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Math Help - What is the basic method for proving the irrationality of a number?

  1. #1
    Member mfetch22's Avatar
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    What is the basic method for proving the irrationality of a number?

    I know to prove the irrationality of \sqrt{2} we simply minipulate the following:

    (\frac{p}{q})^2 = 2

    until we hit the contradiction of not being in reduced terms. I also know that we must take into account that all number can be written in the form 2n or 2n+1 for n=1,2,3...; and the fact that if x^2 is an even number then x is an even number. Heres my question, how do you prove the irrationality of a certain opertation on numbers in general? And for odd square roots?

    Two examples of what I mean:

    [1] How would you prove that if n=1, 2, 3... that \sqrt{n} is always irrational?

    [2] How would you prove \sqrt{3} is irrational? Would you find a way to define numbers in terms of 3's ? Just like with the \sqrt{2}? So all numbers would be of the form 3n, 3n+1, and 3n+2; and would you some how use that fact to prove the irrationality of \sqrt{3}?

    Thanks in advance
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  2. #2
    A Plied Mathematician
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    [1] You wouldn't. It's not true: the square root of 4 is rational.
    [2] I think this proof follows pretty much the same lines as for the square root of 2.
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  3. #3
    Member mfetch22's Avatar
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    Opps. That was a big mistake. For [1] I meant this:

    [1] How would one prove that \sqrt{x} is irrational for all x = 1, 2, 3... such that x \neq n^2 for any natural number n such that n = 1, 2, 3...

    My apologies, that was a very idiotic error in clearity. Hopefully that makes sense now.
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  4. #4
    A Plied Mathematician
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    I think generally that that proof would go along the same lines as for the square root of 2. Try it and see what happens.
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  5. #5
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    Quote Originally Posted by mfetch22 View Post
    I know to prove the irrationality of \sqrt{2} we simply minipulate the following:

    (\frac{p}{q})^2 = 2

    until we hit the contradiction of not being in reduced terms. I also know that we must take into account that all number can be written in the form 2n or 2n+1 for n=1,2,3...; and the fact that if x^2 is an even number then x is an even number. Heres my question, how do you prove the irrationality of a certain opertation on numbers in general? And for odd square roots?

    Two examples of what I mean:

    [1] How would you prove that if n=1, 2, 3... that \sqrt{n} is always irrational?

    [2] How would you prove \sqrt{3} is irrational? Would you find a way to define numbers in terms of 3's ? Just like with the \sqrt{2}? So all numbers would be of the form 3n, 3n+1, and 3n+2; and would you some how use that fact to prove the irrationality of \sqrt{3}?

    Thanks in advance
    For (2), it's the same way how you would prove the irrationality of root(2).

    Assuming that \sqrt{3} is rational, then it takes the form \sqrt{3}=\frac{m}{n}, where m,n\in Z and m and n do not have common factors.

    3=\frac{m^2}{n^2}\Rightarrow n^2=\frac{m}{3}\cdot m

    Since n^2 is an integer, it follows that m/3 is an integer too. Thus, 3 is a factor of m.

    Let m=3p where p\in Z , n^2=3p^2

    p^2=\frac{n}{3}\cdot n

    n/3 \in Z

    We see that both m and n have a common factor of 3 which contradicts our assumption.
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  6. #6
    MHF Contributor Also sprach Zarathustra's Avatar
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    It is a nice exercise to prove following:


    Every square root of a prime number is irrational.
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  7. #7
    Member mfetch22's Avatar
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    Okay, your right, I think I've figured out the proof for all even numbers, but I'm alittle lost for the odd numbers. Here is what I have, please point out any mistakes I made.

    -> We want to prove that \sqrt{n} is irrational, as long as no numbers x = 1, 2, 3.... satisfy x^2 = n.

    [i] If \sqrt{n} is rational, then it can be written in the following form:

    (\frac{a}{b}) = \sqrt{n} thus (\frac{a}{b})^2 = n

    [ii] This leads to the following equation:

    a^2 = nb^2

    a^2 is either even or odd. Lets start with even.

    [Case 1]

    Say a^2 is even, then for some p = 1, 2, 3... we have

    a^2 = 2p = nb^2

    Thus, if a^2 is even, [of the form 2p]

    then a is even, and of the form 2k
    for some natural number k = 1, 2, 3.....

    Thus, we have a = 2k and :

    a^2 = 4k^2 = 2p = nb^2

    This proves nb^2 to be even.

    If n is even, then we'd arrive at the form (for some t=1,2,3...)

    b^2 = 2t

    thus, b = 2g

    for some g =1,2,3....

    Thus a and b have a common divisor, and this is a contradiction.


    Is this correct? For this one case where a = 2k for some k = 1, 2, 3... and n = 2g for some g=1, 2, 3...? I don't know exactly how to go about proving it with the odd numbers, I tried on pen and paper using the form 2k+1 for k = 1, 2, 3... but the "common factor" never seemed to show up. Can somebody give me a hint or show me the right direction?
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  8. #8
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    [1] How would you prove that if that is always irrational?
    When you use the Unique Factorization Theorem, you get a simple proof.

    Let n be a positive integer. We can show that if \sqrt{n} is rational, then n must be a perfect square. From here it follows that if n is not a perfect square, then \sqrt{n} is irrational.

    Suppose that \sqrt{n}=a/b for a, b positive. Then a^2n=b^2 . For any prime q , let q^A, q^N, and q^B be the highest powers of q that divide a, n, and b , respectively.

    Now, because a^2n=b^2 and using the Unique Factorication Theorem, we must have 2A+N=2B and then 2|N. This means that n is a perfect square.

    This way you can even prove a more general result: For any positive integers n, m ; if \sqrt[m]{n} is rational, then n must be a perfect mth power.
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  9. #9
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    An alternative solution.

    There is a nice proof I saw once. It can be generalized to any positive integer that is not a perfect sqaure.

    Soppose that \sqrt{2}=a/b , where a, b are postive integers that are relatively prime. Then \sqrt{2}=\sqrt{2}\cdot1=\sqrt{2}\cdot(ar+bs) for some integers r, s . (Bézout's Identity)

    Because \sqrt{2}=a/b , we have a=\sqrt{2}b and b=a/\sqrt{2}. Then \sqrt{2}=\sqrt{2}ar+\sqrt{2}bs=2br+as, and \sqrt{2}ar+\sqrt{2}bs=2br+as is clearly an integer.

    This means that \sqrt{2} is an integer, but 1<\sqrt{2}<2 ; contradiction.
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