A second proof of 1=2

Mr. John Hush presents a second proof of 1=2 for those who doubted the first one. He noticed that a couple viewers proposed this one so he wishes to share it with everyone. What's wrong with this proof, if anything?!

[sudarshanseshadri]

As an engineering student, I can say that this is a good enough approximation for me.

[skinindagame_]

"Sir, here is your bill..."
"Let me explain why $200 equals $1, Ma'am."

[Ghostly]

This man could prove that the earth is flat

[anantveerbubbles]

0:32 he was so proud of himself for coming up with that 😂😂

[ninotorres]

This is the video everyone will get in their recommended in 5 years
But we got it now
We are... THE CHOSEN ONES

[unknown_guy]

Let’s just hope the Youtube algorithm wont take another 10 years before recommending this video to others

[davidzhu]

I'll try a shot at explaining why this "proof" is incorrect.

I believe that the error lies in the step when you rewrite x^2 as x + x + x ... + x, with x terms. You wrote out some examples: 3^2 = 3 + 3 + 3, and it works, clearly. However, what you failed to address was the cases when x is not an integer. You wrote the expression x + x + ... + x with x terms, but how do you have 2.5 terms of value 2.5 each, or pi terms of pi, or some random irrational number of terms that can't even be expressed in a finite way? You can't, it just doesn't make logical sense. That expression works for only positive integer values of x.

Ok then, let's suppose we take that function with integer x only. Doesn't the rest of the proof hold, that 1 = 2 "when x is an integer" (whatever that means)? No, not really.

The limit definition of a derivative is this: the derivative of a function f at a constant c is equal to the limit as h approaches zero of this expression: (f(c + h) - f(c)) / h.
Now, if c is an integer, which it has to be because that's the only places where the x + x + x +... + x is defined, then that clearly means that c + h is not an integer, because while h is very very small, it's important to note that h is not zero.

Therefore, f(c + h) is not defined, meaning the derivative is not defined, and therefore, you cannot take the derivative of the x + x + ... + x expression.

[usemymemefornoreason]

Next: prove π=0 and eradicate the concept of circles

[davestrrr]

fault: you assume x is discrete, but derivative assumes continuous variable

[natty]

It's 4am and I'm watching a proof that 1 equals 2 instead of doing my math homework

[Elharion]

If you want to know why this is wrong, here it is:
If you have x x’s, and you want to find the slope, you need to consider two things: the value of each x increasing, and the number of x’s increasing. x has a slope of 1, so you do get x for part of it, but for the number of x’s, the number of x’s increases by one for every x+1, so that also gives you an x because the value of that x is 1. So if you add x+x you get 2x. The problem with this proof is that you’re assuming the number of x’s in the set is constant when in fact it is increasing. The dy/dx value of x doesn’t account for that.

[Jake-khpc]

Claim your "here before this go viral"

[thongtranmy]

He is a man of focus, commitment and sheer fucking will

[TheYellowDK]

Hello again! Like the first video, I will once again try to explain the mathematically "correct" reason why this proof is invalid. Like before, this explanation will be very long and detailed. I will need to explain a lot of calculus concepts just to be able to explain the mistake. so thank you for sticking with me. Hope y'all enjoy! :)

To explain why this proof is invalid we have to explain the rigorous definition for derivatives. Two ways of writting it are:
1) f'(a) = lim(b->a) [(f(b) - f(a))/(b - a)]
or
2) f'(x) = lim(h->0) [f(x+h) - f(x)]/h

What does this mean? Well the slope between two seperate points is given by rise over run. On an x-by-y graph thats the difference in y (vertical change) divided by the difference in x (horizontal change). For two points (x1, y1) and (x2, y2) that slope would equal (y2 - y1)/(x2 - x1).

If we swap in standard function notation, y=f(x), and label our points (a, f(a)) and (b, f(b)), then that looks like (f(b) - f(a))/(b - a). Still just the change in f(x) values divided by change in x values. Hopefully this is intuitive enough.

But taking the derivative of a function is looking at the slopes, not between two seperate points, but at a single point. So when (a, f(a)) is the same as (b, f(b)), or in other words, a=b. This obviously creates a problem with our traditional equation for slope because if the two points are the same we get (f(a) - f(a))/(a - a) = 0/0, indeterminate form.

This is why our definition above uses limits. Limits help us simplify indeterminate form (I'll explain why see later). We're basically looking at limit of the slope as the two points get closer and closer to being the same point. In common language, the definition above (#1) is saying that derivative of the function, f, at point a equals the limit of the slope between point b and and point a, as point b gets closer and closer to point a.

