"Prove" 3 = 0. Can You Spot The Mistake?
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- Опубликовано: 28 авг 2024
- Thanks to Lucas from Brazil who sent this problem! This is a proof that 3 = 0 using sneaky algebra that I had never seen before (this is a sneaky mistake, not a simple divide by 0 fallacy). Every step seems correct, but the conclusion is clearly false. Can you find the mistake?
My blog post for this video
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Guy: That will be $500
Me: Let me show you how $500 is $0
@; Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh!!!!
Or they could tell you to pay way more
@@Placeholdernaame yeah😂
Could be 1000$
500 likes = none :)
it would be more interesting to explain why this substitution leads to the extra solution
I agree. But, as some other people have said, step 3 basically substituted a rearranged version of the equation back into itself as if it were a different equation, which ultimately changed the formula.
I think its sorta coz it becomes a cubic, where it was a quadratic before, so you get a 3rd solution.
Like if you take a quadratic and multiply it by x you get another solution of 0
multiply the first equation with (x-1). That adds the root x=1 to the original equation. That becomes x^3-1=0. And then we pretend that root 1 is also a valid solution to the original aequation...
@@martenjanderuiter5474 oo that was smart.
One thing to notice is that the two complex solutions we got are indeed cube roots of 1. We introduced a real solution to x^3 = 1 when we already had two complex solutions to it.
"Prove" 3 = 0. Can You Spot The Mistake?
Yes, i can
3 not equal to 0
You're welcome
Break 350 likes 0 comment Tragedy
Solution of this equation is complex and we work with it here like vith real variable
*refuted*
Shouldn’t “ X to the 2nd power + X +1= 0” get simplified to “X to the third power = -1”?
Then it would be “X times X times X = -1.
Which would make X=-1.
Because (-1)(-1)(-1)= -1.
Nah.
1+-1 +1 not= 0.
The thing here is that you neglected the Fundamental theorem of algebra (twice actually) : a degree n polynomial has exactly n roots. The first polynomial is of degree 2 but when you substituted x + 1 = x*x, you changed the degree of the polynomial (becoming of degree 3) , thus adding a root. (you could do this mistake by multiplying every equation by x, thus adding the root x = 0).
I like this explanation most out of all and I think this best explained the matter. Together with “the substitution is not a if and only if operation”, this problem is explained pretty well.
I definitely agree and I was going to comment something similar to this idea as well.
Thank you, I was not able to immediately spot the mistakes
by far the best and most concise explanation. hopefully its actually correct reasoning!
Very cleaver and ginius dude
3 is close enough to 0, it can be rounded
Similarly e ≈ π ≈ 3
In this way u will round down all our exam marks to 0 dude😂😂😂
ramu what’s e. My maths teacher refuses to tell me
Thomas Roberts probably she don't know
doesn't quite work like that. You could in certain cases argue that 3 is close enough to 1 however you have to be careful with approximating to 0. If you have situations like this 1000+3~0 then yes 3 is much smaller than 1000 and thus we can approximate it to 0 however if you have a situation 3+6 then 3 is not much smaller than 4 and in fact the answer 9 is closer to 10 than 0. Typically we work in orders of magnitude sometimes we take much smaller to just mean a single order of magnitude smaller (something like 40 vs 2, 40 being about an order of magnitude larger than 40 if we are working in tens) sometimes we need greater precision. In physics orders of magnitude typically comes rather naturally. If we have to find a solution to some variable x then we might use perturbation to find this and which essentially involves identifying the orders of magnitude in x. we set x to be x=x0+x1*w+x2*w^2+... where w is some natural constant of the problem. In quantum mechanics it is typically some frequency in the potential and in classical mechanics it could be some characteristic length or time or velocity, etc. then we insert this solution into our equations and we can then remove all of the terms of higher order than w^0 and solve for x0. Then we do the same but we now know x0 and then remove any terms higher than w^1 and can now solve for x1 and so on. This only works if the series converge and then at some point around x6 or x9 we are calculating the 14th decimal of x and typically have stopped way before that because we only require a certain precising for our answer. This method is especially useful when we run into ugly equations with no analytical solution, meaning that there is no way mathematically to find the exact value for x and thus we have to find an approximate answer. Either we need to do this or to just change or remove one of the terms in the equation to simplify it and hope (often with good reason) it doesn't affect the solution too much.
Engineers be like: '3 = 0'.
"Well that's close enough"
The fundamental theorem of engineering.
sin(x)=x
cos(x)=1
@@creepybabby I'm sorry, but I must do it
r/whoooosh
Lol..😂
You are not an Engineer
@@user-ky6vw5up9m I am a mechanical engineer actually
Actually the mistake lies in the assumption that every equation is equivalent to the previous one (). Step 3 is perfectly valid so long as you know that equation 2 implies equation 3, but equation 3 does not imply equation 2.
This means that the statement:
“x^2 + x + 1 = 0, so x^3 = 1”
is valid, and consequently we can say that any solution to the equation x^2 + x + 1 = 0 is also a solution to the equation x^3 = 1.
But because we have not proven the reverse statement:
“x^3 = 1, so x^2 + x + 1 = 0”,
this means that not necessarily all solutions to the equation x^3 = 1 have to be a solution to the equation x^2 + x + 1 = 0 as well
Exactly, yours is the correct answer. This guys simply said "it's not correct because I checked the results and they don't match". Duh, of course they don't match, the point is finding out why.
Either way thank you for being the only person with a brain around here.
@@rayan-xg5kw too much hate dude
nerd
Thank you, I was much less interested in knowing what the answer is, and much more interested in understanding *why* that is the answer. I was disappointed in this video because it made absolutely no explanation as to *why* that was the case. Like, the takeaway i had from this video is "Double check every single algebraic operation you perform on an equation in order to make sure that you didn't accidently introduce a new solution!"
Like, it's great that the quadratic equation offers us a way to double check every step, but if we were working with a formula that didn't have a magical quadratic equation to solve it with, how would we have found the answer?
Actually if I'm not mistaking, the step (x^3=1 ==> x=1) is also false as we do consider complex numbers. (x^3=1 ==> x=exp(2i*k*π/3) with k in {0,1,2}) would've been correct and therefore the original solutions of the problem have to be among those (but are not all of those as you mentionned).
Here is the mistake for those who are still wondering... (x2+x+1) = 0 is multiplied with (x-1) which makes it (x-1)(x2+x+1) = 0, which on expansion gives (x3-1) = 0. So the mistake here is we cannot simply multiply an (x-1) to the quadratic and argue x=1 as its solution. Hope that helps!
As pointed out long ago by others: we can multiply by (x-1). It's just we have to "live with the consequences" which in this case is: set of solutions of the resulting equation is superset of set of solutions of original equation.
When solving equation you want to have bidirectional implication. Generally process of solving equation consists of number of transformations that make the equation trivial. Then, we can say that any solution to the last "version" of equation is also solution to the original "version". However, should any transform be valid in one direction only, we cannot use this last implication as it wouldn't be true. The only true implication would be that solutions of original equation are among solutions of the transformed one.
Thumbs up!
I looked at the thumbnail for 5 minutes to find a mistake
is your youtube profile your reaction
akira yokomoto yes
Ok why did that even take you 5 minutes lmao 😂
@@sketchiipersonality because you got wooooshed
Ultra legends
is it just me or did this video not actually explain why what was happening was happening, just showed what was happeniung with no explanation
There was, there really is a fault there
fyukfy The reason it doesn’t work is because he created a system by adding another equation. The new system is now a different problem.
@@PyroclasticFlow no, it's good to make more equations, do you can see what you are calculating with, it helps you to find solutions in some ways, but you have to write them near the whole thing and write an explanation ne t to it
If you rename the x in the first equation with y. You will two different equations, the x in the first and the second equations are not the same.
