpen how about 2 proofs 1 spoof

@ehess1492 3 жыл бұрын

I wonder if he is setting up the next poll: What is correct spelling of “contradition”? A) contradiction B) contradition

@blackpenredpen 3 жыл бұрын

Nah I am done with that spelling poll

@ehess1492 3 жыл бұрын

@@blackpenredpen so you didn’t spell “contradition” four times on purpose?

@justviewerfromindia2402 3 жыл бұрын

@@blackpenredpen shouldn't be "contradiction"? As the KZfaq closed caption says "contradiction" and u write "contradition". But still best provement 😀

@Mothuzad 3 жыл бұрын

The "con" in "contradition" is like the "con" in "conman". It stands for "confidence". A strong tradition of confidence.

@justviewerfromindia2402 3 жыл бұрын

@@Mothuzad you meant so say as con tradition right? Means Confidence tradition

@spankasheep 3 жыл бұрын

1:45 My teacher, every time he is correcting my assignment ;D

@blackpenredpen 3 жыл бұрын

My favorite part is at 9:00. Get ready to bend your mind!

@zerotwo9607 3 жыл бұрын

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

@sukhps 3 жыл бұрын

@@zerotwo9607 well, explain it boi

@RunstarHomer 3 жыл бұрын

@@zerotwo9607 Yeah, let's see it

@crazystuffofficialchannel4406 2 жыл бұрын

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 😮 😮 😮

@ripjawsquad Жыл бұрын

@@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

@Forced2 3 жыл бұрын

As soon as I saw ln(0) I was like thats negative infinity and you’re going to abuse the heck out of it 😂

@sharpnova2 3 жыл бұрын

and then you abused yourself instead.

@mondherbouazizi4433 2 жыл бұрын

It's cute how people nowadays call ln(0) = -infinity out loud shamelessly. Did you know that there is no such thing as ln(0), and what you are referring to is the limit as x approaches 0 of ln(x). And that's for the simple reason that ln is NOT defined for the real number 0. Also, there is no such a number as "infinity". I remember once a guy was saying "1 to the power of infinity is undeterminate" and I was like 1 is the neutral element in IR and no matter how many times you multiply it by itself, it will always remain 1. Plus, there is no such a number as infinity, and even if it exists, 1 to the power of that thing would be 1. The poor thing was talking about the limit as x approaches 1 to the power of a function of x that approaches infinity. But because his teachers were casually using these "inappropriate" words, he was so full of himself and confident in what he was saying.

@Forced2 2 жыл бұрын

@@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.

@19divide53 2 жыл бұрын

@@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.

@MCNarret 2 жыл бұрын

@@mondherbouazizi4433 infinity is a number. Give me one good reason not to assume a limit when calculating if infinity is unsigned like 0.

@divysaraswat2084 3 жыл бұрын

When he says 1:45 "New Math", I got 2 heart attacks at once.

@mokafi7 3 жыл бұрын

lmao

@JasonOvalles 3 жыл бұрын

But 2=0 so that means you were fine! Right?

@lolpotato 3 жыл бұрын

In kidney

@tgwnn 2 жыл бұрын

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!!!

@vedants.vispute77 3 жыл бұрын

Everyone: Mind Blowing Bprp: Mind Bending

@JediJess1 3 жыл бұрын

StandupMaths: Mind Boggling MindYourDecisions: Mind Your Decisions YGOAbridged: MIND CRUSH!

@anshumanagrawal346 3 жыл бұрын

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

@zerotwo9607 3 жыл бұрын

You can prove it

@anshumanagrawal346 3 жыл бұрын

@@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

@user-en5vj6vr2u 3 жыл бұрын

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

@zerotwo9607 3 жыл бұрын

@@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,

@anshumanagrawal346 3 жыл бұрын

@@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

@fred-ss4ym 3 жыл бұрын

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 ).

@microscopicallysmall 4 ай бұрын

did you comment on the wrong video

@fred-ss4ym 4 ай бұрын

@@microscopicallysmall in the sponsored segment, there is a puzzle for the viewers

@katarzynaguzowska3248 3 жыл бұрын

Thank you that you write everything on board, it is much easier to follow.

@blackpenredpen 3 жыл бұрын

😊

@SlidellRobotics 3 жыл бұрын

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.

