Waste less time on Facebook — follow Brilliant.

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]

Note by Aman Rajput
3 years, 4 months ago

No vote yet
3 votes


Sort by:

Top Newest

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 · 3 years, 4 months ago

Log in to reply

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 · 3 years, 4 months ago

Log in to reply

You are differentiating an equality that holds only for natural numbers ( and natural numbers are discontinuous on a graph). Shourya Pandey · 3 years, 4 months ago

Log in to reply

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 · 3 years, 4 months ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...