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Proof about irrational numbers.

  1. Jan 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that [itex] \sqrt{6} [/itex] is irrational.
    3. The attempt at a solution

    Would I just do a proof by contradiction and assume that [itex] \sqrt{6} [/itex] is rational and then get that [itex] 6q^2=p^2 [/itex] which would imply that p is even so I put in p=2r
    and then multiply it out. then this would imply that q is also even and this is a contradiction because they would have factors in common. I know I skipped some of the steps. Could I also make an argument that [itex] \sqrt{6} [/itex] is [itex] \sqrt{3}\sqrt{2} [/itex] and then say that an irrational times an irrational is an irrational as long as its not the same irrational.
     
  2. jcsd
  3. Jan 21, 2012 #2

    micromass

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    Sounds good.

    Uuh, that isn't true. There are many counterexamples.
     
  4. Jan 21, 2012 #3
    ok thanks for your response. on the second one I cant think of a counterexample off hand. maybe i should think about it more.
     
  5. Jan 21, 2012 #4

    micromass

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    [itex]\sqrt[3]{4}*\sqrt[3]{2}[/itex]

    or

    [itex]\sqrt{2}*\frac{1}{\sqrt{2}}[/itex]

    or

    [itex]\sqrt{18}*\sqrt{2}[/itex]
     
  6. Jan 21, 2012 #5
    [itex]\sqrt{81} = \sqrt[]{3}\sqrt[]{27}[/itex]

    Damn you Micro you're too quick.
     
  7. Jan 21, 2012 #6
    ya but all the counterexample have common factors under the radical. I was saying that there, well I was thinking that there were no common factors under the radical.
    What if i said a prime number that is square rooted times a different prime that is square rooted will be irrational.
     
  8. Jan 21, 2012 #7

    micromass

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    Ya snooze, ya lose!! :biggrin:

    Then the statement is probably true. But did you prove the statement??
     
  9. Jan 21, 2012 #8
    Okay tell me if this works. lets assume we have a multiplication of primes to the nth root.
    and lets assume that it is rational and that they have no common factors .
    [itex] (P_1P_2P_3.......P_r)^{\frac{1}{n}}=\frac{x}{y}[/itex]
    then we take both sides to then power of n and then multiply the y^n over
    and we get [itex] y^n(P_1P_2P_3.......P_r)=x^n[/itex]
    therefor this implies that x^n is divisible by a prime. so we will now write x=aP
    and we get that [itex] y^n(P_1P_2P_3.......P_r)=(aP)^n[/itex]
    and this would imply that y is divisible by some prime in our list. and this would imply that x and y share a common factor which is a contradiction.
    not sure if my last line of reasoning is valid.
     
    Last edited: Jan 21, 2012
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