# Nice proof (1 Viewer)

#### no_arg

##### Member
Let $\bg_white x>0.$.
$\bg_white x^2=x+x+\dots+x$ $\bg_white \qquad$ where the sum on the right has $\bg_white x$ terms.
For example if $\bg_white x=4$ we have $\bg_white 4^2=4+4+4+4.$
Differentiating $\bg_white \qquad$ $\bg_white x^2=x+x+\dots+x$ $\bg_white \qquad$ we have
$\bg_white 2x=1+1+\dots 1$ $\bg_white \qquad$ where the sum on the right still has $\bg_white x$ terms.
Thus $\bg_white 2x=x$ and hence $\bg_white 2=1$.

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#### Drdusk

##### Moderator
Moderator
Let $\bg_white x>0.$.
$\bg_white x^2=x+x+\dots+x$ $\bg_white \qquad$ where the sum on the right has $\bg_white x$ terms.
For example if $\bg_white x=4$ we have $\bg_white 4^2=4+4+4+4.$
Differentiating $\bg_white \qquad$ $\bg_white x^2=x+x+\dots+x$ $\bg_white \qquad$ we have
$\bg_white 2x=1+1+\dots 1$ $\bg_white \qquad$ where the sum on the right still has $\bg_white x$ terms.
Thus $\bg_white 2x=x$ and hence $\bg_white 2=1$.
$\bg_white \text{I believe it's because you can't take the derivative of}\hspace{2mm}x+x+...+x \\ \text{with respect to x because it's summed 'x' times?}$

#### blyatman

##### Well-Known Member
Every step up to and including $\bg_white 2x=x$ seems fine. However, you can't simply divide by $\bg_white x$ much like how you can't divide both sides of $\bg_white x^2=x$ by $\bg_white x$. Doing so is equivalent to dividing by 0, as $\bg_white x=0$ is a solution. The correct way to solve it would be to subtract $\bg_white x$ from both sides, which leaves the solution $\bg_white x=0$. Thus, the only way for a solution to exist is if $\bg_white x\geq0$ (i.e. to include 0 in the domain).

In regards to why this is the case, I'm still trying to get my head around it. One would think that the derivative of both sides would be the same for all $\bg_white x\geq0$ since $\bg_white x^2=x+x+\cdots+x$. However, this definition would start to get a bit iffy for non-integer values of $\bg_white x$. But then that just raises another question: why would this only be true for $\bg_white x=0$, and not all integer $\bg_white x$ values? Regardless, you can't get $\bg_white 2=1$ since the step you took (i.e. dividing by 0) would result in the end of the universe as we know it.

@Drdusk I'm also thinking about your point of having $\bg_white x$ terms. On the surface, this seemed OK to me, but now I'm questioning myself whether this is indeed allowed, since the result of $\bg_white x=0$ seems rather nonsensical.

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#### Carl10101

##### Member
hence
$image=https://latex.codecogs.com/png.latex?\bg_white+2=1&hash=66b765abde49548d8c4576a39296c51e$
.
oh ok

#### Trebla

The main flaw is that the number of terms is also a variable whereas the additivity of differentiation applies when the number of terms is fixed. A similar example is writing:

x = 1+1+1+....+1 (x-times)

Leading to the 1 = 0 fallacy.

Another flaw is that this decomposition assumes x is an integer so obviously there are complications with limits/continuity etc.

#### Drdusk

##### Moderator
Moderator
The main flaw is that the number of terms is also a variable whereas the additivity of differentiation applies when the number of terms is fixed. A similar example is writing:

x = 1+1+1+....+1 (x-times)

Leading to the 1 = 0 fallacy.

Another flaw is that this decomposition assumes x is an integer so obviously there are complications with limits/continuity etc.
Yeah but if you just look at the equation x+x+....+x (x times) then how can you mathematically tell that you cannot do that. Sure you can factorize it and what not to show it but that equation alone doesn't show the flaw which is weird.

