Prove sqrt(6) is irrational

  • Thread starter Klion
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  • #1
Klion
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Alright, heading says it all. This is a nice problem heh.. I can see how to prove sqrt(5) is irrational. I think this method works up to the points where the fact 5 is a prime is used, (ie prime lemma) on 5 which doesn't work so well on 6! hehe

Was thinking of maybe using product of primes somehow but.. hmm dunno

Anyway for sqrt(5) went like this (proof by contradiction)

Prove sqrt(5) is not rational.

Suppose sqrt(5) = a/b, where a,b E Z+ (suppose sqrt(5) is rational)
we'll also assume gcd(a,b) = 1 (otherwise just divide a,b by gcd)
sqrt(5) = a/b <=> 5 = a^2/b^2 or 5*b^2=a^2
so 5|a^2, but 5 is a prime so 5|a Ea' 5*a'=a so 5b^2=(5a)^2 or 5b^2=5^2*a^2 or b^2 = 5*a'^2 so 5|b^2 by prime lemma 5|b 5|a ^ b|b so 5|gcd(a,b)
so 5|1 which is nonsense therefore our initial assumption is incorrect and 5 is in fact irrational.

-Kli
 

Answers and Replies

  • #2
KLscilevothma
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Suppose sqrt(6) is rational

gcd(a,b) = 1

[tex] \sqrt{6} = \frac{a}{b} [/tex]

[tex] 6b^2 = a^2 [/tex]

Now there are 4 possibilities.

1) both a and b are even.
It cannot be true since gcd(a,b)=1

2) a is odd and b is even
It cannot be true too. If a is odd, then left hand side will be even while right hand side will be odd.

3) both a and b are odd.
It cannot be true too and the reason is the same as that in (2)

4) a is even and b is odd.

Let a = 2m and b = 2n + 1

[tex] 6b^2 = a^2 [/tex]

[tex] 6*{(2n+1)}^2 = {2m}^2 [/tex]

[tex] 12m^2 + 12m +3 = 2n^2 [/tex]

left hand side is odd while right hand side is even

Contradiction.

Therefore sqrt(6) is irrational
 
  • #3
HallsofIvy
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"Let a = 2m and b =2n + 1
6b2= a2

6(2n+1)2= 2m2"

No, it should be 6(2n+1)2= (2m)2

"12m2+ 12m+ 3= 2n2"

Now this completely baffles me. Even assuming you accidently switched m and n, it doesn't follow.
From 6(2n+1)2= (2m)2 you get
6(4n2+ 4n+ 1)= 4m2 or
24n2+ 24n+ 6= 4m2 and both sides are clearly even.
 
  • #4
KLscilevothma
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Originally posted by HallsofIvy
24n2+ 24n+ 6= 4m2 and both sides are clearly even.
How about dividing both sides by 2, then we get
12n2+ 12n+ 3= 2m2
where left hand side is odd while right hand side is even.

If we don't divide both sides by 2, left hand side cannot be divisible by 4 while right hand side can.
 
  • #5
mathman
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How about the following (for any n not a square of an integer).

sqrt(n)=a/b, where a and b are integers with no common factors, and b >1.

n=a2/b2

The numerator and denominator still have no common factors, so n is not an integer.
 
  • #6
phoenixthoth
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...for any n not a square of an integer...n is not an integer.

any n in what? is n any nonnegative rational number, then? if it's any nonnegative integer n not the square of an integer, it seems odd to end by saying "n is not an integer."
 
  • #7
mathman
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Sorry for the confusion. I assumed we were discussing integers in general. My discussion is about (positive) integers which are not squares of integers. In n is a square of an integer, then b=1.
 
  • #8
phoenixthoth
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then wouldn't "n is not an integer" prove it?
 
  • #9
uart
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Originally posted by phoenixthoth
then wouldn't "n is not an integer" prove it?

Hi phoenixthoth, Mathman is using proof by contradiction . In the first line of his proof what he means is : Assume that n is a positive integer that is not a perfect square.

He goes on to show that assuming sqrt(n) to be rational, but non integer (see b>1), leads directly to a contradiction of the assumption that n was an integer. Hence the result that the square root of all integers, other than the perfects squares, are irrational. It's a perfectly valid proof.
 
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  • #10
phoenixthoth
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it seems fishy because it was easier to prove it in general for nonsquare nonnegative integers n than for n=6 and it's usually the case that it's easier to do it for n=6 and harder in general. i'm at some point if i feel like it going to check where the assumption that n is not a square was used and if this proof applies to square integers as well. i suppose that's what enables one to say b>1.
 
  • #11
phoenixthoth
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and 4 = 2*2 both of which are prime
 
  • #12
selfAdjoint
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What has 4 = 2 X 2 got to do with that? Some squares are squares of primes and some aren't, like 6 X 6 = 36. How does that affect the proof?
 
  • #13
...

(Repost of 'disappeared' Post with Refinement)

Proof;-

1) Each SQRT(Prime Number) is irrational &

2) SQRT(Prime Number 1) * SQRT(Prime Number 2) is irrational if Prime Number 1 is not equal to Prime Number 2.

