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Math Help - Prove that sqrt(5) is irrational?

  1. #1
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    Prove that sqrt(5) is irrational?

    My work

    Suppose sqrt(5) is rational

    5 = (m/n)^2

    5 = m^2/n^2 or m^2 = 5n^2

    Since m^2 is a multiple of 5, m must also be a multiple of 5? - proof?

    m = 5k for some int k

    5n^2 = (5k)^2

    5n^2 = 25k^2

    n = 5k^2

    n and m are both multiples of 5 this contradicts the assumption sqrt is irrational


    end of work


    My professor didnt like what i wrote in bold. I suppose she wanted me to write a proof out for this? how would I proceed? Is this broken up into cases?


    thank you
    ps. Im sorry I have so many topics in this area. Im trying my hardest to get by
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  2. #2
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    Quote Originally Posted by Thetheorycase View Post
    Since m^2 is a multiple of 5, m must also be a multiple of 5? - proof?
    There's a result called Euclid's Lemma: If a divideds bc and gcd(a,b)=1, then a divides c .
    In particular, if p|ts, where p is prime, then p|t or p|s .
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  3. #3
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    Do you think i should have omitted that line altogtehr then because Euclid lemma comes in the section after?? I thought maybe she wanted me to use the quotient remainder theorem?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Thetheorycase View Post
    Do you think i should have omitted that line altogtehr then because Euclid lemma comes in the section after?? I thought maybe she wanted me to use the quotient remainder theorem?
    you could also prove it by the fundamental theorem of arithmetic (proof by contradiction). If you assume m is not a multiple of 5, it contradicts the uniqueness of prime factorization that the theorem asserts.

    you could also do this proof another way, using the rational roots theorem. have you covered that?
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  5. #5
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    "If you assume m is not a multiple of 5, it contradicts the uniqueness of prime factorization that the theorem asserts."

    This.. Could you show me how it is done?



    would it be broken up into 4 cases?

    1) m = 5k +1
    2) m = 5k + 2
    3) m = 5k + 3
    4) m = 5k + 4
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Thetheorycase View Post
    "If you assume m is not a multiple of 5, it contradicts the uniqueness of prime factorization that the theorem asserts."

    This.. Could you show me how it is done?



    would it be broken up into 4 cases?

    1) m = 5k +1
    2) m = 5k + 2
    3) m = 5k + 3
    4) m = 5k + 4
    well, no. not if i want to use the theorem i asserted.

    We know that each integer is a product of primes, each to some positive integer power, and this product is unique for each integer. we get this from the said theorem. so now, if we take \displaystyle m = p_1^{n_1} \cdot p_2^{n_2} \cdots p_k^{n_k}, where the p_i's are some primes, and the n_j's are some powers, we would see that:

    \displaystyle m^2 = 5n^2 \implies p_1^{2n_1} \cdot p_2^{2n_2} \cdots p_k^{2n_k} = 5n^2

    Now, if we assume to the contrary that \displaystyle m is not a multiple of 5, we would get that in the above equation, there is no factor of 5 on the left hand side, but there is a factor of 5 on the right hand side. this contradicts the fact that each integer has a unique prime factorization, as there is a prime factor on one side that's not on the other (that is, we can represent m with two different prime factorizations. absurd!) this contradicts the uniqueness of prime factorization theorem. hence, m must be a multiple of 5.

    now of course, you can say all that much more concisely.
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