Is it a fallacy ??

Recently, I was doodling in my rough notebook and found something strange:

Let x belong to N (natural no.)

$x^{2}$ = $x^{2}$

$\Rightarrow$ $x^{2}$ = $x+x+x+x+x+x....(x times)$

(Differentiating on both sides)

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

$\Rightarrow 2x = x$

Where was I wrong ??

#MathematicalFallacies

$</code> ... <code>$ $< / c o d e > . . . < c o d e >$ ."> Easy Math Editor

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> 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"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ 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}`	$\boxed{123}$

Comments

Sort by:

Top Newest

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.

Kuldeep Guha Mazumder - 4 years ago

Actually you cannot differentiate the function f(x) = $x^{2}$ 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.

A Brilliant Member - 4 years, 2 months ago

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.

Ashwin Upadhyay - 4 years, 2 months ago

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 Brilliant Member - 4 years, 2 months ago

Very good. He is fooling.

Harendran Bala - 4 years, 2 months ago

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

Vihari Vemuri - 4 years, 2 months ago

I couldn't understand your first reason.

Ashwin Upadhyay - 4 years, 2 months ago

Excel in math and science

Master concepts by solving fun, challenging problems.

It's hard to learn from lectures and videos

Learn more effectively through short, interactive explorations.

Used and loved by over 7 million people

Learn from a vibrant community of students and enthusiasts, including olympiad champions, researchers, and professionals.

Is it a fallacy ??

Comments

Learn from a vibrant community of students and enthusiasts,
including olympiad champions, researchers, and professionals.