The reason is that the original equation was quadratic, so it must have 2 roots (we can find it in imaginary numbers), but since he substituted with -x2 and make 1=x3 he changed the equation from quadratic to cubic, in another word he added one extra root, from 2 roots to 3 roots. That's mean you have changed the solution. So the leason is be careful when you modify equation.
“Can you spot the mistake”
Well yes actually, the mistake is that there’s no cross through the “is equal to” symbol, it should really be: 3≠0
yes
how did you do that
@@gam8052 If you have an Android with Gboard, you can hold on the equals sign to give you a ≠.
≠
You can do it with apple too.
First equation is "Playstation 1"
Second equation is "Playstation 1 mini"
Third equation is "Playstation 2"
Both first and second can play PS1 CDs
The third can play PS1 CDs but can also play PS2 CDs
If you put a PS2 CD on "Playstation 1" or "Playstation 1 mini" it will not work.
That's my understanding of this problem
That is a perfect analogy
Ohhhh, so THAT's where the advertisers got that slogan "Playstation 2, the 3rd place!!!!"
Not a perfect analogy, but close enough I guess
I can prove anything=anything.
First, multiply both side by 0.
Done
So Everything In This World Is Equal 😂
You can't multiply both sides by 0
LOL, thanks bro, now I'll only get A+ in all of my test
"nothing from nothing leaves nothing .... you gotta have something ... if you want to be with me ...."
That doesn't prove "anything = anything". That only proves "0=0" because "anything times 0" results in 0 and not "anything".😛
MindYourDecision: "Prove" 3 = 0. Can You Spot The Mistake?
Me: Oh that's easy, that's me. Problem solved.
F
F
F
F
F
There is a simpler explanation: the logic the false proof used here works only one way, but not the other. It showed that every solution x to the equation x^2+x+1=0 must satisfy the equation x^3=1, but not every x that satisfies the equation x^3=1 is necessarily a solution to x^2+x+1=0.
Yup. x^3 - 1 = (x-1)*(x^2+x+1). That (x-1) factor is why x^3-1 has the x=1 solution but x^2+x+1 does not.
He said that in a roundabout way. You didn't say it efficiently either. Allow me to do it PROPER:
it makes it very easy when you see that he literally just multiplied the original quadratic equation, which had 2 solutions and x=1 wasn't one of them, with (x-1), to turn it into a cubic equation. Then it had x=1 as a solution. Duh.
What I want to say is just that one does not really need to write out every solution of every equation that occurs in the proof in order to understand its falsehood. One does not even need to know about complex numbers. Just a simple understanding of logic would suffice.
I agree. The problematic point is to say "if x^3=1 then x=1" because it is true only among the real numbers. Especially the roots of the original equation are not real, so for them this implication does not hold. Besides "let x be ... " does not say which realm we want the solution. Among the real numbers there is none, so among the real numbers the whole reasoning is moot, void, nonsensical. Among the complex numbers "x=1" does not follow from "x^3=1".
Júlio Gama 1 + 1 = 1? Are you insane?
Since x^3=1 , x has three solutions.
x=1 OR (-1+i√3/2) OR (-1-i√3/2)
As x^2+x+1=0 has unreal roots , x=1 gets eliminated .
And so the mistake is in the fifth step where you considered x=1 as one of the solution.
Ah yes miss..
Being an art student, I am fascinated by your answer
I thought at the begining the equation had not a real solution because x^2+x+1 is all the way bigger than 0 so the following equations are false if we consider x as a real and not as an imaginary number
But x^3 = 1 means that x = 1.
I don't understand any of the logic you use.
f(x) = x^2 + x + 1
f(x) = 0
x^2 + x + 1 = 0, (x + 1 = -x^2)
x + 1 + 1/x = 0
You can pretty much get whatever answer you want as long as you do it wrong. f(x) = x^2 + x + 1 has no roots.
Also one of the solutions to x^3 = 1 is 1.
When you subsstuted x+1 as -x², you should also have done 1/x as 1/(-1-x²) so to get right answer
Finally another one that saw it ^^.
This makes way more sense than the video
There is no such rule, you should've gotten the correct answer regardless. Your substitution doesn't explain anything
@@yuzhmash3390 The substitution does give u the correct answer, which is 0. The x in the substitute -x square does not equal to the true x in x+1. In another word he made his substitution using letter x (which is misleading, he could have used y, since there is already another x), but that x doesn't equal to the true x . Since both x are not equal, he has to finish the substitution, which is converting 1/x into the -x square form, which is 1/(-x square-1).
@@haro1002 Yes, the substitution does give you the correct answer. It's just that Presh's solution has no real mistake in it, you don't have to substitute everything
It would be nice to analyze when and how operations add extraneous solutions, as a practical matter...
When you have x + 1 + 1/x = 0, it's still a quadratic equation because it becomes x + 1 = -1/x -> x(x + 1) + 1 = 0 -> x^2 + x + 1 = 0, that is the original equation. But when you substitute x + 1 with -x^2, it becomes a cubic equation, because -x^2 + 1/x = 0 -> -x^2 = -1/x -> x(-x^2) = -1 -> -x^3 + 1 = 0. Since now it's a cubic equation, it have another solution (1). So the problem is that you can't add degree to an equation by substitution because it will add new solutions that were not true in the previous equation.
@@jan.pierry Thank you! THIS should be the correct answer!
@@jan.pierry So basically, the substitution acts as a "multiplication by x" which creates another solution?
Learned in school (looooong ago) that you must not replace intermediate results from an equation into itself.
You may do that with results from other equations, though.
@@karoshi2
mostly, substitute from an equation that derives from itself will result in getting a number equal to itself
but this question increases the index of the whole equation by one, so another nonsense thing comes out
Or you could say that x^3=1 does NOT imply necessarily that x=1, so that step is false. x could be, for instance, (-1+sqrt(-3))/2.
Exactly... that is actual the wrong step.
Exactly what I thought!
If i'm right x could be 1 or -(-1)^1/3 or (-1)^2/3, but I dont see why it matters. Its obvious that adding a term will give 3 solutions, but where the error occurs ; and why picking the solution among 3 to find the mistake is legit ? It sounds arbitrary
But it DOES imply x=1. It implies other solutions and so the implication cannot be reversed, but this is not of the essence here.
Kudos to you sir
Sorry to point it out but you made a boo-boo here. Having an extraneous solution is perfectly alright and is not the problem here. The actual mistake is when they considered Cube root of 1 = 1. Cube root of 1 has 3 values and the other two values are already shown in this video and they are the values to be considered here, i.e. Cube root of 1 = (-1 + i * root(3)) / 2, (-1 - i * root(3)) / 2, 1. Taking cube root of 1 = 1 is the mistake.
You're too old to be using the term "boo-boo"
@@seadrown6252ikr, sent a shiver down my spine as I read it
"x^2+x+1=0"
me: ey wait a sec thats impossible
this dude: ok so here are complex numbers
since i already knew about complex numbers, i was able to solve it in about 5 mins, but its still a fun excercise. Yes i'm one of those geeks that think maths is FUN
@@edwinvermeulen8187 weird flex but ok
@@Tautviss yes
you mean negative square roots? use pq formula and you will see
@@edwinvermeulen8187
Dude, this is a math channel. Liking math doesn't make you special in the comment section of a video about math.
I spotted the mistake without even watching the video. 3 does not equal 0. Full stop.
rikilii
He was asking for the mistake in his process for how he got 3=0.
are you implying 3 is not equal to zero???
Blox117
It isn’t
He's woke
It is a post-factual world, so your opinion is as valid as mine. Let's compromise and meet in the middle: 0 = 1.5
Okay?
The process of dividing by x, making the stated substitution, then multiplying by x is the same as multiplying the original equation by (x-1).
This is a bit difficult to see, but this is what introduces the extraneous root x=1.
