My Math Forum  

Go Back   My Math Forum > High School Math Forum > Algebra

Algebra Pre-Algebra and Basic Algebra Math Forum

Thanks Tree2Thanks
  • 1 Post By keylewer
  • 1 Post By mathman
LinkBack Thread Tools Display Modes
May 23rd, 2018, 03:00 AM   #1
Joined: May 2018
From: United Kingdom

Posts: 1
Thanks: 1

Question 1=2 or 0/x NOT = 0

I have either proved that 1=2, 0/x for a value of x not equal to 0...
Or I have made a mistake.

I started with e^i*pi=-1
I then squared both sides to get:
Taking the log of both sides allows me to remove the power and multiply by the log for:
S1) 2*i*pi*log(e)=0
To extract the value of i, sqrt(-1), I divide both sides by 2*pi*log(e)
Square both sides for:
+2 both sides:

Thanks from v8archie
keylewer is offline  
May 23rd, 2018, 03:17 AM   #2
Global Moderator
greg1313's Avatar
Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,879
Thanks: 1087

Math Focus: Elementary mathematics and beyond
Your mistake lies in applying logarithmic identities which apply to real numbers to a complex number. Not all of these identities are necessarily true for complex numbers.

See here for more information.
greg1313 is online now  
May 23rd, 2018, 03:21 AM   #3
Math Team
Joined: Jan 2015
From: Alabama

Posts: 3,261
Thanks: 894

Your error is when you take the logarithm of a complex number. A complex number can be written in polar form as $\displaystyle re^{i\theta}= re^{i(\theta+ 2\pi n)}$ since "$\displaystyle e^{i\theta}= \cos(\theta)+ i\sin(\theta)$" is periodic with period $\displaystyle 2\pi$.

Taking the logarithm, $\displaystyle \ln(re^{i(\theta+ 2n\pi)}= \ln(r)+ i(\theta+ 2n\pi)$ for n any integer. That is, ln is multivalued.

Last edited by skipjack; May 23rd, 2018 at 05:45 AM.
Country Boy is offline  
May 23rd, 2018, 03:31 AM   #4
Math Team
Joined: Dec 2013
From: Colombia

Posts: 7,445
Thanks: 2499

Math Focus: Mainly analysis and algebra
Originally Posted by keylewer View Post
I have either proved that 1=2, 0/x for a value of x not equal to 0...
Or I have made a mistake.
This is great, thank you. Too many are unwilling to admit that they might have erred.
v8archie is offline  
May 23rd, 2018, 01:37 PM   #5
Global Moderator
Joined: May 2007

Posts: 6,607
Thanks: 616

A simple way to look at it is $e^{2\pi i}=1=e^0$, but $2\pi i\ne 0$.
Thanks from JeffM1
mathman is offline  

  My Math Forum > High School Math Forum > Algebra

0 or x, broken, e^i*pi, oops

Thread Tools
Display Modes

Copyright © 2018 My Math Forum. All rights reserved.