Find mistake :D

we know that

1 = 1^2

2 + 2 = 2^2

3 + 3 + 3 = 3^2

similarly , x + x + x + x + ..(x times)..... + x = x^2 (assuming x not equla to zero)

now , differentiating both the sides..

1 + 1 + 1 + 1 + ..(x times)...... + 1 = 2x

implies that x = 2x or 1=2 (dividing by x , because x not equal to zero)

have funn :D :D

[I removed "Calvin" from the title as it was not appropriate - Calvin]

Easy Math Editor

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Math	Appears as
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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
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Comments

the definition of x^2 being x+x+...(x times) is valid for natural numbers consider 1.25 how do you apply your definition there (how can you add something 1.25 times)

since your definition is not valid for all rational and irrational it is not a continuous function.hence you cant even differentiate it.

pranav chakravarthy - 8 years ago

x is a variable. When you differentiate with respect to x are you not assuming that x is constant? You say the sum continues x times, so when you differentiate you need to take that into account as well. According to me, your mistake is that while differentiating you took x as some fixed value, which is incorrect.

Aditya Parson - 8 years ago

The trick that u have generalised is for a pure constant which on differentiating gives zero both on LHS and RHS and the equation satisfies but not for any variable like x , So the wrong is taking a variable like x...

Sitesh Pattanaik - 7 years, 12 months ago

You are differentiating an equality that holds only for natural numbers ( and natural numbers are discontinuous on a graph).

Shourya Pandey - 7 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$