'Prove' 3 = 0. Can You Spot The Mistake?

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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[nilswendland]

it would be more interesting to explain why this substitution leads to the extra solution

[grandexandi]

3 is close enough to 0, it can be rounded

[rjkuo]

Guy: That will be $500
Me: Let me show you how $500 is $0

[beebit_]

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).

[Dariolo]

"Prove" 3 = 0. Can You Spot The Mistake?

Yes, i can
3 not equal to 0

You're welcome

[victorvanderdrift]

Engineers be like: '3 = 0'.
"Well that's close enough"

[Coole]

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

[josegeorge]

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!

[roboticreaper]

I can prove anything=anything.
First, multiply both side by 0.
Done

[somxr_]

I looked at the thumbnail for 5 minutes to find a mistake

[fyukfy]

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

[Andryu_d]

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

[rasmusvanwerkhoven]

“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

[vousvxyez]

MindYourDecision: "Prove" 3 = 0. Can You Spot The Mistake?
Me: Oh that's easy, that's me. Problem solved.

[xiaohuwang]

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.

[vimlachoudhary]

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.

[davidbreese]

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)

[tsaan]

"x^2+x+1=0"
me: ey wait a sec thats impossible
this dude: ok so here are complex numbers

[chinmaymathur]

When you subsstuted x+1 as -x², you should also have done 1/x as 1/(-1-x²) so to get right answer

[matsbd]

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.