Forgot password? New user? Sign up
Existing user? Log in
Recently, I was doodling in my rough notebook and found something strange:
Let x belong to N (natural no.)
\(x^{2}\) = \(x^{2}\)
⇒\Rightarrow⇒ x2x^{2}x2 = x+x+x+x+x+x....(xtimes) x+x+x+x+x+x....(x times)x+x+x+x+x+x....(xtimes)
(Differentiating on both sides)
2x2x2x = 1+1+1+1+1+1+1....(xtimes) 1+1+1+1+1+1+1....(x times)1+1+1+1+1+1+1....(xtimes)
⇒2x=x\Rightarrow 2x = x ⇒2x=x
Where was I wrong ??
Note by Ashwin Upadhyay 5 years, 10 months ago
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
_italics_
**bold**
__bold__
- bulleted- list
1. numbered2. list
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"
\(
\)
\[
\]
2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Sort by:
When you are defining x^2=x+x+..+x(x times), you are tacitly defining the domain of f(x) to be N, the set of natural numbers. However, f(x) cannot be differentiable in the domain of N. This is because, if a function f defined from a domain D (a subset of R, the set of reals) to R (the set of reals) has to be differentiable at a real point c, then a necessary criterion is that c has to be a limit point of D and c has to be an element of D itself. In other words, c has to be a member of D such that every arbitrarily small neighbourhood of c has an element of D other than c. But no natural number is a cluster point of the set N. Here lies the fallacy.
Log in to reply
Actually you cannot differentiate the function f(x) = x2x^{2}x2 when you have selected the Domain as Natural numbers . After all , the Natural numbers as a domain will comprise just discrete points , so it'll not be differentiable.
Firstly I differentiated it generally but in that case i couldn't write the no. x times as x could be fraction or 0. Thank you now i have understood.
Anytime . Btw there was a question on this fallacy posted by @Sandeep Bhardwaj sir , I'm not able to find it . I'll give you the link if I'm able to find it .
@A Former Brilliant Member – BTW i just found another note on this
@A Former Brilliant Member – In fact i found a whole set of such apparent fallacies
@Ashwin Upadhyay – Wow . You've been busy :P
Very good. He is fooling.
When you are differentiating on both sides, you assumed that in x times the x is constant. Also you can't talk abt differention, of non continuous function
I couldn't understand your first reason.
Problem Loading...
Note Loading...
Set Loading...
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Sort by:
Top NewestWhen you are defining x^2=x+x+..+x(x times), you are tacitly defining the domain of f(x) to be N, the set of natural numbers. However, f(x) cannot be differentiable in the domain of N. This is because, if a function f defined from a domain D (a subset of R, the set of reals) to R (the set of reals) has to be differentiable at a real point c, then a necessary criterion is that c has to be a limit point of D and c has to be an element of D itself. In other words, c has to be a member of D such that every arbitrarily small neighbourhood of c has an element of D other than c. But no natural number is a cluster point of the set N. Here lies the fallacy.
Log in to reply
Actually you cannot differentiate the function f(x) = x2 when you have selected the Domain as Natural numbers . After all , the Natural numbers as a domain will comprise just discrete points , so it'll not be differentiable.
Log in to reply
Firstly I differentiated it generally but in that case i couldn't write the no. x times as x could be fraction or 0. Thank you now i have understood.
Log in to reply
Anytime . Btw there was a question on this fallacy posted by @Sandeep Bhardwaj sir , I'm not able to find it . I'll give you the link if I'm able to find it .
Log in to reply
note on this
BTW i just found anotherLog in to reply
set of such apparent fallacies
In fact i found a wholeLog in to reply
Log in to reply
Very good. He is fooling.
Log in to reply
When you are differentiating on both sides, you assumed that in x times the x is constant. Also you can't talk abt differention, of non continuous function
Log in to reply
I couldn't understand your first reason.
Log in to reply