Here's my attempt at proving that
$2=1$
. In which step did I
**
first
**
commit a flaw in my logic?

$\begin{aligned} 1^2&=1\\ 2^2&=2+2&\text{(2 times)}\\ 3^2&=3+3+3&\text{(3 times)} \end{aligned}$

**
Step 1:
**
For any positive integer:

$x^2=x+x+\cdots+x\quad\quad\text{(x times)}$

**
Step 2:
**
Now, differentiating with respect to
$x$
:

$2x=1+1+\cdots+1\text{(x times)}.$

**
Step 3:
**
Summing this up, we get

$2x = x.$

**
Step 4:
**
Dividing by
$x$
, we get

$2 = 1.$

4
3
1
2

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(2)

The second step is violating the linearity principle of differentiation operator.The right side of the equation is dependent on variable x, so the diff cannot be applied to each individual x.