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Math Help - [SOLVED] Proving the irrationality of √3.

  1. #1
    Member ilikedmath's Avatar
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    Question [SOLVED] Proving the irrationality of √3.

    I am currently in Introduction to Real Analysis I. This is a homework problem I'm struggling with.

    Here is how I went about my proof of "√3 is irrational":

    Claim: √3 is irrational.
    Proof: Assume √3 is rational. Thus √3 = p/q where p, q are integers and have no common factors. Since √3 = p/q then 3 = p/q. Thus 3q = p. Hence p^2 is divisible by 3 which implies p is divisible by 3 [see (*) proof]. Since p is divisible by 3, p = 3k for some integer k. Now 2q = p becomes
    3q = (3k)
    3q = 9k
    q = 3k. Therefore q and hence q are divisible by 3. C! (contradiction) to p and q having no common factors. Therefore √3 is irrational. QED.

    Here is the additional proof I am including (*):

    Claim: a is a multiple of 3 → a is a multiple of 3
    Proof: Assume a is not a multiple of 3, for some integer a. So a = 3k is not true for any integer k. So a = 3k + 1 or a = 3k + 2 for some integer k.

    Case I: Assume a = 3k + 1.
    a = 9k + 6k + 1
    a = 3(some integer) + 1
    a does not equal 3k for any integer k
    Therefore a is not a multiple of 3

    Case II: Assume a = 3k + 2.
    a = 9k + 12k + 4
    a = 9k + 12k + 3 + 1
    a = 3(some integer) + 1
    a = 9k + 12k + 4
    Therefore a is not a multiple of 3.

    ----
    Any help, suggestions, corrections, and tips are appreciated. Thanks for your time and help!
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by ilikedmath View Post
    I am currently in Introduction to Real Analysis I. This is a homework problem I'm struggling with.

    Here is how I went about my proof of "√3 is irrational":

    Claim: √3 is irrational.
    Proof: Assume √3 is rational. Thus √3 = p/q where p, q are integers and have no common factors. Since √3 = p/q then 3 = p/q. Thus 3q = p. Hence p^2 is divisible by 3 which implies p is divisible by 3 [see (*) proof]. Since p is divisible by 3, p = 3k for some integer k. Now 3q = p becomes
    3q = (3k)
    3q = 9k
    q = 3k. Therefore q and hence q are divisible by 3. C! (contradiction) to p and q having no common factors. Therefore √3 is irrational. QED.
    Perfect

    Here is the additional proof I am including (*):

    Claim: a is a multiple of 3 → a is a multiple of 3
    Proof: Assume a is not a multiple of 3, for some integer a. So a = 3k is not true for any integer k. So a = 3k + 1 or a = 3k + 2 for some integer k.

    Case I: Assume a = 3k + 1.
    a = 9k + 6k + 1
    a = 3(some integer) + 1
    a does not equal 3k for any integer k
    Therefore a is not a multiple of 3

    Case II: Assume a = 3k + 2.
    a = 9k + 12k + 4
    a = 9k + 12k + 3 + 1
    a = 3(some integer) + 1
    ------a = 9k + 12k + 4------
    Therefore a is not a multiple of 3.
    Yes, this is a way of proving it. But you can prove it for any prime number p :

    p divides a iff then p divides a.

    Suppose the prime factorisation of a :

    a=\prod_{i=1}^n p_i^{\alpha_i} (where p_i are all prime numbers dividing a and p_i \neq p_j \quad \forall i \neq j)
    We have :
    a^2=\prod_{i=1}^n p_i^{2 \alpha_i} (using basic exponents rules)

    Any prime number dividing a has an even power. So if p divides a, at least p also divides a.
    So p divides a.




    Huh...my proof is experimental, so if you have any remark on this...
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  3. #3
    Member ilikedmath's Avatar
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    Thumbs up thank you

    thanks for looking at this and for the proof of primes there.
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