It's easier to see what's happening if you instead let f = x^2 + x + 1, then do the same steps.
f = x^2 + x + 1
divide by x
f/x = x + 1 + 1/x
From the first equation, we have x + 1 = f - x^2, so doing the substitution for x + 1 gives
f/x = f - x^2 + 1/x
rearrange to get
f - f/x = x^2 - 1/x
factor out f
(1 - 1/x) f = x^2 - 1/x
multiplying by x then produces
(x - 1) f = x^3 - 1
Letting f = 0, then we have x^3 - 1 = 0 or x^3 = 1, which is the final equation stated in the video, however,
we've introduced the extraneous root by (effectively) multiplying by (x-1)
Best explanation!
I hope he will pin your comment, finally somebody actually explained it, thanks.
a real explanation at last
nice explanation
@@gavinhua9753 the explanation is literally self evident. Why would it have a solution of 1 if we weren't multiplying by (x-1) to an equation that equals 0. Did we all collectively forget that 0 times anything equals 0??
The last step x^3 = 1 is totally valid! Taking any of the solutions and cubing them will give 1. The error is assuming that x^3 = 1 means that x must be 1.
The last step is actually 1 = x^3 (x-1)/(x-1), so x can't be 1
Neglecting complex numbers ain't we x^3=1 not always 1 my guy
I'm not sure if I'm right, but the denominator 3 was for the whole equation and not just 1.
If x^3 = 1 then x has to be 1. There are no other solutions.
@@physicssimulator2656 but one of the solutions are 1 though.
I edited this comment cause why not
Not when both sides tend to infinity :D
😮😮😮😮
@@thomastcheu3990 shhh it has to be child friendly
@@thomastcheu3990 If you multiply both sides with 0 often enough, it still works
It works also when it tends to infinity because it's a perfect zero not tends to zero ( indeterminate form) because anything multiplied by zero (perfect) gives zero
When the solution of x^3 = 1is sought, there are two more solutions that just x=1, you have x^3-1=0, which factors as (x-1)(x^2+x+1)=0. Again, here, we see how x=1 has been added as a solution, because the other factor is still the original constraint.
To me, this is the best and cleanest explanation!
I needed this explanation!
im 14 years old, i can understand this comment's explaining
plus the video now i hate math
@@rightangle9211 - If you are 14 and watching videos like this, then I suspect you are better at Math than you think! Math is a field where you are always going back to the basics, even when you get to the top levels. Stick with it, and work hard!
@@lialos ok yes
it is easy to spot the mistakes on those sorts of ""proofs"". Check if they divide by 0, check if they back tracked the equation and plugged values from that same equation on itself. Bam.
Yeah, and in this case the substitution was particularly "special", since they forgot to substitute the 1/x part in x+1+1/x with x+1=-x^2 in some way, hence the substitution was invalid.
Since I couldn't find a comment correctly pointing out the full explanation for why this comes up I decided to post my own.
So there are 2 mistakes going on from line 2 to 3.
The one I found less obvious and I did not initially understand but many people correctly pointed out in the comments that when you substitute x+1=-x^2 you really make two substitutions at once because when you substitute with a squared variable like -x^2 you can get solutions +/ - abs(sqrt(-x^2)). So if the substitution was done correctly we would get the additional solutions +/ - whichever we solve x for with our substitution.
Now clearly that did not happen here and that's because there was a 2nd mistake made in going from line 2 to 3 which was immediately obvious to me but I haven't found anybody else in the comments I briefly scanned pointing this out.
When you substitute x+1=-x^2 you have to apply this substitution for all x in the equation, otherwise you'll be dealing with 2 differing variables both called x.
But the substitution is only applied to (x+1) - "the left x" - from line 2 to 3 when it should really also be applied to (1/x) - "the right x" - which you can easily do by letting (1/x) equal (1/((x+1)-1)) which then leads to line 3 being -x^2+(1/(-x^2-1)).
This will lead to the behaviour expected from my paragraph above that we get the 4 complex solutions +/ - x_1 and +/ - x_2 from which only the two solutions with the negative real part which are also shown in the video correctly solve our equation from the starting point.
However, the incorrect substitution is a much more severe mistake as it introduced a completely different solution and I think it would be important for people to notice what should be a glaringly obvious mistake.
Thank you!
bump
well the mistake lies in considering x as a solution of the equation, the given equation has no solution hence it is useless
@@siddhantamallick6837 It's a polynomial and by the fundamental theorem of algebra must have a solution. In fact it has 2 solutions in the complex numbers. It merely has no solutions in the real numbers.
Simple way to understand it : we use substitution for at least two equations. We put value obtained from one equation to the other equation. In this case, we have only one equation. Division of one equation through x doesn't make it a different equation. So basically in the whole process, we are working with one single equation. This equation can be written in different forms. We are taking value from one form and putting that value in the same equation but different form. We are thinking this is substitution. But it is not. In substitution we deal with two different equation. Here we are using one single equation but with different forms.
The easier way to explain this: when you reach the conclusion x^3 = 1, the algebra is fine and correct. However, what this says is that the solutions to the original polynomial satisfy x^3 = 1, but not every number that satisfies x^3 = 1 is a solution to your polynomial. So there are 3 possible candidates, and 2 definite solutions. Since complex solutions to polynomials must come in conjugate pairs, the complex candidates are the correct solutions and x = 1 is not.
Yes, this is the correct answer to the problem. The assumption that x^3 = 1 implies x=1 is incorrect and missed the other two solutions, which are the ones that match the original equation and must be checked at each step. Most fundamentally, the derivation took a second-degree problem and created a third-degree, which would imply that an extra solution (may) have been inserted that must be checked and eliminated to satisfy both (all) equations.
Yeah its a cyclotomic polynomial. It's roots are identified with the elements to its galois group over the rationals. This is the order 2 cyclic group.
Simple way to understand it : we use substitution for at least two equations. We put value obtained from one equation to the other equation. In this case, we have only one equation. Division of one equation through x doesn't make it a different equation. So basically in the whole process, we are working with one single equation. This equation can be written in different forms. We are taking value from one form and putting that value in the same equation but different form. We are thinking this is substitution. But it is not. In substitution we deal with two different equation. Here we are using one single equation but with different forms.
The mistake was me not paying attention in algebra class
Aidan Artichoke yes
2 + 2 = 4 - 1 = 3 ≠ 0 quick maths
Wrong
@Universe 123 I suspect it's just a joke, but I could be mistaken.
Someone's been smoking trees again
It's not correct maths, but quick maths. Learn the difference.
2 + 2=4 - 1=3
2 + False - False = 2
A variant of the immortal line “left as an exercise for the reader”
THE MISTAKE IS HERE :
x³ = 1 ⇒ x = 1 ❌ Not necessary
because :
x³ - 1 = (x-1)(x² + x + 1)
so x = 1 OR x² + x + 1 = 0
Bro I didn't understand
hello , it's wrong to say that
[ x³ = 1 ⇒ x= 1 ] ❌❌❌
let's see why :
(-½ ± i √3/2)³ = 1
and [ -½ ± i √3/2 ] ≠ 1
So x³ = 1 ⇒
x = -½ + i√3/2 or x = -½ - i√3/2 or x =1
so x can be ≠ 1
I think this is a much better solution than what presh said (no offence). Well done Saadallah
@@neelnarlawar thank you :)
That's not the problem, the problem is that x^3=1 does not imply x=1 since we are working with *complex* numbers, assuming x was real was the problem. Observe the two actual (complex) solutions are indeed cube roots of 1, so x^3=1 is perfectly correct. Also observe that x^3-1 can be factored as (x-1)(x^2+x+1), the first factor has root 1 and the second factor has our two complex roots.
Also your explanation was of no use, if you're trying to solve an equation you can't detect a false step by saying that you just created a new solution because you don't know the set of original solutions. When you make deductions you are making implications, which means that your solution x must verify the new equation but not the way around (i.e. that all solutions of the new equation are solutions of the original one). This is the classic mistake that confuses implications with *equivalences* , each of those steps are only implications. The equation x^3=1 has 3 solutions of which 2 are the original ones, so the moral is that making implications gives you the list of *candidates* for solutions, but you should always check by substituting and in this case you would have discarded the 1.