The second definition is a more succinct way of writing the same thing. Write a = x and b = x+h, and now instead of taking the limit of the slopes between a & b as b->a, you take the limit of slopes between x+h & x as h->0 (h is the difference between the two points, so h->0 is the same thing as the points becoming the closer to the same). Same idea, different ways of writing it. But I'll be using the second definition from now on because its easy to work with.

So Let's see how this works with f(x) = x^2:
f'(x) = lim(h->0) [(x+h)^2 - x^2]/h = lim(h->0) [(x^2 + 2xh + h^2) - x^2]/h
The x^2 and -x^2 cancel out so that gives us:
f'(x) = lim(h->0) [2xh + h^2]/h

Here's where limits come in. (2xh + h^2)/h when h=0 is indeterminate form. If h=0 then you get 0/0. The temptation might be to cancel the h's on top and bottom so that (2xh + h^2)/h = 2x + h.
But you can't do that because h=0 and you're not allowed to divide by 0.

But with our definition, we're not looking the value for h=0, we're looking at limit as h approaches 0. In other words, we're not looking at the actually value of the slope when h exactly equals 0 (which is indeterminate), we're looking at how the function behaves as h gets closer and closer to 0. And here we can cancel out the h's on the top and bottom.

Why can we do this? Well, it's true that (2xh + h^2)/h does not equal 2x + h when h=0 (first one equals 0/0, second equals 2x, there not the same). But the limit of (2xh + h^2)/h as h->0 does indeed equal the limit of 2x + h as h->0 because the two funtions behave the exact same way.

[To make sense of why this is true, here's a different example. Consider the expressions 1 and x/x. Are they exactly equal? Does x/x = 1 exaclty? Well no, because when x=0, then x/x = 0/0 (indeterminate form) and therefore does not 1. But for every other value of x, then x/x does equal 1. And furthermore the behavior of x/x as x->0 is the still same as the behavior of 1 as x->0, so lim(x->0) x/x = lim(x->0) 1.]

So bring this all back to our f(x) = x^2 derivative:
f'(x) = lim(h->0) [2xh + h^2]/h = lim(h->0) 2x + h
And as h gets closer and closer to 0, 2x + h gets closer and closer to 2x. So:
lim(h->0) 2x + h = 2x.
And voila! f'(x) = 2x.
The derivative of x^2 is 2x (shown as rigourously as possible).

Now that we know how to calculate derivatives rigorously, lets do the derivative for f(x) = (x+x+...+x) {x times} with the definition f'(x) = lim(h->0) [f(x+h) - f(x)]/h.

f'(x) = lim(h->0) [(x+h)+(x+h)+...+(x+h) {x+h times} - (x+x+...+x) {x times}] / h
Let separate and combine the x terms and h terms (including distributing the subtraction):
f'(x) = lim(h->0) [(x+x+...+x) {x+h times} - x-x-...-x {x times} + (h+h+...+h) {x+h times}] / h

He have x+h many positive x's and x many negative x's, so if we simplify we should get h many positive x's.
In other words (x+x+...+x) {x+h times} - x-x-...-x {x times} = (x+x+...+x) {h times} So:

f'(x) = lim(h->0) [(x+x+...+x) {h times} + (h+h+...+h) {x+h times}] / h

Now we know by the rules of multiplication that x*h = (x+x+...+x) {h times}. And so by the same logic (and because multiplication is commutative) then x*h also = (h+h+...+h) {x times}
So if we substitute (h+h+...+h) {x times} for (x+x+...+x) {h times}

f'(x) = lim(h->0) [(h+h+...+h) {x times} + (h+h+...+h) {x+h times}] / h
if we put all our h's together we get
f'(x) = lim(h->0) [(h+h+...+h) {2x+h times}] / h
Cancel h's on top and bottom to get
f'(x) = lim(h->0) (1+1+...+1) {2x+h times}
As h -> 0, 2x+h -> 2x, so:
f'(x) = (1+1+...+1) {2x times}
And of course 2x many 1's added together equals 2x, so:
f'(x) = 2x

So it doesn't matter if you use f(x) = x^2 or f(x) = (x+x+...+x) {x times}, the result (when calculated rigourously) is the same: f'(x) = 2x

So where does this "proof" go wrong? There's a difference between adding up series of x's a constant number of times versus a variable number of times. When he differentiates over the (x+x+...+x) {x times} in the video, he's derivating over the terms but not over the number of terms.

