2 legit proofs & 1 false proof
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- Опубликовано: 1 авг 2024
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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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Thank you,
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
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 RUclips 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
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
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.
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.
@@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.
1:45 My teacher, every time he is correcting my assignment ;D
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!!!
Everyone: Mind Blowing
Bprp: Mind Bending
StandupMaths: Mind Boggling
MindYourDecisions: Mind Your Decisions
YGOAbridged: MIND CRUSH!
Thank you that you write everything on board, it is much easier to follow.
😊
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
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
@@microscopicallysmall in the sponsored segment, there is a puzzle for the viewers
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
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.
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.
1:10
Me: Oh, My favorite number is 14....
bprp: *17*
Me: Oh, Nevermind.
I was searching for this comment 😅😅😅😅
Could this become a regular series? I would definitely watch it.
Congrats on 700k subs
@@HrsHJ Same here... are you on instagram?
Love this format! Do more of these please
“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!"
Proving a proof just blows my mind
You just proved that the proof is wrong
I have studied about indeterminate forms but today I have studied many new things from your video, I highly appreciate your work.
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
Cancelling In (0) from both sides was wrong.
⭕
The contradiction is this video has 3 legit proofs
Lol true that
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.
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!
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
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?
7:02 That bird in the background is golden
In 2nd one,
You could take
0°=2°
Then,
0=2 which is not possible.
Therefore, 0° is not equal to 1.
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 '_'
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 was so happy to see how happy he was when he found the contradiction
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.
the end is quite satisfying with the proof by contradiction of proof by contradiction
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!
Hey Blackpenredpen , I have an interesting proof that I want you to see, is there somewhere I can send it?
Wouldn’t 3 be wrong since after you removed the star you got 2^1 = 2 which is true?
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.
@1:41
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?
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.
But i have a hard time with 2^x/=0 proof cause if i plug in -infinity, isn't it 0 then?
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 have a doubts sir, if we cannot divide any number by 0 then how u divide 0 both side to get 1=17?
700k huh, congrats man 👊
Very interesting!
One of the best explainer of mathematics on RUclips...
7:18 yummy!!!
You tackle some very important topics!
Awesome! 🤩🤩🤩
Thanks!
@@blackpenredpen You're welcome!
The moment Ln showed up I had a bad feeling. Anything with euler cant be trusted to act normal
the bonus part is amazing.
Does the 0^0 not equal one proof assume you can subtract ln0 from both sides
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?
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!
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.
I do not understand the bonus part. Why do we have to prove that 0^2 is not equal to 0?
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
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.
But wouldn't taking the logarithm of 2^☆ also be undefined therefore making the proof invalid?
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
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 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.
Is the third proof incorrect?
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!
I thought the mistake was subtracting ln0 on both sides since it's undefined
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?
yeah but is it ever 0 for any real number?
@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)
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
Why in proof 1 is assuming 0/0=1 okay?
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.
SIR IS BODMAS IS ALWAYS APPLICABLE?
Aren't we taking log of 0 in the third proof as well?
ln(0) is not defined so the 2nd is invalid.
What happens if we define 0/0 as the set of all real numbers(IR)?
Then IR*1=IR*17 which means... that if you multiply every real number with 17 you will end up with all real numbers again?...
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?
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
Hey can you solve cos²(x)+sin²(x )=2 where x is complex?
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. 🤷🏻♂️
17 is actually my favourite number and I was really surprised when you guessed it correctly O.O
both ln(0) and the star approach negative infinity, so aren’t you doing infinity - infinity in both the second ans third
On the first equations aren't we assuming that 0/0 is 1? I thought that was invalid
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 really love your derivatives merchandise
So 0^0 = 1 as the “contradiction” you showed is a contradiction in itself?
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.
Why 0^0 is not the same with 0/0?
But in first proof we assume that 0/0 = 0/0, which is not true? And that video from this channel in recommended on that topic also assumes that. What am I missing?
Isn't the 3rd one incorrect? 1 to the power of 5 and 1 ttpo 1 are the same, but 1 != 5.
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
I usually are amazed by the things you do bc they are out of my mind but this i actually noticed bc we just had log(x) in school
But the second proof immediately goes wrong because you can't work with 0^0, which is undefined?
Many mathematicians will tell you it is not undefined. Furthermore, there is no good argument that it is undefined, while 0^0 = 1 can be easily argued to follow from the definition of exponentiation itself, among other things.
The second proof, that 0^0 = 1 implies 0·ln(0) = 0, and therefore, a contradiction, is actually one very common argument used to attempt to demonstrate that 0^0 should be undefined. However, the proof is notably known to be fallacious, because it assumes ln(x^y) = y·ln(x), which is not true in general, and if working with real numbers, then it requires that x > 0.
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
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
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
3 because the log function has many values, so both star and star+1 could be solutions
The second one you can't do ln(0•ln(0)) since ln(0) is undefined
What would go wrong it we defined ln(0) = -infty? Is the problem then that ln(0) has no additive inverse?
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
i thought the first proof was incorrect because even though you divide by 0, you assume that 0/0=1 which was not stated in your proof beforehand
7:15 You misspelled undefound
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.