I agree with your assertion that x^3=1 does not imply x=1, and it is more obvious to see in step 5 than it is in step 3. However, x^3=1 and -x^2+1/x=0 are equivalent statements. You could also say that -x^2+1/x=0 does not imply x=1. So the error actually occurred in step 3 as Presh identified. His explanation may not have been as clear as it could have been, but it is no less correct.
Yes but that is not an error, it is indeed true that x^2+x+1=0 implies -x^2+1/x=0, the only false implication is the last one. Of course, no one talked about equivalences (and that might be a common error, interpreting those steps as equivalences rather than implications).
Eduardo 2 equivalent statements, should not have the same set of solution?
Yes, the problem is that most of the times people don't make equivalent statements, like here, they make implications, which are one directional, i.e., the new statement must be verified, but it does not necesasrily imply the previous one.
I can't agree with you more,implications are not equivalences.Thanks for making me think further.
3=0
just move the lines of = to make a >
3>0
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Hi Presh. Thanks for the nice video.
In your response, you based your critique on the fact the you know the answer before hand to discard a step, rather than identifying what is algebraically wrong with the step itself. I am not convinced with the reasoning, unfortunately.
I think the mistake is in the x^2 + x + 1 = 0, it can't be equal to 0, x^2 + x + 1 will always be positive, even if x = 0 then it would be 0^2 + 0 + 1 = 0, it will always positive, if you make the x negative then it will be: -x^2 - x + 1 = 0, and it would be x^2 - x + 1 = 0 which is positive too, so there is no answer which satisfy this equation equal to 0
You're correct: D = b^2-4ac = 1-4 = -3; -3 < 0 => this equation has no solution
@@cffex3858 We are considering complex numbers. The imaginary unit is i and has the property that i^2 = -1. This means it is a root to an equation like x^2 + 1 = 0. Presh shows the two solutions. The mistake is not what Presh stated, though. It is the last step, in which he states if x^3 = 1 then x = 1, and that is definitely not true (consider one of the solutions to x^2+x+1). It is true that if x^2+x+1=0, then x^3=1, as the other steps shown.
@@bartholomewhalliburton9854 how else would you solve x^3 = 1.
x = 1 seems obvious to me.
This video would have you believe that the usual method of solving equations involving square roots is bad. This is not the case. It is fine to introduce extraneous solutions to your equation(and sometimes this is necessary in order to find the solutions you are actually interested in), you just need to check that the solutions you find are actually solutions to your original equation.
The mistake is not introducing an extraneous solution. The mistake is in concluding that x^3=1 requires x to be 1.
You have an excellent point. The first step was ambiguous because x is not quantified (is x real? complex?), but as soon as we know that x is complex, then x^3 = 1 should have 3 complex solutions
Hey even x^8 = 1 has 8 solutions, only two of them are real
@@diffusegd
Yeah, of 8 solutions only 2 are real, and only 2 are imaginary. It's enough to give you a complex
I would say that the mistake is in concluding that the solution must be real, when in fact it can't be real.
That's it
Hi Presh
Actually, You just take ANY equation. Divide x both sides to make a new equation. Subtract the older from the new and you will generate a new solution i.e. x=1 which is wrong. In the question, substitution is equivalent to subtraction of the older from the new equation.
Details:
Let an equation be f(x) = g(x) where f(x) and g(x) are any two functions of x.
divide it by x both sides
f(x)/x = g(x)/x
Subtract from the older equation:
f(x) - f(x)/x = g(x) - g(x)/x
or
f(x) (1 - 1/x) = g(x) ( 1 - 1/x)
Now x=1 satisfies this, but that's just because you created it forcefully, which is wrong.
just keep it simple
YOU CAN NOT SUBSTITUTE THE SAME EQUATION TO ITSELF you end up with 1=0 solutions
Yaman Sanghavi This is wrong, x=anything works under these circumstances As you've stated the functions are equal
ANIKHTOS Also wrong
@Good Memory, Bad Chess I am afraid you are mistaken. When someone says that f(x) = g(x) is an equation, it means it need not satisfy for all x. In fact, solving an equation is a task to find the x at which these 2 functions are equal. For example suppose I have an equation: Cos(x) = Sin(x). We know this satisfies at x=pi/4. But, this doesn't mean that sin(x) is the same function as cos(x). Functions are equal if and only if they are equal for all x. Whereas an equation can be satisfied only at some values of x.
Lovely explanation.
I did the math myself and rediscovered the cube roots of unity I had studied a decade ago! Thanks for the video, I don’t know if I should be proud of my math or be ashamed of my memory!
x^3 = 1 has 3 solutions, 2 of them are the complex solutions to x^2 + x + 1 = 0 , and the other is 1. However, you must reject 1 since it obviously doesn’t satisfy the originally equation of x^2 + x + 1 = 0
Yeah it is 1,w,w²
The way I understand this is that the step turns the quadratic equation into a cubic equation which has 3 solutions, containing the two from the original equation and the extra one.
That's it.
Thank you, that for once is an actual explanation. The video is pretty bad in saying "just be careful!" without ever explaining why it does what it does.
Yes. That is why, when you solve an equation starting by giving you x a solution, you always have to check reciprocally if what you got is indeed a solution.
And the other two are complex numbers which are commonly known as cube root of unity(omega)
There's nothing wrong with deriving a cubic equation from a quadratic. The mistake was picking one of three solutions to the cubic and then plugging it back into the original.
3 x 3=9
0 x 0=0
9 not equal to 0
bruh moment
(x-x)(x-x)=3x(x-x)
0x=3x
0=3
3=0 XD
You cannot decide by x when you are 0x is undifintable
Nvsb Channel and also you cannot divide by (x-x) cause it's zero
Or is it?
@@user-fw4nj6pf5b the right hand side also becomes 0.
The problem is in the very first step actually. To see this, multiply by x instead of dividing. This creates the additional "solution" x=0. Plug back in and get 1=0.
Also we can use brahmagupta's relation (he is the person who found the quadratic formula) of variable's to coefficent in the first equation x^2 + x + 1 = 0. He stated that if b^2 - 4ac < 0 then the quadratic equation has no real roots/no real solution. Hence from there we can conclude that x is not possible to be 0
Not possible to be 1 either... when x =1 then 3 =0 ....
Simple way to understand it : we use substitution for at least two equations. We put value obtained from one equation to the other equation. In this case, we have only one equation. Division of one equation through x doesn't make it a different equation. So basically in the whole process, we are working with one single equation. This equation can be written in different forms. We are taking value from one form and putting that value in the same equation but different form. We are thinking this is substitution. But it is not. In substitution we deal with two different equation. Here we are using one single equation but with different forms.
But, but, but...why did that substitution introduce an extraneous solution? It looks valid on the surface, and without further explanation you have only given half an answer to the problem. (It is apparent that in that step the equation was changed from a quadratic to a cubic, and that is where the extra solution came from, but unless you explain that you have given your viewers no tools to understand how to avoid similar mistakes in their own derivations.)
That substitution subtly injects a new solution into the equation. We reduced the situation of x + 1 = -x^2 = -1/x to just -x^2 = -1/x. By eliminating that detail, you allow for a solution that didn't originally hold-particularly, the solution that _doesn't_ satisfy the original restraint that either side also equals itself plus 1.
Here's a more straightforward example of the same thing. Suppose we have x(x-1) = 1. Then x = 1/(x-1) and x = x^2 - 1 . From here, we can conclude that 1/(x-1) = x^2 - 1, which clearly has x = 0 as a solution. However, we still need both the lefthand and righthand side to equal x, and x=0 doesn't fit the bill here. (The solutions to x(x-1) = 1 do, though!)
