Waste less time on Facebook — follow Brilliant.
×

Differentiation show 2=1?

Given that \(6\times6 = 6+6+6+6+6+6\), or equivalently
\(x\cdot x = x + x + x +x +x+x...\), where \(x= 6\).
Differentiate both sides with respect to \(x\), we get
\( \dfrac{d}{dx} (x^2) = \dfrac d{dx} x + \dfrac d{dx} x + \dfrac d{dx} x + \dfrac d{dx} x + \dfrac d{dx} x + \dfrac d{dx} x... \)
\(2x = 1 + 1 + 1 + 1 + 1 + 1 ...= 1\cdot x \)
\(2x = x \)
\(2 = 1\). why?

Note by Choi Chakfung
1 year, 8 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. 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"
MathAppears 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} \( \boxed{123} \)

Comments

Sort by:

Top Newest

You can't differentiate an equation until it's an identity. Also, both sides must be defined in the neighbourhood of the point at which you're differentiating. As these statements don't hold for the above equation, the proof is flawed.

Deeparaj Bhat - 1 year, 8 months ago

Log in to reply

2x = 1 +1+1+1+1+1......(2x times) 2x =2x Not 2x=x So 2 not same with 1

Ivan Tanuwijaya - 1 year, 8 months ago

Log in to reply

A link to solution

Brilliant Logo

Brilliant Logo

Joel Yip - 1 year, 8 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...