#### ultra908

##### Active Member
Yeah but if you just look at the equation x+x+....+x (x times) then how can you mathematically tell that you cannot do that. Sure you can factorize it and what not to show it but that equation alone doesn't show the flaw which is weird.
I guess its like trying to do d/dx (x^x) as x*x^(x-1). Bcos "x times" is also varying, you can't just diff it normally.

#### Trebla

Yeah but if you just look at the equation x+x+....+x (x times) then how can you mathematically tell that you cannot do that. Sure you can factorize it and what not to show it but that equation alone doesn't show the flaw which is weird.
The decomposition step is perfectly valid for integer values of x.

The invalid step is the differentiation step because it is not recognising that the number of terms is also a variable.

#### stupid_girl

##### Active Member
The decomposition step is perfectly valid for integer values of x.

The invalid step is the differentiation step because it is not recognising that the number of terms is also a variable.
If the decomposition step is only valid for positive integer values of x, then it's simply not continuous and therefore not differentiatable.

#### InteGrand

##### Well-Known Member
$\bg_white \noindent It also works for non-integer but \textbf{constant} multiples of x, e.g.$

\bg_white \begin{align*} \frac{d}{dx}(\pi x) &= \frac{d}{dx}(\underbrace{x + x + x + \cdots + x}_{\pi \text{ times}}) \\ &= \underbrace{ \frac{d}{dx}(x) + \frac{d}{dx}(x) + \frac{d}{dx}(x) + \cdots + \frac{d}{dx}(x)}_{\pi \text{ times}}\\ &= \underbrace{1 + 1 + 1 + \cdots + 1}_{\pi \text{ times}} \\ &= \pi. \,\checkmark \end{align*}

$\bg_white \noindent You can replace \pi with any real constant a.$

$\bg_white \noindent It only fails when the number of terms is variable.$

#### no_arg

##### Member
$\bg_white \noindent It also works for non-integer but \textbf{constant} multiples of x, e.g.$

\bg_white \begin{align*} \frac{d}{dx}(\pi x) &= \frac{d}{dx}(\underbrace{x + x + x + \cdots + x}_{\pi \text{ times}}) \\ &= \underbrace{ \frac{d}{dx}(x) + \frac{d}{dx}(x) + \frac{d}{dx}(x) + \cdots + \frac{d}{dx}(x)}_{\pi \text{ times}}\\ &= \underbrace{1 + 1 + 1 + \cdots + 1}_{\pi \text{ times}} \\ &= \pi. \,\checkmark \end{align*}

$\bg_white \noindent You can replace \pi with any real constant a.$

$\bg_white \noindent It only fails when the number of terms is variable.$
I think you can have x pies but you can't have pi x's!

#### ultra908

##### Active Member
I think you can have x pies but you can't have pi x's!
aren't they the same?

Moderator

oh lol pies yum

#### dan964

##### what
Try differentiating the RHS by first principles:
\bg_white \begin{align*}\frac{dy}{dx} &= \lim_{h \rightarrow 0}{\frac{1}{h}\left(\underbrace{(x+h)+(x+h)+\dots+(x+h)}_{(x+h) \text{ times}} - \underbrace{x+x+\dots+x}_{x \text{ times}} \right)} \\ &= \lim_{h \rightarrow 0}{\frac{1}{h}\left(\underbrace{h+h+\dots+h}_{x+h\text{ times}}+ \underbrace{x+x+\dots+x}_{h \text{ times}} \right)} \\ &= \lim_{h \rightarrow 0}{\left(\underbrace{1+1+\dots+1}_{x+h \text{ times}}+ x \right)} \\ &= \lim_{h \rightarrow 0}{2x+h} \\ &= 2x \end{align*}

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#### jyu

##### Member
False statement (the sum on the right has x terms), because x is not necessarily a natural number.

If x are natural numbers, x^2 is not differentiable.

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