Exactly. A full solution to the problem would be to explain WHY that step introduced an extraneous solution. Normally we expect extraneous solutions to pop up if, for instance, we square both sides of an equation. But here, you're merely substituting one quantity for an equal quantity. Generally, if you substitute equals for equals, you don't change the solution(s) to an equation. Yet here, it clearly does. It would be interesting to answer why.
Simple: It went from a quadratic equation to a cubic one. That's all you need to do to introduce a new solution. So that's why when solving polynomials, you really want to make sure, if you add a new solution to solve a problem, it's noted, or at least a repeat of the other solution(like x(x-1) going to x(x-1)(x-1) is no problem).
If you want the mathematical proof what just happened, here is a comment by yaman sanghavi:
You just take ANY equation. Divide x both sides to make a new equation. Subtract the older from the new and you will generate a new solution i.e. x=1 which is wrong. In the question, substitution is equivalent to subtraction of the older from the new equation.
Details:
Let an equation be f(x) = g(x) where f(x) and g(x) are any two functions of x.
divide it by x both sides
f(x)/x = g(x)/x
Subtract from the older equation:
f(x) - f(x)/x = g(x) - g(x)/x
or
f(x) (1 - 1/x) = g(x) ( 1 - 1/x)
Now x=1 satisfies this, but that's just because you created it forcefully, which is wrong.
It takes "x+1" and substitutes it with "-x²". This is not a valid substitution for x+1+1/x = 0, the equation x+1+1/x = 0 does not imply anywhere that x+1 = -x². If anything, then it impies that x+1 is -1/x. But we still can't simply substitute x+1 with -1/x, even though it must be equal to it, as we can only do that if this would be true IN ALL CASES, not only in THIS SPECIFIC ONE. Also, the second step actually has introduced a mistake already, it would correctly be x+1+1/x = 0/x since we had divided by x.
you can not use the same equation to substitute to itself
if you do you end up with 1=0 solutions
We can see this another way. The final result is 1=x^3, but it can be rewritten as
0 = x^3-1 = (x-1)(x^2+x+1) = -x(-1+1/x)(x^2+x+1)
This factorization is no accident.
If you look at the steps it took to reach the 3rd step, it's equivalent to multiplying by that 2nd factor.
That final factor of x is introduced in the final step.
As can be seen, the x=1 root is introduced by the (-1+1/x) factor.
pierrecurie sir your explanation is best one
pierrecurie ,by your explanation he is multiplying the equation by x-1 and after he is saying that x=1 is not satisfying the equation ,what is that lol video hahahahahahw
But why isn't 0 a solution then??
@Bernhard Bish I mean 0 = x^3-1 = (x-1)(x^2+x+1) = -x(-1+1/x)(x^2+x+1) right? So why doesn't plugging in 0 work? I know it's not a solution, I'm just curious. Is it because 1/x creates a problem?
@Bernhard Bish Oh right. One can't get the second equation if x=0 was a solution, since it's obtained by diving by x. Ok I get it. Thanks
That’s not actually where the error is. The only error is that x^3 =1 does not imply that x=1.
?
@@claudiosayian4311 ?
2:50 Presh, you got it wrong.
So this proof is basically (with y being x^2+x+1):
Assume y = 0.
Then it follows that y equals either 3 or 0 - correct, that's the third line that you highlighted as wrong but isn't.
Then it follows that y = 3. - this is the fifth line that's actually wrong.
Therefore 0 = 3.
The third line does follow from the first two lines. It's not equivalent, as it obviously introduced extra solutions. But it does follow. As in "A implies (A or B)"
The fifth line is the problem, there are 3 solutions on the left, but only 1 on the right. So that line says: if x is either 1 or something else (2 complex numbers), then x = 1. Basically doing "(A or B) implies B" (which is wrong).
I usually like your videos but your explanation here is pretty bad.
The real problem is that x^3 = 1 doesn't necessarily imply x =1. That's the mathematical mistake.
The 3rd line is not a problem. There is a correct mathematical implication between lines 2 and 3 (but of course not an equivalence).
The 3rd line will give all the POSSIBLE solutions for the equation but you'll have to check which ones are solutions of the first equation too.
There was a mistake make, you can't substitute only a part of the equation.
You can substitute whatever you like as long as you know what you're doing. If you assume that x+1 = -x^2 then you're allowed to subsitute -x^2 for x+1.
@@bobfr4806 well, if you substitute the 1/x as well by the same formula, you will get the right answer.
No, the mistake is moving the equivalence of x+1=-x^2 to the second equation, but that turns the first equation into x^2-x^2=0, so now you shouldn't divide by 0 because x can now be 0
I think the original video is fine.
He is pointing out the first line where equivalence (if and only if property between lines) breaks down. That happens at first point where the solutions are no longer the same. That happens exactly where he said it happens.
Your comment about the solutions for x^3=1 is valid in isolation, but it is not the first time the erroneous solution x=1 was introduced.
x^3 = 1 is true but it doesn't mean x = 1
actually it'll be like this;
x^3 = 1
x^3 -1 = 0
(x-1)(x^2+x+1) = 0
so, yes there are three values that satisfy x^3 = 1 but there are only two values that satisfy x^2 + x + 1 = 0 and x=1 isn't one of them.
what you wrote doesn't change anything, those 3 values are 1 and the 2 complex, non-real values he showed, there is nothing wrong in saying x^3=1 gives us x=1 , but by the time you reach that step, the mistake has already been done.
Pabhanuwat Pongsawad x²+x+1=(x³-1)/(x-1) so we can not take x=1 as it make denominator zero
You forgot about the other part x-1=0
x=1
@@khalilrahme5227 well , you know that x^2 = -x -1 , wich can only mean that -x-1 has to be greater than 0 , wich means x
@@khalilrahme5227 Using the phrase "gives us" tells us that you do not understand (mathematical) logic.
I am a medical student and my maths days are long behind me...and i still love watching this
Actually, the equation is wrong from the beginning because the left side of its is always greater than 0. In fact, the side is greater or equal to 3/4
that's because you assumed x is a real number, while in reality both solutions are complex numbers
Actual mistake is in line 5 because 1=x^3 does not implicate x=1 in complex numbers
Your explanation with additional solutions for x is a little bit inaccurate
Mistake is in the first line since x^2 will be greater than or equal to x, so x^2+x will always be a positive number, and adding 1 to a positive number won't equal to 0.
The game was rigged from the start.
@@PunkZombie1300 x^2+x+1=0 has two complex solutions, another real one (x=1) is added in line 3, so this equation is not equal to our original equation but implicated by it
Mistake is in the last line in which it says "3=0"
@@PunkZombie1300 Try x = 0.5
That's what I was thinking too!
The equations do have different solution sets, so they are different, but the question "Why are the solution sets different even though all the algebra was correct?" will give any mathematical student the perception that the algebra only works some of the time. That's dangerous grounds as an educator since you want your student to understand the root of the error. (Pardon the pun).
The error was assuming that x^3 = 1 implies x = 1. The writer of the book forgot we were looking at a polynomial ring over the field of complex numbers.
~~
First, we know the Fundamental Theorem of Algebra can be stated that every single variabled nth degree polynomial with complex coefficients has as most n complex roots. This means the original equation has at most two complex roots.
But when you go to solve x^3 = 1, you should be looking for three roots. Doing the algebra in the complex field, you can find them.
x^3 = 1
x^3 = e^(2n(pi)i)
x = e^(2n(pi)i/3)
x (elementof) {1, e^(2(pi)i/3), e^(4(pi)i/3)}.
You truly get three roots, but only two of them are correct since our algebra gave us a bad value. It is the mathematician's job to check the solutions and determine the validity of the algebra.
But this does pose an interesting question for further study: which algebraic tricks are afforded to a mathematician to guarantee preservation of an equation's set of solutions?
you can not substitute the same equation to itself
thats 101 for algebra
if you use the same equation to itself you end up like this
X^3 = 1 => X=1 is a true statement when X is real. Although there are many people saying it was a mistake in the video, it really wasn't. Also the only wrong step was concluding that X=1 is a solution of the original statement, because this was logically false.
