'Prove' 2 = 1 Using Calculus Derivatives - Can You Spot The Mistake?

This is a false proof of why 2 = 1 using a bit of differential calculus. Can you figure out where the mistake is?

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[NoamMohr]

You cannot write x^2 = x + x + x +... (x times) unless x is an integer. And if you restrict x to integers, you do not have a continuous function and cannot take a derivative. More generally, since the number of terms changes as x changes, when you take the derivative (rate of change), you have to take the derivative of both all those individual terms and the parenthetical (x times), since that is also changing with the value of x.

[ceceliapassarella]

Two equals one in any good relationship

[GrxyFn]

I found the mistake







2 doesn’t = 1

[shubhammrawatt]

then I can prove 2=0
2x=1+1+1...1
again differentiating
2=0
hence proved😂😂😂

[PinguinCrazy]

The real challenge here is looking at the description apparently.

[iurigrang]

actually, the problem is that the equation on the top only holds to x natural, you can't take a derivative of a function of naturals.

[thetrueleviticus]

i asked my teacher

turns out she watches you aswell lol

[dhidhi]

Ok, here is what I got:

1. x^2 = x + x + ... + x (x times).

Say the sum on the right hand side has n term (hence, we know n = x). It becomes:
2. x^2 = nx

Taking the derivative:
3. 2x = n
4. 2x = x (because x = n).
5. 2 = 1

Where is the mistake? 
Well, when we go from (2) to (3). That's because n is NOT a constant. n varies as x varies. Therefore, (2) is equivalente to:
x^2 = (x)x

Using the chain rule:
2x = 2x

The mistake is that we cannot evaluate
d(nx)/dx = n
if n is dependent of x.

[venumanikanta]

x^2 = will not hold for negative integers

[matthewhowey]

I found this really interesting, it challenges us to make sure that we are understanding calculus properly and not just using the formulas mechanically. When we talk about the derivative of x^2, we are talking about the RATE OF CHANGE of x^2 with respect to x. The key thing is that on the right side of the equation, when x changes it is not just the value of each member of the sum which changes. It is the total number of x's which changes AS WELL. When he gets the derivate result 'x', it is as if he changes the value of each 'x' in the sum itself, (..x+x+x..),   but keeps the x in the '...x times' constant. E.g. when x = 5, he would be taking the derivative of 5x not x^2. And so the derivative of 5x would be 5, the same value as x in this case. (And yes it is impossible to take a derivative when we restrict ourselves to the natural numbers,  because it would be impossible to take the limit of delta-x as it approaches 0; but I think it is possible to see the flaw in the proof without mentioning this.) I'm not sure I found his explanation of the flaw very satisfying; he shows a different way of getting the right answer but doesnt' really explain the flaw in the x+x+x+... method.

[liamskeen]

The video is exactly 4:20. Just wanted you all to know.

[gnarlydyn]

Only when x is integer can we see x^2 same as x+x+...+x (x times), so in the equation two functions y=x^2 and y=x+x+...+x (x times) are defined only in a set of integer and we can't simply take the derivative and compare.

...right?

[ilcyclista]

The equation x^2 = x + x + x... x times obscures the fact that the right side secretly has a function of x within it. We can replace the sum of x's as nx, where n is the value of x. By taking the derivative and setting 2x = n, we are ignoring the chain rule, as the number of x's on the right depends on the value of x. Instead it should be 2x = n'(x)*x + n(x). But n(x) = x and n'(x) = 1, so: 2x = 1*x + x, 2x = x + x, 2x = 2x.

Another way you can think of it is seeing the right hand side as lim n-> x sum( i=1 to n) of x. A bit of algebra shows that the right side is then: x (lim n->x sum(i=1 to n) 1 ), we compute the sum and find: x (lim n->x  n), take the derivative and we get:1*(lim n->x n) +x*(lim n->x n)'. We then take the limit and find:x + x*(x)' = x + x = 2x.

[blazingfire]

Yeah I'll come back when I've learned what a derivative is.

[leunghowah]

The mistake simply happened when he takes the derivative. Derivative looks at the change of the equation with respect to some variable x. So the trick is that "(x times)" on the rhs would be affected when x is changed.

[michaelhazellful]

Power rule isnt generally used for basic addition problems. Power rule is used for multiples and to return back to the parent function. If we are doing x^2, we must also verify it is x * x or x(x). Derivative of x^2 is 2x and obviously it would be 2x on the other side. Power rule is simply bringing down the exponent(n) and after that doing n-1

[rituchandra]

you cannot write x*x = x+x+x...x an x number of times because that is true only for Natural numbers, u cannot say that for real numbers. for eg. sqrt(2)*sqrt(2) = sqrt(2)+sqrt(2)... a sqrt(2) number of times?? no... that doesn't make sense! and if x*x=x+x+x...x then u cant differentiate the function since functions of natural numbers are discrete, not differentiable.

[Simoendi]

take care when you use derivatives. becaus a funktion can only be derived if the funktion is continous, but as you setup x^2 = x + x + .... + x (x times) -> x has to be a integer, and so it isnt continous...

[rosebuster]

Yes, the number of terms you add has to be a constant to make it okay to take derivatives of each term separately. It certainly cannot depend on the variable used in the differentiation.

[ayushmishra]

you can not differentiate to {each x times) as the number of terms is also a variable
.
.
.
.
.
.
and on differentiating the rate of change in that variable will be neglected