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Prove that it is irrational?

  1. Jun 15, 2009 #1
    Well, there is a problem, i have solved/proved it, but i am not sure whether it is correct.
    THe problem is this:

    Using unique factorization into primes prove that there are no integers a and b such that [tex]a^2=30b^2[/tex], and thus show that [tex]sqrt{30}[/tex] is irrational.

    Proof:using unique factorization of any integer greater than 1 or less than -1, we can factor any such integer into the product of powers of distinct primes, or simply into a product of primes.

    [tex]a^2=30b^2=>b^2|a^2=>b|a=>\exists k,a=kb[/tex]

    Let:

    [tex]a=p_1p_2...p_r; b=q_1q_2...q_s[/tex]

    [tex]a^2=30b^2=>30=\left(\frac{b}{a}\right)^2=k^2=>\sqrt{30}=k[/tex]

    Now from the unique factorization theorem again:[tex]\sqrt{30}=k=d_1d_2...d_n=>30=d_1^2d_2^2...d_n^2[/tex]

    =>

    [tex]2*3*5=d_1^2d_2^2...d_n^2=>2|d_1^2d_2^2...d_n^2=>2|d_i^2=>2=d_i[/tex]

    but this would contradict the unique factorization theorem, and thus this contradiction shows that such a, and b do not exist.

    Is this about correct, or there is another way around it?
     
  2. jcsd
  3. Jun 15, 2009 #2
    i don't know what's wrong with latex?
     
  4. Jun 15, 2009 #3

    MathematicalPhysicist

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    Gold Member

    Recall how do you prove sqrt(2) is irrational: a^2=2b^2 (where gcd(a,b)=1) now 2 divides a^2 and thus 2 divides a and thus 4 divides a^2, so a=2k, 4k^2=2b^2 => b^2=2k^2 so also 2 divides b^2 and thus divides b, which means that gcd(a,b)>1 which is a contradiction, the same method is used here as well.
     
  5. Jun 15, 2009 #4
    Actually you can use the unique factorization thus:

    [tex]
    a^2=30b^2
    [/tex]

    There are an even number of 2's on the left, but an odd number on the right. Contradiction.
     
  6. Jun 15, 2009 #5
    well, yah i thought about this one, but since they asked to use the unique factorization of a number into primes, that part through me off, and i didn't know whether the same method is applied here.
     
  7. Jun 15, 2009 #6
    Which method? Why can't you use that one-line proof?
     
  8. Jun 15, 2009 #7
    THis is pretty much what my proof eventually shows, that we will have more 2's in one side than on the other.
     
  9. Jun 15, 2009 #8
    Here i was referring to mathematical physicists's post. I know how to show that a nr is irrational using that methodology.
     
  10. Jun 15, 2009 #9
    Yes but you can say this immediately. There's a lot of unnecessary stuff in your proof.
     
  11. Jun 15, 2009 #10

    JasonRox

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    Homework Helper
    Gold Member

    Just look at the prime factors on both sides... done.
     
  12. Jun 16, 2009 #11
    yeah i got it! thnx for the input!
     
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