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Prove sqrt(6) is irrational

  1. May 8, 2014 #1
    1. The problem statement, all variables and given/known data
    Prove that [itex]\sqrt{6}[/itex] is irrational.


    2. Relevant equations



    3. The attempt at a solution
    [itex]\sqrt{6}[/itex] = [itex]\sqrt{2}[/itex]*[itex]\sqrt{3}[/itex]

    We know that [itex]\sqrt{2}[/itex] is an irrational number (common knowledge) and also this was shown in the textbook.

    So, let's assume [itex]\sqrt{6}[/itex] and [itex]\sqrt{3}[/itex] are both rational.

    [itex]\Rightarrow[/itex] [itex]\sqrt{2}[/itex]*[itex]\sqrt{3}[/itex] = [itex]\frac{a}{b}[/itex]

    Also since, [itex]\sqrt{3}[/itex] is assumed to be rational it can be written as [itex]\frac{c}{d}[/itex] [itex]\Rightarrow[/itex] [itex]\sqrt{6}[/itex] = [itex]\sqrt{2}[/itex]*[itex]\frac{c}{d}[/itex] = [itex]\frac{a}{b}[/itex] [itex]\Leftrightarrow[/itex] [itex]\sqrt{2}[/itex] = [itex]\frac{d}{c}[/itex]*[itex]\frac{a}{b}[/itex] which is a contradiction since a, b, c and d are integers and [itex]\sqrt{2}[/itex] is not a rational number but [itex]\frac{d}{c}[/itex]*[itex]\frac{a}{b}[/itex] would be a rational number. Therefore, we can conclude that [itex]\sqrt{6}[/itex] is an irrational number.

    Would this solution be correct? Or rather is this solution enough?
     
    Last edited: May 8, 2014
  2. jcsd
  3. May 8, 2014 #2

    Mentallic

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    Not quite. Your proof has assumed something that isn't even true. [itex]\sqrt{3}[/itex] is irrational as you probably know, and you've only assumed it to be rational. If we consider that we didn't know whether it were rational or irrational, then we would need to include both cases in our proof.
     
    Last edited: May 8, 2014
  4. May 8, 2014 #3
    So that is needed? But if [itex]\sqrt{3}[/itex] is irrational then we have the product of two irrational numbers? Which per definition is irrational since the product of two different irrational numbers is irrational...?
     
  5. May 8, 2014 #4

    phyzguy

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    The product of two irrational numbers can be rational: [itex]\sqrt{2} \times \sqrt{2} = 2[/itex].
     
  6. May 8, 2014 #5
    Yeah I know, that's why I wrote two different irrational numbers...
     
  7. May 8, 2014 #6
    [tex](\sqrt{2}+1)(\sqrt{2}-1)=1[/tex]
     
    Last edited: May 8, 2014
  8. May 8, 2014 #7
    Exercise: Prove that given any rational number ##r\neq0## there are uncountably many pairs ##x,y## such that ##x## and ##y## are irrational, ##x\neq y##, and ##xy=r##.
     
    Last edited: May 8, 2014
  9. May 8, 2014 #8
    I wouldn't even know where to begin...

    I'm in first year university...and there is a biiig different between high school and university over here. I basically aced every test in math during my high school years and I'm having trouble with this...
     
  10. May 8, 2014 #9
    Here's a hint. The proof that ##\sqrt{6}## is irrational has a proof that is very similar to the classic proof that ##\sqrt{2}## is irrational.
     
  11. May 8, 2014 #10
    Reframe the question: given a rational ##r\neq 0 ## and an irrational ##x##, is there a number ##y## such that ##xy=r##? Is this number rational or irrational?
     
  12. May 8, 2014 #11
    I know how to prove ##\sqrt{2}## is irrational but I wanted to check if there was any way to just use the fact ##\sqrt{2}## is irrational and that ##\sqrt{6}## = ##\sqrt{2}##*##\sqrt{3}## to prove that ##\sqrt{6}## is irrational...
     
  13. May 8, 2014 #12
    No because if that was true then the irrational ##x## could be written as a rational, right?
     
  14. May 8, 2014 #13

    Ray Vickson

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    Theorem: if ##n > 0## is an integer and ##\sqrt{n}## is not an integer, then ##\sqrt{n}## is irrational. Google 'irrational square root'.
     
  15. May 9, 2014 #14
    Given an irrational number ##x## and a rational ##r\neq 0##, is ##rx^{-1}## a number? If it is a number, is it a rational number or an irrational number?

    Just to be clear, this line of questioning has very little to do with the problem in the original post. It is directed towards the incorrect assumption that products of different irrationals must be irrational.
     
  16. May 9, 2014 #15
    Here is a start with proof by contradiction:
    let's assume that the square root of 6 is rational.

    By definition, that means there are two integers a and b with no common divisors where:

    a/b = square root of 6.

    So let's multiply both sides by themselves:

    (a/b)(a/b) = (square root of 6)(square root of 6)
    a2/b2 = 6
    a2 = 6b2

    But this last statement means the RHS (right hand side) is even, because it is a product of integers and one of those integers (at least) is even. So a2 must be even. But any odd number times itself is odd, so if a2 is even, then a is even.

    Can you go from here?
     
  17. May 12, 2014 #16
    Yes, thank you. I believe I mentioned I know that I can prove it in the same manner I can prove sqrt(2) is irrational. I just wanted to know if I can prove it using the fact that sqrt(6) = sqrt(2)*sqrt(3) and the fact that sqrt(2) is irrational...
     
  18. May 12, 2014 #17
    Can you try proving it by stating that rational numbers eventually have a periodic decimal expansion and irrationals do not?
     
  19. May 12, 2014 #18
    Irrationals are not closed under multiplication, so I do not think you can prove it like that.
     
  20. May 12, 2014 #19

    Ray Vickson

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    I already suggested to you in post #13 what you need to establish, and how to find it using Google. Have you even tried?
     
  21. May 18, 2014 #20
    Sorry Ray, totally forgot about that suggestion. Will look it up though! Thanks!
     
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