@blackpenredpen 3 жыл бұрын

I have done those proofs already 😊

@mrpie3055 2 жыл бұрын

You can also prove the law of cosines from vector rules

@mrocto329 2 жыл бұрын

@@mrpie3055 If you mean the dot product formula, that won't work as the dot product formula uses law of cosines AFAIK

@ameerunbegum7525 3 жыл бұрын

1:10 Me: Oh, My favorite number is 14.... bprp: *17* Me: Oh, Nevermind.

@chanuldandeniya1824 3 жыл бұрын

I was searching for this comment 😅😅😅😅

@Bodyknock 3 жыл бұрын

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.

@elithanathileoathbound3772 3 жыл бұрын

Could this become a regular series? I would definitely watch it.

@thesinglemathnerd 3 жыл бұрын

“We cannot divide by 0.” *laughs in wheel theory*

@angelmendez-rivera351 3 жыл бұрын

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.

@OrangeC7 3 жыл бұрын

@@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!"

@aswinr9676 3 жыл бұрын

Proving a proof just blows my mind You just proved that the proof is wrong

@drpeyam 3 жыл бұрын

Ooooooh fancy!!!

@pmathewizard 3 жыл бұрын

2 days ago

@viral724pathak 3 жыл бұрын

@@pmathewizard 🧐

@aswinr9676 3 жыл бұрын

Wait how??

@HershO. 3 жыл бұрын

Video uploaded 1 hour ago and comment 2 days ago *Illuminati sounds play* LOL

@redrobbie7977 3 жыл бұрын

Peyam teach us your ways.

@cormalan9894 3 жыл бұрын

Love this format! Do more of these please

@HrsHJ 3 жыл бұрын

Congrats on 700k subs

@Angel_Sony 3 жыл бұрын

@@HrsHJ Same here... are you on instagram?

@apocalipseleaguepl9248 3 жыл бұрын

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.

@MikehMike01 3 жыл бұрын

sounds like the halting problem

@gurkiratsingh7tha993 3 жыл бұрын

I have studied about indeterminate forms but today I have studied many new things from your video, I highly appreciate your work.

@josir1994 3 жыл бұрын

The contradiction is this video has 3 legit proofs

@blackpenredpen 3 жыл бұрын

Lol true that

@Zyx3ds18 3 жыл бұрын

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.

@huhbooh 5 ай бұрын

I thought proof 3 was incorrect because it had a multiplication with 0

@mathsman5219 3 жыл бұрын

Cancelling In (0) from both sides was wrong. ⭕

@roberternest7289 Жыл бұрын

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.

@kazuhoshiinoue2695 3 жыл бұрын

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...

@_Ytreza_ 2 жыл бұрын

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 '_'

@isaacormesher2851 2 жыл бұрын

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!

@LOLHeadVideos 2 жыл бұрын

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

@cube7353 2 жыл бұрын

In 2nd one, You could take 0°=2° Then, 0=2 which is not possible. Therefore, 0° is not equal to 1.

@fantiscious 2 жыл бұрын

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.

@Nebula_ya 2 жыл бұрын

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?

@kxngkvde 2 ай бұрын

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.

@lolerishype 2 жыл бұрын

7:02 That bird in the background is golden

@Imran-Shah Жыл бұрын

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!

@jacksonpercy8044 3 жыл бұрын

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.

@HeraldoS2 3 жыл бұрын

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.

@josiproak739 2 жыл бұрын

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

@Dyllon2012 2 жыл бұрын

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.

@petrie911 2 жыл бұрын

This can be rectified by instead noting that 2^x is strictly increasing on R.

@angelmendez-rivera351 3 жыл бұрын

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.

@SyberMath 3 жыл бұрын

7:18 yummy!!! You tackle some very important topics! Awesome! 🤩🤩🤩

@blackpenredpen 3 жыл бұрын

Thanks!

@SyberMath 3 жыл бұрын

@@blackpenredpen You're welcome!

@MrRyanroberson1 3 жыл бұрын

the end is quite satisfying with the proof by contradiction of proof by contradiction

@SpaceWithSam 3 жыл бұрын

Very interesting!

@andrewkoo6806 3 жыл бұрын

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.

@blackpenredpen 3 жыл бұрын

Perfect explanation!

@bm-br3go 3 жыл бұрын

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.