@Supremebubble. x^2 + x + 1 = 0 has no real solutions so x is complex. Thus x^3=1 implies x=1 is false.
@ANIKHTOS Claiming you can't substitute an equation into itself is effectively stating that recursion doesn't exist. Computer programming wouldn't be what it is without it.
You should look up Continued Fractions. Some of the most beautiful things are born by self-substitution.
@Supremebubble You are correct to say x^3 = 1 implies x = 1 if x is real, however there was no condition forcing x in the set of real numbers. To assume x is a real number without justification is err.
Took me a second to figure it out, but I got it: x^3 = 1 does not imply that x = 1. It is true that both solutions of x^2 + x + 1 satisfy the property x^3 = 1, but the solutions are complex-valued. This is an important lesson in understanding the direction of implications, and in treating polynomials with respect in the complex plane.
EDIT/ADDENDUM: I gave my solution before watching the rest of the video. Presh's solution is another way to interpret the problem, and it's technically correct too. When you "create another equation" from the first equation, that can mean two different things. It could mean that you start with equation A, and then from equation A, you get the _equivalent_ equation B (whereby equation A and equation B describe exactly the same solutions). Or it could mean that you start with equation A, and then you conclude that the solutions to equation A must satisfy equation B. In other words, you could either have "Equation A holds if and only if equation B holds", or you could have "If equation A holds, then equation B holds." I personally use the former method when I'm doing scrap-work, and the latter method when I'm doing proof-work.
This is true :D
Also add that x^3-1 factorize into (x-1)(x^2+x+1)
Your explanation is the correct one. I'm sorry I don't agree that "Presh's solution is another way to interpret the problem, and it's technically correct too".
Nowhere the problem states that every equation is equivalent. It said the first line implies the second line that implies the third one etc... To introduce a new equation that has an extra solution is not a mistake as long as you don't assume that the equations are equivalent.
I absolutely agree with you, and I also think that yours is the correct answer, not the one given in the video. It's important to have in mind that "x is a solution to the original equation", and x stills verifies the next equations, so everything is right up to the fourth step.
Trucmuche I agree: thinking that the equations are equivalent is an interpretation mistake, but not a proof mistake as that is not stated.
FYI: The two valid solutions are actually the third roots of unity, excluding 1.
I disagree with your answer to the problem. There is nothing wrong with the algebra that leads us to the conclusion that if x^2+x+1=0 (1) then x^3=1 (2); this can be seen by the fact that both the roots of (1) are also roots of (2). The mistake in the "proof" is to assume that the logic goes both ways and that all roots of (2) are roots of (1). Also, x^3=1 does not imply x=1since we are considering complex numbers, but I wouldn't consider that the actual problem with the "proof".
This guy has no regards for cube roots of unity
In my opinion, the error is in the line 1 = x^3 --> x = 1. Up to this point, all the statements are true; for example, -x^2 + 1/x = 0 is a true statement about x given how we have defined x. However, 1 = x^3 --> x = 1 is not a true statement, because 1 = x^3 has two other solutions, which happen to be the two possible values of x.
X^3 = 1 => X=1 is a true statement when X is real. Although there are many people saying it was a mistake in the video, it really wasn't. Also the only wrong step was concluding that X=1 is a solution of the original statement, because this was logically false.
The thing is 1 has three cube roots. 1 is one of them. The other two are imaginary roots being -1/2 + (i√3)/2 and -1/2 - (i√3)/2. 1 doesn't seem to satisfy that initial equation, but the other two imaginary roots do.
the mistake is saying if x^3=1 then x=1. there are 3 solutions. so it's not on the 3rd step, it's on the fifth step. you gotta find all the solutions (not just one) and try them all in the original equation.
When -x^2 was substituted in place of x+1, the expiation becomes a cubic equation which simplifies to x^3-1=0 which has all three cube roots of 1 as the solutions. x^2+x+1 is a factor of x^3-1.
You have actually multipled the equation by (x-1), because
(x-1)(x²+x+1)=0
=> x³=1
So you have manually added the x=1 as solution.
x³=0 has 3 roots 1,(-1±3i)/2.
I think I understand where the 1 comes from. The substitution done is the same thing as setting the two equations equal to each other (if you subtract both sides, the x^2 becomes negative, and x + 1 cancels out). Now, since one side of the equation is the other divided by x, x can be 1 because both sides will remain equal. x can also still be either of the solutions because both sides of the equation will become 0.
I agree that line 3 creates the extra solution, but I don't think that's the "mistake." The mistake is in fact in line 5, where (a) we consider only one solution of the equation x^3 = 1, and (b) assume this solution must be valid in the original equation. Note that the first statement is "let x be a solution of...". So there is nothing wrong with line 3: every equation that follows is still valid when you plug in the two original solutions of x, and there is certainly nothing wrong with making a simple substitution! It's as if I said "let x=1; square both sides, and we have x^2=1. This equation has the solution x = -1, but x also equals 1, so -1 = 1." Of course this is nonsense, but there is nothing wrong with squaring both sides, just with assuming new solutions must equal the old ones.
Exactly what I was going to say. If x is a solution to x^2+x+1=0, then x^3=1 is correct. What is not correct is that those two equations have the same solution set, but his wording focuses on a single number (either one of two possible complex numbers), not the full solution sets. The flaw in the reasoning is most definitely the assumption that x^3=1 implies x=1.
You can not substitute form the same equation
if you use the same equation again and again you will end up with this.
Finally someone else who knows what's going on :)
Every comment here including yours is wrong
It's more to the beginning
I notice that you can also obtain the 3rd equation from the 2nd by multiplying both sides by (1 - x). This might give an intuitive idea of why it introduces x=1 as an extraneous solution.
but with x = 1 you get 0 = 0.
and you cannot obtain the 2nd equation from the 3rd solution because that would mean that you are diving it by (1 - x), and therefore for a case where x = 1 you are dividing by zero
The 3rd step was fishy because the equation went from a quadratic to a cubic.
you can not substitute the same equation to itself
you will end up with 1=0 events
no it's fine. it's usually just useless and you end up with 2=2 events.
Donald Johnson You spotted the issue but used the wrong words.
Not seeing how I used the wrong words. The 3rd step formula is a cubic given X not = 0.
How about all these mathematical modalities are bullcrap since they do nothing for man's journey inward, nothing for finding oneself! I have an engineering degrees and I got it on partial credit since not one "math" genre makes since to me.. Laws of thermo dynamics are BS too. All just limitations... I can create empty parking spaces...the best parking spaces with my mind...what do ya call that in math terms?
I actually assumed the mistake was when we obtained the solution for x^3 = 1 as x = 1, since x^3 = 1 means the solutions are the cube root of Unity, which are the two solutions to the quadratic alongside 1, rather than just 1. I usually assume that when I see paradoxical equalities like these, it involves ignoring imaginary/complex numbers wherever they arise.
You seem pretty smart, can I ask why when we substitute x+1=-x² it adds 1 more solution?
I think I am missing some basics on imaginary numbers (I'm in 10th grade but please explain it in the matematically correct way, without making it more simple)
@@HandleNotAvailabIeThe one extra solution was always there, but you can avoid reaching it if you follow the steps in this video. Even the original equation x^2 + x + 1 = 0 has imaginary solutions, and you'll see them if you use the Quadratic Formula to work them out. And when you get to x^3 = 1, we cannot simply say the solution was x = 1 because we're not mentioning the imaginary solutions here, which leads us to the paradox. If you can get your hands on some mathematics books from 11th or 12th grade, you might get a more rigorous explanation for "cube root of unity", sometimes called the "nth root of unity". Aside from 1 itself, there are some imaginary values whose cube is 1.
