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September 11th, 2012, 06:58 PM   #1
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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?
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September 11th, 2012, 07:33 PM   #2
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Re: Proof for there does not exist a unique x that is an ..

Personally, I would use reductio ad absurdum (contradiction) for this.
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September 11th, 2012, 07:44 PM   #3
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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.
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September 11th, 2012, 08:03 PM   #4
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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?
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September 11th, 2012, 09:18 PM   #5
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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....
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September 11th, 2012, 09:20 PM   #6
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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 co-prime, 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 co-prime has been contradicted.
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September 11th, 2012, 09:34 PM   #7
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Re: Proof for there does not exist a unique x that is an ..

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 is a multiple of 9 since there is a three already multiplying but at least has to be a multiple of 3 which is a common factor with 9 so the conclusion of absurdity is safe here!
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September 11th, 2012, 09:46 PM   #8
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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!
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September 11th, 2012, 10:04 PM   #9
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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
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September 11th, 2012, 10:08 PM   #10
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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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