@carpedm9846 2 жыл бұрын

The moment Ln showed up I had a bad feeling. Anything with euler cant be trusted to act normal

@victorsouza3709 2 жыл бұрын

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

@victorsouza3709 2 жыл бұрын

Or how about... 0/0 = # Such that # is something that is not in the real numbers It looks like it is coherent

@pedroribeiro1536 3 жыл бұрын

700k huh, congrats man 👊

@louisduhamel4040 2 жыл бұрын

I was so happy to see how happy he was when he found the contradiction

@jaymercer4692 3 жыл бұрын

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.

@xwtek3505 2 жыл бұрын

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

@hoffmanmustardoil6191 3 жыл бұрын

One of the best explainer of mathematics on KZfaq...

@dimitriosb.3242 3 жыл бұрын

Hey Blackpenredpen , I have an interesting proof that I want you to see, is there somewhere I can send it?

@filthypete13 3 жыл бұрын

I was able to figure out which proof was wrong but my reasoning was slightly different (and probably wrong) I was thinking that you couldn't take the ln (0 x inf) because 0 x inf would be undefined or nonsensical.

@manamtiwari 3 жыл бұрын

I am just amazed how the heck did you read my mind 17 is my favourite number ❤️

@gnorthey 2 жыл бұрын

Mine is too i almost dropped my phone when i heard him say that!

@Thror251 2 жыл бұрын

the bonus part is amazing.

@electra_ 2 жыл бұрын

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

@PoundersPlatinum 2 жыл бұрын

Wouldn’t 3 be wrong since after you removed the star you got 2^1 = 2 which is true?

@prayrenayadav5466 3 жыл бұрын

I really love your derivatives merchandise

@TheMahri77 2 жыл бұрын

@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?

@novidsonmychanneljustcomme5753 3 жыл бұрын

You always used to write "contradition" during the video, just as a hint. 😉 (This does not make the mathematical content worse of course. :))

@stlemur 3 жыл бұрын

He forgot the +C

@myrus5722 3 жыл бұрын

Edmund Schluessel You win

@novidsonmychanneljustcomme5753 3 жыл бұрын

@@stlemur Lol there's no better way to express this. 👍🏻😁

@artsmith1347 3 жыл бұрын

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.

@novidsonmychanneljustcomme5753 3 жыл бұрын

@@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. 🤷🏻‍♂️

@prathampatel1740 3 жыл бұрын

1st suspicion: "wait, isn't 0^0 actually 1" 2nd suspicion: "ln0"

@MrYesman43 3 жыл бұрын

How do you get from 2^(1+*) = 2^* to 1+* = *? You say you can take the logarithm but since 2* = 0, log(2*) = log(0) which is the same error which was made in "proof" 2.

@applimu7992 3 жыл бұрын

3 because the log function has many values, so both star and star+1 could be solutions

@ramusai8031 3 жыл бұрын

Excellent

@butter5014 3 жыл бұрын

Dear BPRP and others, please help me with this problem: 12x^2 - (x^2)(y^2) + 11y^2 = 223 , solve for x,y out of Z (whole numbers). Thanks in advance!

@Ben-rd3mg 2 жыл бұрын

Does the 0^0 not equal one proof assume you can subtract ln0 from both sides

@user-pi1jf2fu3f 3 жыл бұрын

How are you sure the zero will simplify in the first proof? We don't necessaraly have to suppose 0/0 = 1, do we? In the third proof, how could use the logarithm when you assume 2^☆ = 2^(☆+1) = 0?

@adamp9553 3 жыл бұрын

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.

@alexatg1820 3 жыл бұрын

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.

@alexatg1820 3 жыл бұрын

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

@varunsbharadwaj1359 3 жыл бұрын

@blackpenredpen for proving why cant we divide by zero, we can take the same no. say 'a' and 0*a = 0*a and then we divide both sides by zero, we get a=a which is legit ryt? (ps ik division by zero is meaningless but still correct me if i am wrong)

@jorgelenny47 3 жыл бұрын

I thought the mistake was subtracting ln0 on both sides since it's undefined

@Slijee 3 жыл бұрын

This might sound stupid but if you get a number that is approaching negative infinity, and raise 2 to that power, won't that number approach 0 because it's getting closer and closer to 1/infinity?

@danielyuan9862 3 жыл бұрын

yeah but is it ever 0 for any real number?

@Inspirator_AG112 3 жыл бұрын

*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.