@@rohitchaoji thanks a lot, I looked it up and now I get how it's possible
Note the range of powers of x in the first equation is 0,1, and 2 (squared), the second equation is -1,0,and 1, while the 3rd equation expands the range with 2,-1. Expanding the range of powers in a equation by such things as squaring or other substitutions is dangerous.
We have done all the manipulation assuming that "x" is the solution to the original equation. The last equation x^3=1 has 3 solutions....x=1, and two imaginary solutions. Now since x=1 is not a solution to the original equation, then why the hell we are putting x=1 in the original equation???? 🐯🐯🐯
So math is this hard. I cant understand anything
Because _normally_, rearranging wouldn't introduce any new solutions, but here, the equation was artificially raised from quadratic to cubic, so you get another solution which satisfies the cubic, but not the original quadratic.
jack sparrow if you spend a few hours every day with math, things start to make sense there
Jakub Frei thanks
jack sparrow U Welcome :)
You substituted the equation with the same equation! This is redundant!
Should have got 1=1
no it was fine, the problem was he only changed a part (namely x+1) of the equation for the other but left the 1/x as is. That is the mistake. The substituted version would be -x^2 + 1/(-x^2-1)
@@hiddevanzalm272 yeah, in this case the x in -x square are not the same as the x in x+1. I calculated that the x in -x square equal to i, and the x in x+1 (the true x) is equal to 0 when u use x+1=-i square(-x square)
@@hiddevanzalm272 but the question is incorrect though, the true x is equalled to 0, while the question state that x doesn't equal to 0
@@hiddevanzalm272 thanks a lot. I only understood the mistake with your comment.
Mistake was in line that x^3=1 only means x=1
It actually means
x^3-1=0
using formula of a^3-b^3=(a-b)(a^2+b^2+a*b)
x^3-1= (x-1)(x^2+1+x)=0
from this we can get either x=1 or x^2+x+1=0 and as we know x^2 + x +1=0
but x=1 was actually not a solution for the base equation we started with.
So when we do any kind of substitution we have possibility of getting any extraneous solution.
Actually, I would argue that the mistake came in the step when you asserted x^3=1 THEREFORE x=1. The step where you added x^2 in is actually fine, because you defined x to be a solution to x^2+x+1, so it IS true that x^2-1/x=0. The problem was going from (any of those equations) to the conclusion that x=1.
Good point.
What a nonsense....
X³ =1 that means it is an cubic equation which has three solutions 1, w, w²... Except 1 other two w & w² satisfy the equation x²+X+1=0
Hence a quadratic equations has two roots only
X²+X+1=0 has two roots w&w²
And 1, w, w² are roots of equation x³=1
Also called cube roots of unity
Thanks.
Hi, I think the wrong deduction is actually in the step x^3=1 => x=1, as some people pointed out already. The substitution you mentioned is a correct deduction, (though not an equivalent statement). It does not matter, that an additonal solution is introduced. Love your videos!
Diesse nutz hah got em
well for x^3=1, x would have to be 1 since -1^3 would be -1
@@deadfox5489 Only if we don't go into complex territory; because in the complex world, there are n roots to a polynom of degree n (here, three)
In this case the problem is that we assume the equivalence () of every step, but we lost it at step 2 (like he kind of explained but not really).
If the equivalence was kept, any solution that we find for the last equation should be a solution for the first, including 1.
That's why generally we prefer to only use implication (=>) instead of equivalence, and then check the reciprocal (
For me the solution is wrong because you just can't replace x+1 = -x^2 in second step if x+1 is not placed in brackets. Just my thoughts. Too easy, but it may be right problem solution
When you multiply both sides of an equation by a variable expression you potentially create extraneous solutions.
Even if the expression is not variable.
Try multiplying both sides of any equation by 0, and you'll find out everything is a solution of the resulting equation, but not the first
@@xhantTheFirst 0 is the important number here. Whenever the variable expression is 0 is the extraneous solution.
I'm going to have to say that while that step introduces an extra solution, it isn't really an invalid algebraic manipulation, so I can't call it erroneous.
The real problem is near the end, where one goes from x^3 = 1 to x = 1. _That_ is the actually invalid step, because x^3 = 1 is not sufficient to show that x = 1, and indeed, the other two cube roots of 1 are the correct solutions to the original equation.
Yeah you see, in high school maths (I assume it is high school maths problem) the distinction between the solution set and one particular solution does not always get clear. And by artificially creating a cubic equation out of a quadratic one you certainly expand the solution set. That is also what President Skywalker wanted to tell us.
I think 1 = x^3 is 1 = x because 1^3 = 1 and so is 1^3 = x^3 the same as 1 = x^3
Namooo No, because [ (-1 + SqRt(-3))/2 ]^3 = 1, yet the ( -1 + SqRt(-3) )/2 is not 1, so you cannot conclude x = 1 from x^3 - 1 = 0. You can only conclude a disjunction, not any particular part of the disjunction. Namely, there is a theorem in algebra that if ab = 0, then you can only conclude the disjunction a = 0 or b = 0, but you cannot conclude either of the two as definitive without some other equation that the other is false.
@@angelmendez-rivera351 Thx
@@angelmendez-rivera351 from x^3 = 1 you can conclude x=1 if x is a part of R. Why can't you? There is no other possible solution (with real numbers).
The problem is that there are 3 roots to x^3=1. x=1 is only one of the roots. If we factor x^3-1=0 we get (x-1)(x^2+x+1)=0. Leaving us with 1 and the two original solutions to the 1st quadratic. This is not allowed.
Yes, but that's not the problem
De sampige kasser No, that actually IS the problem. He decided to conclude x = 1 from x^3 = 1, which you can never do unless they tell you x is real, which was obviously not the case in this video. x^3 - 1 = (x - 1)(x^2 + x + 1) = 0. He arrived at this proposition from x^2 + x + 1 = 0, which also implies x =/= 1. Now x=/= 1 and (x - 1)(x^2 + x + 1) = 0 in conjunction imply x^2 + x + 1 = 0, not x = 1. He basically just failed formal logic.
The reason of the error: equivalent transformations are only allowed in linear equations, where the highest power of x is 1. The degree n of such an equation is 1.
Generally there are n solutions to an equation with degree n. Here we have the special case n=2 solutions in a quadratic equation that cannot be substituted by ONE value.
In line #5 the expression 1=x³ has degree n=3, consequently we have 3 solutions of which one solution is real (i.e. 1), the remaining 2 solutions are the complex conjugate solutions of the initial equation x²+x+1=0.
One important thing here is when you get x^3 = 1, it doesn't mean x=1. There are 3 possible cube roots for 1, which you can google more to find out. Thus, it means out of 3 possible cube roots, at least one cube root follows the earlier equation. I did some manual hit and trial, and clear enough, you get the same results.
If you use x=1, i.e. the first cube root of unity, the original equation x^2+x+1 becomes 3, which is not equal to 1.
If you use x= (1+ sqrt(3)i)/2, i.e., the second cube root of unity, x^2+x+1 becomes 2, which is not equal to 1.
If you use x= (1- sqrt(3)i)/2, i.e., the third cube root of unity, x^2+x+1 becomes 1, which is equal to RHS.
The method, if you deep dive properly, yields a perfectly logical explanation. We just forget the x^3=1 has 3 possible solutions, not just 1.
When you look away from the board for 5 seconds
When multiplying or dividing an equation, the equation remains true but not always remains the same.
Since the first equation has no real solution isn't the problem coming from the fact that we divide by a complex number ?
Maxime Bardiau No, not quite. The problem is that we multiplied by x - 1. You see, if you have an equation and you multiply by x, you change the equation itself, which means it has new solutions. x^3 - 1 has factorization (x - 1)(x^2 + x+ 1). The solutions of x^2 + x + 1 = 0 are also solutions of x^3 - 1 = 0, but not vice versa, because of the extra factor. There is an algebraic theorem that states that if ab = 0, then you know at least one of the two is 0, but you cannot claim to know which one it is, so you cannot claim a = 0, and you cannot claim b = 0, you can only claim “a = 0 or b = 0”, and as such, you cannot claim a = b. You could only claim a = b if you knew “a = 0 and b = 0”, which you could not know because I already stated you could not even know which of the two is the case. Not unless you had extra equations. So you cannot conclude x = 1 from x^3 = 1.