What does that mean? Well, consider the equation f(x) = 100x. This can be written as f(x) = (x+x+...+x) {100 times}. And in this situation it's true that f'(x) = (1+1+..+1) {100 times} = 100. This is because as x changes the number of additions stays the same, so you only have to differentiatee over the terms.
But when you're adding up a variable number of x's, as in x many x's, this does work not the same way.
For the function f(x) = (x+x+...+x) {x times}, as x increases you're increasing both the value of terms being added and number of the terms being added, and you have to differentiate over both.

To explain, remeber what the slope of a function is supposed to mean. It's the rise over run. It's the ratio you increase the output (y, f(x)) for every increase of input (x). So if the slope is 100 that means as you increase x by 1, you increase y by 100.
So if you go from x=3 to x=4, you go from y = (3+3+...+3) {100 times} = 300 to y = (4+4+..+4) {100 times} = 400.
If you go from say x=100 to x=101, you go from (100+100+...+100) {100 times} = 10, 000 to (101+101+...+101) {100 times} = 10, 100.

More generally, if you go from x to x+1, you go from (x+x+...+x) {100 times} to (x+1)+(x+1)+...+(x+1) {100 times}, which can be alternatively written as (x+x+..+x) {100 times} + (1+1+...+1) {100 times}. So here the difference, the increase in y, is (1+1+...+1) {100 times} or 100.

Here the increase comes from the terms themselves. But for the function f(x) = (x+x+...+x) {x times}, the increase comes from both the terms themselves and the number of terms.

here, if you go from x=100 to x=101, you go from (100+100+...+100) {100 times} to (101+101+...+101) {101 times}, which can be thought of as (101+101+...+101) {100 times} + 101.
Here you can see you're not just increasing the value from 100 to 101, but you're also adding a whole addition 101 on the end.

Generally, if you go from x to x+1, you go from (x+x+...+x) {x times} to (x+1)+(x+1)+...+(x+1) {x+1 times}. Which can be alternatively written as (x+x+...+x) {x+1 times} + (1+1+...+1) {x+1 times}
And further more simplified to (x+x+...+x) {x times} + x + (1+1+...+1) {x times} + 1. Which is the same thing as x^2 + x + x +1, or maybe more familiarly x^2+2x+1, the standard expansion of (x+1)^2

So that's the trick. He's not using the rigorous definition of derivatives, he's using a shortcut. It's a shortcut that usually works fine, so it seems like it should be correct. But this time it ignores the fact that you also have to derivate over both the terms themselves and the number of terms.

TL;DR - The method he uses to get the differentiate for x+x+...+x {x times} differentiates over the x terms themselves, but not over the number terms, which in this case is also a variable and needs to be differentiated over too.

[morganga]

Another way of putting it: he has persuaded you that one of the xs in x * x is a constant, then differentiated, then persuaded you the constant is the variable x again.

[rahulthapamagar]

2021 will no longer suprise me
Proof of 1 = 2

[aayush]

I can’t wait this to go viral and gets millions of views 🤩

[markmetalen]

This is what is obtained whenever mathematical defintions aren't followed: derivatives aren't properly defined on number sequences so this 'proof' doesn't hold for all real values of x. Taking a limit that approaches a certain value isn't possible in an infinitely small interval around an integer, to which the proof applies, because integers aren't elements from a continuous set (i.e. there is no such thing as an interval defined around a pure integer). In order to define derivatives, taking limits need to be possible. Also, just try to write the righthand side (stating that x*x=x+x+x....+x, x times) down for an example fractional value like x=23/72 or for a transcendental value like pi, that's simply impossible. This means that the second part of the so called proof is clearly incorrect.

[zapking]

Okay my guess as to why this is not correct is that you can’t keep the “x amount of times” part of x+x+x… x amount of times when differentiating. This is because the amount of times you add x to itself also changes with respect to x, and just saying “oh now it’s 1+1+1… x amount of times” doesn’t take into consideration that change. It’s the same reason that the derivative of x*lnx isn’t lnx, 1, or 1/x. It’s because those answers don’t take into account the change in some expression with respect to x (for the first answer, it neglects the change in lnx, the second answer neglects the change in x, and the third answer neglects the fact that their product changes at a different rate than the two individual functions).

[Kuro-pvly]

if 1 = 2, and 2 = 1 + 1, so 2 = 4, bc 4 is 1 + 1 + 1 + 1, but 1 = 2, oh no, this will never end