@Grass89Eater 2 жыл бұрын

3:d proof is also incorrect (or incomplete). It could be multiple solutions.

@therattleinthebook397 2 жыл бұрын

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.

@gurkiratsingh7tha993 3 жыл бұрын

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)

@pardeepgarg2640 3 жыл бұрын

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)

@pardeepgarg2640 3 жыл бұрын

I am in 11 grade , and I leave maths 2months ago due to pressure to gain marks

@lukandrate9866 3 жыл бұрын

I think I shouldn't look here

@gurkiratsingh7tha993 3 жыл бұрын

@@pardeepgarg2640 okay I will try

@gurkiratsingh7tha993 3 жыл бұрын

@@pardeepgarg2640 okay, I know, indian educational system is not very good.

@gianglai7346 2 жыл бұрын

17 is actually my favourite number and I was really surprised when you guessed it correctly O.O

@toaj868 3 жыл бұрын

But wouldn't taking the logarithm of 2^☆ also be undefined therefore making the proof invalid?

@MikehMike01 3 жыл бұрын

I am no expert but I think the problem in proof 2 is that he subtracted ln(0) to both sides. You can’t add/subtract an undefined quantity. Meanwhile, log() is defined as the inverse of the exponent, so something like e^ln(0) = 0

@MCNarret 2 жыл бұрын

A proof by contradiction doesnt necessarily prove the statement if the solution is conditional, where cannot becomes cannot always.

@GodzillaFreak 2 жыл бұрын

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

@b77vedantmore51 2 жыл бұрын

I have a doubts sir, if we cannot divide any number by 0 then how u divide 0 both side to get 1=17?

@ernestschoenmakers8181 Жыл бұрын

But i have a hard time with 2^x/=0 proof cause if i plug in -infinity, isn't it 0 then?

@sowndolphin5386 2 жыл бұрын

2:15 Phone calculators: and we took that personally

@cansomeonehelpmeout 2 жыл бұрын

What would go wrong it we defined ln(0) = -infty? Is the problem then that ln(0) has no additive inverse?

@Firefly256 2 жыл бұрын

Because you can’t use infinity. you can’t define ln(0) if the 0 is exactly 0, but if the 0 is not zero (for example the smallest thing bigger than 0, infinitesimal), then you can say it’s -inf So lim ln(x) as x->0 = -inf, but ln(0) is undefined

@pixel0818 Жыл бұрын

both ln(0) and the star approach negative infinity, so aren’t you doing infinity - infinity in both the second ans third

@l-a6167 3 жыл бұрын

On the firts demostration, if you assume that we can divide Real set becomes to IR={0} a real boring set!

@simonwillover4175 3 жыл бұрын

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.

@EgorTimatkov 2 жыл бұрын

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?

@jeejjaaj7136 3 жыл бұрын

Is ln(0) possible on the 2nd proof ?

@jeejjaaj7136 3 жыл бұрын

I didnt see the end of the video oooooooooo

@ZrJiri 2 жыл бұрын

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.

@connorm9176 3 жыл бұрын

The second one is bad right? because he also secretly makes another assumption that it is valid to take the ln of 0, which it is not

@isaacaguilar5642 3 жыл бұрын

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.

@utalbor1426 3 жыл бұрын

ln(0) is not defined so the 2nd is invalid.

@volcarona8401 2 жыл бұрын

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.

@prunusserrulata6303 2 жыл бұрын

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

@Firefly256 2 жыл бұрын

Because 1/0 can result in negative infinity or positive infinity defending on whether you take it as 0- or 0+

@gt8846 3 жыл бұрын

SIR IS BODMAS IS ALWAYS APPLICABLE?

@lool8421 3 жыл бұрын

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

@danielyuan9862 3 жыл бұрын

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.

@rhc1560 3 жыл бұрын

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.

@neilkrieger5334 2 жыл бұрын

To me the second proof is wrong because 0⁰=1 (its the number of functions from the empty set to the empty set, and in that case the function that does nothing is defined) and also because even though ln(0⁰) is defined you can't simply write ln(0⁰) = 0 ln(0) =0 because ln(0) doesnt exist, (even though it makes some sense since for P a polynom with P(0)=0 P(x)ln(x) tends to 0 in 0, the second point is when we substract ln(0) to both side, in essence we are writting infinity minus infinity which is also not defined (even in R U {±infinity})