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Math Help - Help with my proofs

  1. #1
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    Help with my proofs

    If you could check my 2 proofs and help me start on one please.

    1a) If n^3 is a multiple of 2, then n is a multiple of 2. Domain of n is all integers.

    My proof:

    An integer n is even if there exists an integer k such that n = 2k.
    n = 2k, n^3 = (2k)^3 = 8k^3 = 2(4k^3)
    4k^3 is an integer therefore n^3 is even, thus making it a multiple of 2.
    By the definition of n = 2k, n is also a multiple of 2.



    1b) \sqrt[3]{2} is an irrational number.

    My proof(by contradiction):

    \sqrt[3]{2} is a positive number such that its cube is 2.

    Assume \sqrt[3]{2} is rational.
    \exists integers P, Q such that \sqrt[3]{2} = \frac{P}{Q}, fully simplified.

    (\sqrt[3]{2})^3 = (\frac{P}{Q})^3  \rightarrow  2 = \frac{P^3}{Q^3}  \rightarrow  2Q^3 = P^3

    (Corollary: If n^3 is even, then n is even. (proven in 1a))
    An integer n is even if there exists an integer k such that n = 2k.

    2Q^3 = (2k)^3 = 8k^3  \rightarrow Q^3 = 4k^3 = 2(2k^3)

    2k^3 is an integer, therefore Q is even, along with P.

    Since P and Q are even, they share a common factor. Since we defined that  \frac{P}{Q} was fully simplified, we have a contradiction.



    Now the one I am having a hard time with is
    2) Prove or disprove the following proposition: If x and y are positive integers such that x > y + 1, then x^2 - y^2 is not prime.


    Thanks in advance
    James
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by jrh1337 View Post
    If you could check my 2 proofs and help me start on one please.

    1a) If n^3 is a multiple of 2, then n is a multiple of 2. Domain of n is all integers.

    My proof:

    An integer n is even if there exists an integer k such that n = 2k.
    n = 2k, n^3 = (2k)^3 = 8k^3 = 2(4k^3)
    4k^3 is an integer therefore n^3 is even, thus making it a multiple of 2.
    By the definition of n = 2k, n is also a multiple of 2.
    Huh
    But here you proved that if n is even, then n^3 is even... while you're asked to prove the converse.
    Actually, it's more complicated, you proved that if n^3 is in a form 2k, with k a very special number, then n is even. Hehe that was unfair

    So...what you can do is to prove the contrapositive :
    "if n is odd, then n^3 is odd"

    Let n=2k+1.
    So n^3=(2k+1)^3= \text{ even or odd ? }


    1b) \sqrt[3]{2} is an irrational number.

    My proof(by contradiction):

    \sqrt[3]{2} is a positive number such that its cube is 2.

    Assume \sqrt[3]{2} is rational.
    \exists integers P, Q such that \sqrt[3]{2} = \frac{P}{Q}, fully simplified.

    (\sqrt[3]{2})^3 = (\frac{P}{Q})^3  \rightarrow  2 = \frac{P^3}{Q^3}  \rightarrow  2Q^3 = P^3

    (Corollary: If n^3 is even, then n is even. (proven in 1a))
    An integer n is even if there exists an integer k such that n = 2k.

    2Q^3 = (2k)^3 = 8k^3  \rightarrow Q^3 = 4k^3 = 2(2k^3)

    2k^3 is an integer, therefore Q is even, along with P.

    Since P and Q are even, they share a common factor. Since we defined that  \frac{P}{Q} was fully simplified, we have a contradiction.
    It's your method, you keep it because it's okay !


    Now the one I am having a hard time with is
    2) Prove or disprove the following proposition: If x and y are positive integers such that x > y + 1, then x^2 - y^2 is not prime.
    Difference of two squares ^^

    x-y=(x-y)(x+y)
    (note that x-y < x+y since x and y are positive integers)
    The only way for it to be a prime is that x-y=1 and x+y is a prime number.
    But... It is said that x > y+1; that is to say x-y>1

    Got it ?


    Thanks in advance
    James
    Blop.
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  3. #3
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    Thanks for the reply. I shall work on it some more after class.

    But doesn't showing n is even indirectly show it is a multiple of 2?
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  4. #4
    Moo
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    Quote Originally Posted by jrh1337 View Post
    Thanks for the reply. I shall work on it some more after class.

    But doesn't showing n is even indirectly show it is a multiple of 2?
    Yes it does.
    But it's not the problem here.


    You found an expression for n^3, starting from n=2k (that is to say even). But you want to prove that n is even, you don't know it yet !

    If you want to do the direct proof, you will have to assume that 2 divides n^3. There is a theorem (not sure) that says "if a prime number p divides n^m, then p^m divides n^m". But I don't think you're supposed to use it (and you'll have to prove it).

    So do the contrapositive, starting from what I've showed you, and it'll be okay



    Nota Bene : the contrapositive of A \Rightarrow B is \neg B \Rightarrow \neg A, and if the contrapositive is true, then the original statement is true too. And conversely.
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  5. #5
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    Thanks Moo! I was able to do the problem I couldn't start, and fix my proof today.
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