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September 11th, 2012, 05:58 PM  #1 
Senior Member Joined: Jan 2010 Posts: 324 Thanks: 0  Proof for there does not exist a unique x that is an ...
Proof of: There does not exist a unique x that is an element of the rationals such that x squared equals 3. Would contradiction be the way to do this or trichotomy like how you would for there exists a unique x >0 such that x squared equals 2? 
September 11th, 2012, 06:33 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Proof for there does not exist a unique x that is an ..
Personally, I would use reductio ad absurdum (contradiction) for this.

September 11th, 2012, 06:44 PM  #3 
Senior Member Joined: Nov 2011 Posts: 595 Thanks: 16  Re: Proof for there does not exist a unique x that is an ..
Hi, Ok I have never been very good at this so please don't take this for granted, might be wrong, just an attempts here. So if is rational ,there exists p and q with gcd(p,q)=1 such that implies that . Since 3 is prime can not divide 3, so divides which means there exists an integer a such that which means which is absurd (a is an integer). I guess this shows generally (IF it is correct....) that with p prime is always irrational. 
September 11th, 2012, 07:03 PM  #4 
Senior Member Joined: Jan 2010 Posts: 324 Thanks: 0  Re: Proof for there does not exist a unique x that is an ..
Mark, elaborate. I would do x squared does not equal three (he hasnt defined square root so i cant use it in the proof). Then do what? 
September 11th, 2012, 08:18 PM  #5 
Senior Member Joined: Nov 2011 Posts: 595 Thanks: 16  Re: Proof for there does not exist a unique x that is an ..
So is there something wrong in the proof I wrote? Would be nice to at least acknowledge you saw it, especially when people spend a bit of their time to try to help I hope it is not just a problem that "he hasnt defined defined square root" because you can of course simply start the proof saying: Let us assume a rational x=p/q such that so that and then catch back above.... 
September 11th, 2012, 08:20 PM  #6 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Proof for there does not exist a unique x that is an ..
I would begin by assuming there is a rational number whose square is 3: where p and q are coprime, i.e., is in lowest terms. This implies: which means must be a multiple of 3, but since p is squared, it must be a multiple of 9, which implies must also be a multiple of 9, and so our assumption that p and q are coprime has been contradicted. 
September 11th, 2012, 08:34 PM  #7  
Senior Member Joined: Nov 2011 Posts: 595 Thanks: 16  Re: Proof for there does not exist a unique x that is an ..
I think this one also works! Just when you say Quote:
 
September 11th, 2012, 08:46 PM  #8 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Proof for there does not exist a unique x that is an ..
Yes, you are right, I meant to type a 3, but typed 9 instead! I'm glad you caught that! 
September 11th, 2012, 09:04 PM  #9 
Senior Member Joined: Nov 2011 Posts: 595 Thanks: 16  Re: Proof for there does not exist a unique x that is an ..
No problem And both proofs would work the same for any prime number instead of three 
September 11th, 2012, 09:08 PM  #10 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Proof for there does not exist a unique x that is an ..
Yes, and hopefully we are not as stricken as the followers of Pythagoras... 

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