3) SQRT(6) = SQRT(2) * SQRT(3)

4) 2 & 3 are two different Prime Numbers.

i.e.

5) SQRT(6) is irrational.

This completes the proof...


kx21

https://www.physicsforums.com/member.php?s=&action=getinfo&find=lastposter&forumid=80
 
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  • #14
phoenixthoth
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Originally posted by selfAdjoint
What has 4 = 2 X 2 got to do with that? Some squares are squares of primes and some aren't, like 6 X 6 = 36. How does that affect the proof?
this was in reference to a deleted post.
 
  • #15
phoenixthoth
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Originally posted by kx21
...

(Repost of 'disappeared' Post with Refinement)

Proof;-

1) Each SQRT(Prime Number) is irrational &

2) SQRT(Prime Number 1) * SQRT(Prime Number 2) is irrational if Prime Number 1 is not equal to Prime Number 2.

3) SQRT(6) = SQRT(2) * SQRT(3)

4) 2 & 3 are two different Prime Numbers.

i.e.

5) SQRT(6) is irrational.

This completes the proof...


kx21

https://www.physicsforums.com/member.php?action=getinfo&find=lastposter&forumid=111&


assumption #1 is as clear as conclusion #5. some people would not take #1 without proof.

selfadjoint, this post in its original form was about six words long and the point in bringing up 4=2x2 is that it should have been stated that the two primes must be different; it was stated in the second form. this argument would have to be messaged before it would work for all n. the main thing there would be to prove/say a nonsquare has not all even powers in its prime factorization.
 
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  • #16
uart
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1) Each SQRT(Prime Number) is irrational &

2) SQRT(Prime Number 1) * SQRT(Prime Number 2) is irrational if Prime Number 1 is not equal to Prime Number 2.

The problem is that you never really establish the truth of either point 1 or point 2 anywhere in the proof kx21


I still think that the best proof is as follows.

Let [tex]\sqrt{n} = \frac{a}{b}[/tex]

Then [tex] a^2 = b^2 n[/tex] <- Eqn1

From the fundamental therom of arithmetic (that is the uniqueness of prime factorization) it is clear that any integer squared contains only even powered prime factors.

Thus the LHS of Eqn1 can contain only even powered prime factors whereas the RHS of Eqn1 will contain odd powered prime factors whenever n contains any odd powered prime factors.

So LHS = RHS is possible if (and only if) n contains only even powered prime factors. This is of course equivalent to the requirement that n be a perfect square.

In other words, either n is a perfect square or [tex]\sqrt{n}[/tex] is irrational.
 
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  • #17
*** a 'M' Generalization with a bonus Link Phi, e & i... ***

Theorem- The irrationality of SQRT(N)

SQRT(N) is irrational if

N = p1*p2...*pk, where

pi i =1 to k are different Prime Numbers.


For instances,

1) Is SQRT(6) irrational?

...

It's irrational as 6 = 2 * 3

[?]


2) Is SQRT(30) irrational?

A simple Number Test,

with a Bonus Link about Phi (Golden Ratio), e, pi for all:-

http://superstringtheory.com/forum/extraboard/messages12/376.html [Broken]

Happy exploring & have a wonderful memory & time...

kx21

https://www.physicsforums.com/member.php?s=&action=getinfo&find=lastposter&forumid=80
 
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  • #18
uart
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Theorem- The irrationality of SQRT(N)

SQRT(N) is irrational if

N = p1*p2...*pk, where

pi i =1 to k are different Prime Numbers.

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. ;-)
 
  • #19
Originally posted by uart

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. ;-)

...

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...


[?]


kx21
https://www.physicsforums.com/member.php?s=&action=getinfo&find=lastposter&forumid=80
 
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  • #20
phoenixthoth
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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.
 
  • #21
oen_maclaude
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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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  • #22
HallsofIvy
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Okay, but that requires proving the "rational root theorem" which is surely harder to prove in general that specifically that sqrt(6) is irrational.
 
  • #23
oen_maclaude
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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. :wink:
 
  • #24
modmans2ndcoming
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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.
 
  • #25
modmans2ndcoming
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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)
 
  • #26
matt grime
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Originally posted by 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.

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.
 
  • #27
modmans2ndcoming
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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.
 
  • #28
matt grime
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Originally posted by 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.


and the product of pi and 1/pi is?
 
  • #29
modmans2ndcoming
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umm...1/pi is not an integer so it can't be a prime :-)
 
  • #30
matt grime
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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.
 
  • #31
modmans2ndcoming
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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.
 
  • #32
matt grime
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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.
 
  • #33
NateTG
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if you multiply the root of 2 distict prime numbers together, you end up with an irrational number
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
 
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  • #34
modmans2ndcoming
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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.
 
  • #35
modmans2ndcoming
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yes, proof by contradiction is probably the best meathod for this since it is so short and uncomplicated.
 

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