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Math Help - Help with mod and odds

  1. #1
    polish13
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    Help with mod and odds

    Questions

    Prove or give counter examples for the following problems

    1) For any integer n, (n^2 + 2n)mod4 is equal to 0 or 3.

    2) Suppose m, n, and d are integers, and m mod d = n mod d. Prove that dl(m - n) (says d divides (m minus n).

    3) If n^3 is odd, then n is odd

    4) If a, b, and c ate integers and a^2 + b^2 = c^2, then at least one of a, b, and c is odd

    5) Prove that the cube root of 4 is irrational

    Please help
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by polish13
    3) If n^3 is odd, then n is odd
    Either n \equiv 0 \pmod{2} meaning that n is even or n \equiv 1 \pmod{2} meaning that n is odd. So assume that n is equivalent to 1. Then n^3=1*1*1=1. Thus n^3 \equiv 1 \pmod{2}.

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by polish13

    1) For any integer n, (n^2 + 2n)mod4 is equal to 0 or 3.
    Either n is even or it is odd. If n is even then n is equivalent to 0 or 2 (mod 4). In the first case n^2=0*0=0, and 2n=2*0=0. So n^2+2n is equivalent to 0+0=0 (mod 4). In the second case n^2=2*2=4->0 (mod 4) and 2n=2*2=4->0 (mod 4), so n^2+2n is equivalent to 0+0 (mod 4).

    Now assume n is odd. So n is equivalent to 1 or 3 (mod 4). In the first case n^2=1*1=1 and 2n=2*1=2. So n^2+2n=1+2=3 (mod 4). In the second case n^2=3*3=9->1 (mod 4) and 2n=2*3=6->2 (mod 4). So n^2+2n=1+2=3 (mod 4).

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by polish13

    4) If a, b, and c ate integers and a^2 + b^2 = c^2, then at least one of a, b, and c is odd
    Using my previous two posts, you should find it easy to say that if n is even then n^2 is even and if n is odd then n^2 is odd.

    So all we need to consider is if a^2, b^2, or c^2 are even or odd. Let's take the (essentially three) cases:
    1)a^2, b^2 are both even
    2)a^2 is even, b^2 is odd (equivalent to a^2 is odd, b^2 is even)
    3)a^2, b^2 are both odd.

    Let's take 1) last.
    2) a^2+b^2 is even+odd = odd. So c^2 is odd.
    3) a^2+b^2 is odd+odd = even. So c^2 is even. Note that we already have two odds in the problem so we are done.

    1) a^2+b^2 is even+even = even. So c^2 is EVEN! This implies that c is even. This means that your statement is FALSE! (Example: a=6, b=8, c=10).

    -Dan
    Last edited by topsquark; March 6th 2006 at 04:17 AM. Reason: Goof
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by polish13
    5) Prove that the cube root of 4 is irrational
    I'm not sure of which method you need to use here. I can't say I know of one using modular Mathematics because we're dealing with the possibility of rational solutions to the equation x^3-4=0. However, there is a simple proof using the Rational Roots test:
    If x^3-4=0 has a rational root it will be of the form: \pm \frac{1}{1}, \, \pm \frac{2}{1}, \, \pm \frac{4}{1}. As none of these possibilities work, x has no rational solution. Thus \sqrt[3]{4} is irrational.

    -Dan
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  6. #6
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    Quote Originally Posted by polish13
    Questions
    5) Prove that the cube root of 4 is irrational

    Please help
    Based on Euclid's proof,
    Assume that,
    \sqrt[3]{4}=\frac{p}{q}
    Then,
    \frac{p^3}{q^3}=4
    Thus,
    p^3=2^2q^3
    Notice when you use the fundamental theorem of arithmetic the right hand has a number of prime factors not divisible by 3, while the left hand side does. Thus, an impossibility because of the uniquess of factorization into primes.
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