Proof for there does not exist a unique x that is an ...

FreaKariDunk · September 11th, 2012, 05:58 PM

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?

MarkFL · September 11th, 2012, 06:33 PM

Personally, I would use reductio ad absurdum (contradiction) for this.

Dougy · September 11th, 2012, 06:44 PM

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 $\sqrt{3}$ is rational ,there exists p and q with gcd(p,q)=1 such that $\frac{p}{q}=\sqrt{3}$ implies that $p^2=3q^2$ .
Since 3 is prime $p^2$ can not divide 3, so $p^2$ divides $q^2$ which means there exists an integer a such that $q^2=ap^2$ which means $3a=1$ which is absurd (a is an integer).
I guess this shows generally (IF it is correct....) that $\sqrt{p}$ with p prime is always irrational.

FreaKariDunk · September 11th, 2012, 07:03 PM

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?

Dougy · September 11th, 2012, 08:18 PM

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 $x^2=3$ so that $\frac{p^2}{q^2}=3$ and then catch back above....

MarkFL · September 11th, 2012, 08:20 PM

I would begin by assuming there is a rational number whose square is 3:

$\frac{p^2}{q^2}=3$ where p and q are co-prime, i.e., $\frac{p}{q}$ is in lowest terms.

This implies:

$p^2=3q^2$

which means $p^2$ must be a multiple of 3, but since p is squared, it must be a multiple of 9, which implies $q^2$ must also be a multiple of 9, and so our assumption that p and q are co-prime has been contradicted.

Dougy · September 11th, 2012, 08:34 PM

I think this one also works!
Just when you say

Quote:

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

actually I think it does not imply $q^2$ is a multiple of 9 since there is a three already multiplying $q^2$ but at least $q^2$ has to be a multiple of 3 which is a common factor with 9 so the conclusion of absurdity is safe here!

MarkFL · September 11th, 2012, 08:46 PM

Yes, you are right, I meant to type a 3, but typed 9 instead!

I'm glad you caught that!

Dougy · September 11th, 2012, 09:04 PM

No problem

And both proofs would work the same for any prime number instead of three

MarkFL · September 11th, 2012, 09:08 PM

Yes, and hopefully we are not as stricken as the followers of Pythagoras...

September 11th, 2012, 05:58 PM	#1
FreaKariDunk 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?
$FreaKariDunk is offline$

September 11th, 2012, 06:33 PM	#2
MarkFL 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.
$MarkFL is offline$

September 11th, 2012, 06:44 PM	#3
Dougy 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 $\sqrt{3}$ is rational ,there exists p and q with gcd(p,q)=1 such that $\frac{p}{q}=\sqrt{3}$ implies that $p^2=3q^2$ . Since 3 is prime $p^2$ can not divide 3, so $p^2$ divides $q^2$ which means there exists an integer a such that $q^2=ap^2$ which means $3a=1$ which is absurd (a is an integer). I guess this shows generally (IF it is correct....) that $\sqrt{p}$ with p prime is always irrational.
$Dougy is offline$

September 11th, 2012, 07:03 PM	#4
FreaKariDunk 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?
$FreaKariDunk is offline$

September 11th, 2012, 08:18 PM	#5
Dougy 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 $x^2=3$ so that $\frac{p^2}{q^2}=3$ and then catch back above....
$Dougy is offline$

September 11th, 2012, 08:20 PM	#6
MarkFL 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: $\frac{p^2}{q^2}=3$ where p and q are co-prime, i.e., $\frac{p}{q}$ is in lowest terms. This implies: $p^2=3q^2$ which means $p^2$ must be a multiple of 3, but since p is squared, it must be a multiple of 9, which implies $q^2$ must also be a multiple of 9, and so our assumption that p and q are co-prime has been contradicted.
$MarkFL is offline$

September 11th, 2012, 08:46 PM	#8
MarkFL 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!
$MarkFL is offline$

September 11th, 2012, 09:04 PM	#9
Dougy 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
$Dougy is offline$

September 11th, 2012, 09:08 PM	#10
MarkFL 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...
$MarkFL is offline$

Thread Tools
$Show Printable Version$ Show Printable Version $Email this Page$ Email this Page
Display Modes
$Linear Mode$ Linear Mode $Hybrid Mode$ Switch to Hybrid Mode $Threaded Mode$ Switch to Threaded Mode

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
unique factorization domain, zero of a polynomial , proof	rayman	Abstract Algebra	2	October 19th, 2012 06:38 AM
0/0 does not exist?	Dart Plegius	Algebra	4	June 19th, 2012 12:37 PM
Prove that lim (x -> 0) (1/x^2) does not exist	scherz0	Real Analysis	10	October 24th, 2009 01:21 PM
Does the following limit exist?	BUBLIL	Complex Analysis	1	December 12th, 2008 08:15 AM
Does this graph exist?	tinatran2	Applied Math	1	May 9th, 2007 08:16 PM