# A proof that $1=2$. May I know why it’s false? [duplicate]

im not good at formatting. May I know why it’s false

$x^2= x+x+x+\ldots(x\; \text{times})$

apply derivative on both sides

$=> \frac{d}{dx}(x^2)=\frac{d}{dx}(x+x+x+\ldots(x\; \text{times}))$

$=> 2x=1+1+1+\ldots(x\; \text{times})$

$=> 2x=x$

$=> 2=1$

• How do you make sense of $x+x+x+..(x\text{ times})$ when $x$ is not a whole number? – alex.jordan Apr 2 '13 at 4:59
• But the real problem is that your right-hand derivative has not considered the "$x$ times". The right-hand side depends on that; so the derivative should do something with it. – alex.jordan Apr 2 '13 at 5:00
• Duplicate of math.stackexchange.com/q/237817/264 – Zev Chonoles Apr 2 '13 at 5:01
• @ZevChonoles Thanks. – Pedro Tamaroff Apr 2 '13 at 5:02
• @ZevChonoles Thanks – Civa Apr 2 '13 at 5:03

You cannot add '$x$', $x$ number of times without knowing what $x$ exactly is. Hence the differentiation on the RHS has no real meaning.

This fails for many cases like:

$1)$ '$x$' is not an integer.

$2)$ Dependency of $x$ on $x$ which you are not considering when differentiating on RHS.

This has been asked twice or thrice already. Two problems might be pointed out

$(1)$ "$x$ times" is ill-defined for non integer values of $x$; and even so

$(2)$ "$x$ times" is also a function of $x$, which you're ignoring