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Irrational Numbers

  1. Feb 12, 2008 #1
    I know that √2 is irrational (and I've seen the proof).

    Now, what is the fastest way to justify that 2√2, 2-√2, 17√(1/2) are irrational? (they definitely "seem" to be irrational numbers to me) Can all/any these follow immediately from the fact that √2 is irrational?

    Thanks!
     
    Last edited: Feb 12, 2008
  2. jcsd
  3. Feb 12, 2008 #2

    Dick

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    The product of an irrational and a nonzero rational is always irrational. Prove this, it's not hard. The sum of an irrational and a rational is always irrational. Prove this, also not hard. This last one is not completely trivial. You just have to think about prime factorization. It's basically the same proof as that the sqrt(2) is irrational.
     
  4. Feb 12, 2008 #3
    Let me try the last one,
    Suppose
    17√(1/2)=m/n
    => n17=2 m17
    By unique prime facotrization, 2 must occur as a prime factor of n (since the right side has a "2"), so on left side, 2 to the exponent a multiple of 17
    On the right side, 2 to the exponent (a multiple of 17) + 1
    Contradiction!
    So 17√(1/2) must be irrational. <====Is this proof correct?


    On the other hand, I am still having trouble with understanding why 2√2 and 2-√2 are irrational. How can you know for sure that
    "The product of an irrational and a nonzero rational is always irrational"
    & "The sum of an irrational and a rational is always irrational" ?
    I have no idea how to prove those...

    Thanks for explaining!
     
    Last edited: Feb 13, 2008
  5. Feb 13, 2008 #4

    NateTG

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    Assume, by contradiction that [itex]2\sqrt{2}[/itex] is rational, then you can write it as:
    [itex]\frac{a}{b}[/itex]
    with integers [itex]a,b[/itex], but then
    [itex]\sqrt{2}=\frac{a}{2b}[/itex]
    contradicts that [itex]\sqrt{2}[/itex] is irrational.
     
  6. Feb 13, 2008 #5

    Dick

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    You aren't quite getting the 2^(1/17) proof either. Look up and understand the sqrt(2) proof. You HAVE to say "assume 2^(17)=m/n is in lowest terms", i.e. gcd(m,n)=1. Now you say 2*n^17=m^17. So m is even. So 2^17 divides 2*n^17. So 2 divides n. So n is even. Now what?
     
  7. Feb 13, 2008 #6
    But I don't see where is it going wrong in the logic of my proof, would you (or anybody else) mind pointing it out, please?

    Proof:
    Suppose 17√(1/2)=m/n
    => n17 = 2 m17
    Now factor n and m into primes, say
    n=p1a1...pkak
    m=q1b1...qkbk
    where pi, qj are distinct primes
    Then
    p117a1...pk17ak=2 q117b1...qk17bk
    If these are equal, then since 2 occurs in the prime facotrization on the right side, it must occur on the left side (by uniqueness of prime factorization), so 2=pi for some i. Also, 2 MAY occur in one of qj
    So left side we have 217ai for some integer ai
    Right side we have 2 x 217bj for some integer bj
    So it's impossible for these 2 numbers to be equal. Factorization into primes is unique, but 217ai cannot be equal to 2 x 217bj)
    So 17√(1/2) must be irrational.


    And my method also seems to work in a more general situation like proving that 3√(2/7) is irrational, too. But the method of proving that √2 is irrational doesn't seem to work here...

    Please let me know if I am wrong anywhere.

    Thanks!
     
    Last edited: Feb 13, 2008
  8. Feb 13, 2008 #7

    Dick

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    I believe you. 17*ai=17*bi+1 doesn't work. 1 isn't divisible by 17. But it really is the same proof as for sqrt(2). In that case you get 2*ai=2*bi+1. Still doesn't work. 1 isn't divisible by 2 either.
     
  9. Feb 14, 2008 #8
    Is there a neat way to prove that 3√(2/7) is irrational?
     
  10. Feb 14, 2008 #9

    NateTG

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    You might as well just move on to a generic proof for all roots of rational numbers.
     
  11. Feb 18, 2008 #10
    In wikipedia, there is a method using the concept of unique prime facotraztion to prove that a number is irrational, so I think my proof on 17√(1/2) being irrational is also correct.

    http://en.wikipedia.org/wiki/Square_root_of_two#Proof_by_unique_factorization

    17√(1/2)=m/n
    But I noticed that there are two BAD cases: m=1 or n=1. No prime factorizations exist, so the step in my proof
    "...Now factor n and m into primes, say
    n=p1a1...pkak
    m=q1b1...qkbk
    where pi, qj are distinct primes"
    is not valid. How can I actually prove that it also works in these 2 cases?

    Can somebody help me, please?
     
  12. Feb 18, 2008 #11

    Tom Mattson

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    The proof? There isn't just one!

    Are you aware of the following result?

    Theorem: The Rational Root Theorem (RRT)
    Let [itex]a_nx^n+...+a_1x+a_0=0[/itex] be a polynomial equation with [itex]a_i\in\mathbb{Z}[/itex] for all [itex]i[/itex], and with [itex]a_n,a_0\neq 0[/itex]. Then [itex]p/q[/itex] be a rational root for the equation with [itex]p[/itex] relatively prime to [itex]q[/itex] and [itex]q\neq 0[/itex].

    Then:

    (i) [itex]p[/itex] divides [itex]a_0[/itex], and
    (ii) [itex]q[/itex] divides [itex]a_n[/itex].

    Now consider the equation [itex]x^2-2=0[/itex], which clearly has [itex]\sqrt{2}[/itex] as a root. By RRT, the only possible rational roots are -2,-1,1,2. But none of them works!

    Therefore, [itex]\sqrt{2}[/itex] is irrational.

    I don't know about the fastest way, but a viable way would be to construct polynomial equations as I have done above, and apply RRT. Here, I'll do another one for you.

    Let [itex]x=2\sqrt{2}[/itex]. Then [itex]x^2=8[/itex], or [itex]x^2-8=0[/itex].

    Clearly, this equation has [itex]2\sqrt{2}[/itex] as a root. By RRT, the only possible rational roots to that equation are -8,-4,-2,-1,1,2,4,8. But none of these works. Therefore, [itex]2\sqrt{2}[/itex] is irrational.

    Comprede?
     
  13. Feb 21, 2008 #12
    Sorry about that, but I thought that was only one way before, since only one proof in presented in my class...but now I am aware of many proofs...
     
  14. Feb 21, 2008 #13
    Does anyone have any idea how to deal with the case m=1 or n=1? I am stuck here...
     
  15. Feb 21, 2008 #14
    sorry for being offtopic and interfering.. but how can one proove that √2 is irrational?
     
  16. Feb 21, 2008 #15

    Gib Z

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  17. Feb 21, 2008 #16

    Dick

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    (1/2)^(1/17)=1/n -> 2=n^(17). That's ridiculous. 2 isn't the 17th power of any integer. (1/2)^(1/17)=m/1 -> 2*m^(17)=1. Also ridiculous. 1/2 isn't the 17th power of any integer. Was that really so mysterious?
     
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