@@angelmendez-rivera351 I agree, but I wouldn't say multiplying by x-1 was the problem as theres no harm in doing that.
The math is still correct is my point there.
The error was assuming that x was 1 given that x^3=1.
Meanwhile, JEE Advanced students reacting to the level of the question: "Am I a joke to you?" 😂
Why???
There is the concept used called Root Gain,Some times we change the equation in such a way that it's solutions increases i.e.
x=-1
We square both side,
x²=1,and now x=±1,and we all know it's not possible 1=-1,Hence Root gain.
Thank you from india🇮🇳
awesome explanation, thanks.
The mistake is the statement "if x^3=1 then x=1". No x can be one of 3 solutions and 2 of them satisfy the initial condition.
Also you can get x^3=1 by just multiplying both sides with (x-1)
mustafa azyoksul This is more complicated than you construe it. Yes, you could just obtain the final equation by multiplying the original by x - 1, but this fails to illustrate why the substitution he performed creates a new solution to the equation, and why does it amount to actually multiplying by a monomial. In other words, the substitution is not innocent, and the confusion is why it is not. Students are always incorrectly taught in schools that if you derive one equation from another, then this new equation always has the same solutions as the original, and this video proves this is not the case, but it fails to explain why this is the case. Re-explaining it as "the substitution amounts to multiplying by x - 1" also fails to explain it. The real explanation is that the substitution changes the degree of the polynomial equation as the substitution itself has two degrees of freedom. And by the fundamental theorem of calculus, changing degrees changes the solutions. This is why you cannot conclude a derived equation has the same solutions as the original. You can only do this for same-degree polynomial equations. In other words, this works only if you apply linear transformations to equations, or unary invertible operations to equations.
Haven't seen the answer yet. My guess is that it lies in the initial equation. x^2 + x + 1 cannot equal 0 unless x is an imaginary number. You can prove this by trying the quadratic formula and finding that the discriminant is a negative number. So plugging -x^2 in for x + 1 doesn't work cause they ain't actually equal cause x has an i in it
I saw a lot of cool explenations in comments, but didn't see this one:
Original equation has most exponent 2, so it has 2 answers, while in equation x^3-1=0 there are 3 answers, because of exponent 3.Also as others said: x^3-1=0, (x-1)(x^2+x+1); we proved that answers from x^2+x+1 are also answers for x^3=1, but not reverse;
I knew something weird was going on with that substitution that introduced the problen, though I couldn't figure out what the mistake was
But what is the cause of the creation of the extraneous solution?
Darshan Krishnaswamy I would like to know this too, why does substituting something that is equal create another solution
The transformation from a quadratic equation to a cubic equation will inevitably cause an equation with only 2 roots to create a 3rd root. But I don't know at which point of the substitution is "illegal".
+Anonymous71475
Not necessarily. You could double the roots. You could multiply the quadratic by (x - a) where a is one of it's roots, and the resulting cubic would have the same roots (only 2).
Though in this case it'd be a cubic with complex coefficients which looks fishy.
Substituting an equation into itself creates extraneous solutions. An extreme example: let x be a solution of the equation x=x+1 (a). It follows x+1=x, so I can substitute the right-hand side of (a) by x and get the equation x=x, which is solved by any x, while my original equation has no solution.
The actual error in the video is the implication at the end: x³=1 does not imply that x=1. It is true that x=1 is a solution of x³=1, but that only means x=1 ⇒ x³=1. You can't just reverse an implication without giving a reason.
Going by what other comments have said, the 'substitution' is effectively the same as multiplying the whole thing by (x-1). Since (x-1) is now a factor, x=1 is now a root of the new equation. If you multiply both sides by (x+8) you would generate x=-8 as a new root. Basically, when you make the equation into a higher-order polynomial, you add roots to do so (or increase the order of a pre-existing root).
What you should really do is check that your solution to x^3=1 is a valid solution of the original equation first, which it isn't.
There is a difference between the first "=" and the followings. The first one is an equation and not an identity, which means it represents a condition, and not a general rule. On the substitution he assumes that x^2 + x + 1 = 0 for every x, but that's not true, it's just an equation and not an identity.
When bag at her tw and the Wi-Fi sucks! Lush her on at the ok her to go to sleep now hsfst 👍 good 👍👍 I mean what
you're right
My answer is basically the same but easy to understand. The last implication that 1=x^3 ==> x=1 is false. From the problem statement it is obvious that we can't exclude complex roots, and x^3=1 has three roots (fund. thm. of algebra). Asserting x=1 implies taking and using only the extraneous and real root, i.e. using the false assumption that x is purely real.
I could tell which step introduced the problem, I was reasoning that to use that substitution, you first needed to divide it by x like we did for the original equation.
You can not substitute form the same equation
if you use the same equation again and again you will end up with this.
our teacher used to say 2x2 could be 4, 5, 6, maximum 7
Also can be π
@@Nyo-ho π = 3.14
That's almost true (maximum is wrong). It depends on your number system and your unit. For example 2 persons who are mothers and 2 persons who are daughters can be 3 persons. i.e. 2x2 = 3 (grandmother, mother and granddaughter). Moreover, if the units are different we may have 2x2=96 (2x 2(days) = 96 (hours)).
@@JMehri Purrfect!!
The value of x is cube root of unity but it is a complex no. This can be spotted from the first equation as its discriminant is less than 0.
This value of x also satisfies x^3=1. That means the cubic equation is true but the root is not 1 , there are 2 roots which are complex.
The complex no. which is cube root of unity is famously known as omega
densch123 that complex no. is pretty easy to find. The solution of x^2+x+1=0 are the required to complex no.
These complex no.s are pretty famous btw
I spotted the error in the first step - since, we know that x^2 + x MUST be equal to -1 (by subtracting 1 from both sides of the equation). But, it CAN'T be equal to -1 because a negative number multiplied by a negative number will be POSITIVE. So, for ANY value of x, x squared will necessarily be positive. This basic fact renders the original expression in error.
That is true, there is no real solution. But the equation is still valid in complex numbers and has two answers.
X^2+X+1 not equal to 0 because any value of X (X= real number ) it gives a positive answer
Quickest answer i red on the comments
At a first glance on step 3, my inner engineer said "Well, dude! There got to have something completely insane here."
Yes! Though I don't think I have an engineer but more of a scientist. LIke... "HOw exactly did you end up there?" right?
I spotted the mistake: (theres no meaning for me to lie since it gives no benefit)
before he substitude x+1=-x^2
1/x should be changed to 1/x+1-1
so when he substitute it, 1/x becomes -(1/x^2+1)
so its -x^2-1/x^2+1=0
x^2+1/x^2+1=0
1=-(x^2)·(x^2+1)
I don't have time to simplify that but Ill edit it soon to show further proof. Idk if Im right but I will see after I watch the second part of this vid
edited:wow the vid didnt even give us the full proof just gave us the theriotical proof
I kind of identified it as the line where you wrote x^3=1 implies x=1... That’s where the error is. After all, the two solution you wrote, if you take their cubes, you will still get 1.
It’s like solving for time and we get t^2=1 ... we will obviously ignore t=-1.
But you got to the crux of it better than I did. So thanks for that.
Simple way to understand it : we use substitution for at least two equations. We put value obtained from one equation to the other equation. In this case, we have only one equation. Division of one equation through x doesn't make it a different equation. So basically in the whole process, we are working with one single equation. This equation can be written in different forms. We are taking value from one form and putting that value in the same equation but different form. We are thinking this is substitution. But it is not. In substitution we deal with two different equation. Here we are using one single equation but with different forms.