My friend showed me 1=2 proof by using derivative and I am kind of stumped. Reddit r/calculus

We start with x=1+1+..+1, with x many 1s. Then we can say x^2=x+x+...+x, with x many x's. Then we differentiate both sides and get 2x=1+1+...+1. which implies 2x=x and then 2=1. No, it's not because x=0 and we have just divided by 0. There's an even BIGGER problem for this false 1=2 proof.

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[Mike__B]

usually most 1=0 or 2=1 type of problems I have seen involve either infinities and treating them as whole numbers or dividing by zero. This one with derivatives is a new one for me.

[BigDBrian]

why overcomplicate the setup?
start with x = 1+1+...+1 (x times)
differentiate both sides
1 = 0+0+...+0 (x times)
1 = 0
oh my god!!

[matthiasbergner]

1 = 2, proof: We are in the ring that contains only one element.

[alexpopblock]

isnt it just that from the very beginning, x=1+1+...+1 x times implies that x is a constant number and so squaring it is also constant, however you differentiate it later as if it was a variable and so you get a constant, not zero?

[cdkw]

Thank you for picking up my doubt sir, you are my role model and this means a lot to me.

[Roboticgladiator]

I think X is more like a constant in this case. It's also discontinuous.

[JESUS_CHRlST]

Absolutely true and also nice to just think of it as you can’t differentiate with respect to a constant

[chrisdaley]

We can also look at the formal derivative in some integer system(which is common in p-adic analysis for instance) and the first step is actually fine. The bigger problem is the differentiating step since we're differentiating the individual x's and ignoring the fact that we're also working on a sum with x terms. The derivative distributes over finite linear combinations but not necessarily variable sized linear combinations.

[michalchik]

I don't think that this is the real reason why there's a problem though it's close to it. The real reason why it's a problem is that you go from one side which is a variable that has a rate of change to the other side where the X's have been erroneously replaced by constants so the derivative doesn't reflect the fact that x has to change. It's a bit like on the right side we're calling one of the X's a constant and then taking the derivative of the constant. Another way to look at it is that we lose the product rule because we lose one of the variables.

[fangliren]

This is, essentially, the difference between the functions f(x) = kx for some fixed constant k, and f(x) = x^2, when we examine both at the value x=k

[laszloliptak]

The big issue here is that when you take the derivative of the sum, it is equal to the sum of the derivatives only for fixed number of terms, while here the number of terms varies.

[Sam_]

I am the friend. Btw I have Great respect for you and you make really interesting and mind provoking videos

[davidw]

Suppose we generalized this by starting with X=1+1+1...+1+R, where R is the fractional part of X. Then we would have a continuous function over the real numbers. However, when you differentiate, you have to account for the fact that R is a function of X. Where R is differentiable, dR/dX = X. So that's where the other X comes from.

[briansheldon]

Interestingly if you incorporate non-integers into the proof we recover 2=2 instead of 2=1.

Here is how you can do it:
1. Instead of writing x as 1+1+...+1 x times we'll write it as 1+1+...+1 {floor(x) times} plus an additional term of x-mod-1 (for example 20.4 would be written as 1+1+...+1 {20 times} plus 20.4-mod-1 which correctly yields 20.4)
2. Next you can plug it in to get
x^2 = x*x
x^2 = x*(1+1+...+1 {floor(x) times} + x-mod-1)
x^2 = x+x+...+x {floor(x) times} + x*(x-mod-1)
3. Now we differentiate both sides (note that d/dx of x-mod-1 is 1)
(d/dx)x^2 = (d/dx)(x+x+...+x {floor(x) times} + x*(x-mod-1))
2x = 1+1+...+1 {floor(x) times} + (d/dx)(x*(x-mod-1))
2x = 1+1+...+1 {floor(x) times} + x-mod-1 + x*(d/dx(x-mod-1))
2x = 1+1+...+1 {floor(x) times} + x-mod-1 + x
4. Then we remember that 1+1+...+1 {floor(x) times} + x-mod-1 is just x to finally yield:
2x = x + x
2 = 2

(I definitely have issues with the differentiation step since it treats the number of terms as constant as opposed to a function of x but it is interesting that only changing x's representation to include non-integers restores some truth to the proof, whether on accident or not)

[davidsousaRJ]

I think the first step is not wrong, if you tell since the beginning that x is some positive integer. In that case, the derivative step will be wrong, because derivatives operate on functions, and "x" and "x²" are not functions, they are positive integers.

[InvaderMixo]

The first line isn't wrong; it is given. You cannot use the derivative operator after the first line, since the first line forces X to be a positive integer, where every positive integer N can be written as sum_i=1^i=N 1.

[pglink]

Let x = 1.
Take the derivative with respect to x on both sides.
1 = 0.
Add 1 to both sides.
2 = 1.
Easy.

[secret]

1:54 The error is here, because the limits of your sum are dependent on x, so it's not a simple derivative as shown here.

[xshortguy]

the derivative of "adding something x times" is "adding 1 x times, plus adding x once". It's just the product rule. Of course there's the other difficulties presented here as well.

[arkadiosvotessocialist]

I feel like the problem here starts from first concepts: Calculus is not in itself about numbers—it's about functions, and how they change. Here, X is a constant of unknown value, not a function definition, so taking its derivative just gives a zero on both sides.