
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
April 3rd, 2017, 06:45 AM  #1 
Member Joined: Aug 2016 From: Used to be Earth Posts: 64 Thanks: 14  1=1 Incorrect Proof
1=1 1²=1² Since, (1)²=1=1², we have 1²=(1)² So, taking square root on both sides gives 1=1. Obviously, this is wrong, but where is the mistake? I suppose it has to do with the square root of (1)², but why is that an issue? (1)² is greater than 0, so afaik, its square root shouldn't cause any trouble. Thanks for the help. 
April 3rd, 2017, 06:59 AM  #2  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,021 Thanks: 666 Math Focus: Physics, mathematical modelling, numerical and computational solutions  Quote:
 
April 3rd, 2017, 07:18 AM  #3  
Member Joined: Aug 2016 From: Used to be Earth Posts: 64 Thanks: 14  Quote:
So, if I follow, then it's because my last equation is assuming a onetoone relationship when in fact it was a "manytomany" relationship? So, instead, the correct implication should be 1²=(1)² => +1=+1 ? I'm not sure I clearly get this, so please bear with me if what I'm saying is nonsense...  
April 10th, 2017, 01:40 PM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,524 Thanks: 641 
Strictly speaking the square root function returns the positive root. That is, , not . That's the reason we need to write the when solving the equation (). If we wrote only we get only the positive root. is the same as and results in . Last edited by Country Boy; April 10th, 2017 at 01:44 PM. 
April 10th, 2017, 04:59 PM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,517 Thanks: 910 Math Focus: Elementary mathematics and beyond 
$$\sqrt{a^2}=a$$

April 12th, 2017, 02:59 AM  #6 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,277 Thanks: 204 
As greg1313 has pointed out, perhaps we should not be too quick to 'strike out' the exponents on both sides of the equation ... $1^2 = (1)^2 \ \ \ \ \ \ \ $< TRUE $\sqrt{1^2} = \sqrt{(1)^2} \ \ \ \ \ \ \ \ $ < TRUE Now be careful ... $1 = 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ < TRUE $1 = 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ <TRUE There is no paradox. Last edited by skipjack; April 12th, 2017 at 03:26 AM. 

Tags 
incorrect, proof 
Search tags for this page 
Click on a term to search for related topics.

Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Proof this question is incorrect (Unsolvable)  Rxyzan  Physics  4  January 7th, 2015 06:10 AM 
Is this incorrect?  CherryPi  Calculus  56  February 2nd, 2012 02:13 AM 
A075441 has incorrect data  duz  Number Theory  7  February 9th, 2010 06:06 PM 
probability of incorrect conclusion? confused..  bob.dob  Algebra  1  October 29th, 2008 12:50 PM 
two different answers in limits but one is incorrect  conjecture  Calculus  3  July 16th, 2008 05:12 PM 