Prove sqrt(6) is irrational

uart

While correct that theorem is only "half baked" as there are very many integers n for which sqrt(n) is irrational but yet n cannot be expressed as such a simple prime product as stated by the therom (that is, with all prime factors to unity power only).

Why not do it properly as per either mathmans or my own post, and establish irrationality for sqrt of all integers which are not perfect squares ?


Also, if you're going to just quote a therom that has already established primeness of certain square roots (as opposed to establishing it from something more fundamental like the uniqueness of prime factorization) then why pick such a "half baked" theorem in the first place? Why not use a better theorem to begin with.


Theorem : Either n is a perfect square or sqrt(n) is irrational.

Then all you need to do is to say. "Since 6 in not a perfect then by the above theorem then the sqrt of 6 is irrational." QED.

Really, if you are going to be lame then at least do it properly. ;-)

kx21

[8)]...

Let me think...

Are there any 'Topological Defects' hidden in your Theorem.

Eureka!!!

Your Theorem is invalid if...

n = -1.

Proof:-

Given SQRT(-1)= i * i

This Completes the proof...


[a)] [?]


kx21
http://www.physicsforums.com/member....ter&forumid=80

phoenixthoth

it was clarified earlier that n is a nonnegative nonsquare integer. i is irrational if irrational means not rational. but irrational usually refers to R\Q. technically, this doesn't violate an or statement though i think what was mean by adding the word "either" was exclusive or, Xor.

oen_maclaude

square(6) is irrational.

first of all, we let x = sqrt(6).
squaring both sides we have xsquare is equal to 6.

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HallsofIvy

Okay, but that requires proving the "rational root theorem" which is surely harder to prove in general that specifically that sqrt(6) is irrational.

oen_maclaude

well, i am sorry but i haven't heard of the prime lemma. however, in my convinience, it is better to prove the statement using the rational root theorem. the only thing that you need here is the rational root theorem. [;)]

modmans2ndcoming

proove that the SR of 2 is Irrational, and the SR or 3 is irrational (they are prime so they have no factors other than 1 and itself therefore the root is irrational)

since those two roots are irational, an Irrational times an irrational is an irrational, therefore, root 6 is irrational.

modmans2ndcoming

your proof using the rational root theorim is incomplete.

you need to point out that since 6 is a real, Positive number, the SQRT of 6 cannot be complex, so the roots of X^2 can not be complex. (the complex root theorim says complex roots come in pairs, so it is important to rule those root out)

matt grime

certainly not true that last bit. It is easily possible for the product of two irrationals to be rational, sqrt(2) and, er, sqrt(2) for instance.

modmans2ndcoming

sorry, I was not percise enough in my proof and left some speculation.

I should have said " the product of 2 distinct irrational numbers is irrational.

matt grime

and the product of pi and 1/pi is?

modmans2ndcoming

umm...1/pi is not an integer so it can't be a prime :-)

matt grime

you didn't state that it *had* to be only to do with integers or primes. You just said that the product of two (distinct) irrationals is irrational.

If you don't like pi, then sqrt(2) and 2sqrt(2) are two distinct irrationals involving only integers and whose product is rational. Until such time as you accurately state all your hypotheses and what that impies you can't discount counter examples. So phrase the statement to remove ambiguity.

modmans2ndcoming

well first off, the discussion was in the universe of Natural numbers since we are discussing primes.

so here it is.

6 is a natural number. six can be composed as 2*3. 2 and 3 are primes, so the SQRT of a prime is irrational since a prime has no factors other than itself and 1. if you multiply the root of 2 distict prime numbers together, you end up with an irrational number, and since SQRT(6) can be composed as SQRT(2*3) and since we know that the SQRT of 2 and 3 are irrational and that the product of the SQRT of 2 distict prime numbers is irrational, SQRT(6) is irrational.

there, I have used all the language in one context.

matt grime

Thats fine. At no point did you state that your deductions were dependent on being only taking roots of primes though. And the result you gave had been proven on page 1 of the thread by two or three people without these imprecisions.

NateTG

Of course, this is probably what the original poster was trying to prove.

If you want to do something like this, it's easier and more understandable to do this:

Assume the fundemental theorem of algebra:
Each natural number has a unique prime factorization.

And then prove the theorem:
Any number whose prime factorization has any odd exponents has an irrational square root.

Proof:
Assume by contradiction that
[tex]\sqrt{n}=\frac{a}{b}[/tex]
where [tex]\frac{a}{b}[/tex] is a reduced fraction.
We can square both sides of the equation to get:
[tex]n=\frac{a^2}{b^2}[/tex]
Now, we know that
[tex]n=mp^{2k+1}[/tex]
where [tex]m[/tex] is not divisible by [tex]p[/tex] for some prime [tex]p[/tex] by the hypothesis.
So we have
[tex]mp^{2k+1}=\frac{a^2}{b^2}[/tex]
so
[tex]b^2mp^{2k+1}=a^2[/tex]
Now we have 2 cases:
case 1:
[tex]p[/tex] divides [tex]a[/tex]
Then [tex]b[/tex] is not divisible by [tex]p[/tex] since the fraction is in reduced form.
Then [tex]a=p^i a'[/tex] where [tex]a'[/tex] is not divisble by [tex]p[/tex].
so
[tex]b^2mp^{2k+1}=(p^ia')^2[/tex]
[tex]b^2mp^{2k+1}=p^{2i}{a'}^2[/tex]
Now the prime factorization of the LHS will have an odd power of [tex]p[/tex] ([tex]2k+1[/tex]) and the RHS will have an even power of [tex]p[/tex] ([tex]2i[/tex]). So by the fundemental theorem of algebra they cannot be equal. Hence this case is contradictory.
case 2:
[tex]p[/tex] does not divide [tex]a[/tex]
Now we have:
[tex]b^2mp^{2k+1}=a^2[/tex]
Where the LHS is clearly divisible by [tex]p[/tex] and the RHS is clearly not divisble by [tex]p[/tex] so this case is also contradictory.

Since the assumption that [tex]\sqrt{n}[/tex] can be written as a reduced fraction leads to contradictions, it must be incorrect.

Since [tex]\sqrt{n}[/tex] cannot be written as a reduced fraction, it cannot be written as a fraction, so it must be irrational. QED

modmans2ndcoming

actualy, it seemed to me that the OP was trying to proove that SQRT of 6 is irrational.

the OP did not say if they had prooven that root 2 and root 3 were irrational.