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Math Help - More Proofs

  1. #1
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    More Proofs

    I have three proofs that I am not really sure how to do. I will provide you with what I have so far, if you could please fill in the holes that would be great.

    1) Prove that the product of two consecutive integers is even.

    Pf: Let x,y be integers such that y = x + 1. Then xy = x(x+1) = x^2+x. There exists some integer c such that c = 2|(x^2+x).

    2) For x > 0 and y>o, prove that x < y if and only if x^2 < y^2

    => If x < y then x^2 < y^2.

    Pf: Let x and y be integers such that y = x +5. x^2 = x^2 and y^2 = (x+5)^2 = x^2 + 10x + 25.

    <= if x^2 < y^2 then x < y

    Pf: Let x^2 and y^2 be integers such that y^2 = x^2 + 1.

    3) For n>2 prove or dispirove that n^2 -1 is composite.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by algebrapro18 View Post
    I have three proofs that I am not really sure how to do. I will provide you with what I have so far, if you could please fill in the holes that would be great.

    1) Prove that the product of two consecutive integers is even.

    Pf: Let x,y be integers such that y = x + 1. Then xy = x(x+1) = x^2+x. There exists some integer c such that c = 2|(x^2+x).
    i don't think that does the job

    use the fact that for consecutive integers, one has to be even and one odd. so without loss of generality, assume the first integer is even. then the integers are given by 2n and 2n + 1 for some integer n. clearly 2n(2n + 1) = 2(2n^2 + n) is even. this completes the proof
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by algebrapro18 View Post
    2) For x > 0 and y>o, prove that x < y if and only if x^2 < y^2

    => If x < y then x^2 < y^2.

    Pf: Let x and y be integers such that y = x +5. x^2 = x^2 and y^2 = (x+5)^2 = x^2 + 10x + 25.

    <= if x^2 < y^2 then x < y

    Pf: Let x^2 and y^2 be integers such that y^2 = x^2 + 1.
    i have no idea what you are doing here. i will prove one direction, the other is similar and i leave it to you.

    we show that if x^2 < y^2 then x < y (for x,y > 0)

    assume x^2 < y^2

    => x^2 - y^2 < 0

    => (x + y)(x - y) < 0

    => x + y < 0 or x - y < 0

    => x < -y or x < y

    clearly we must have x < y, since x < -y => x < 0

    this completes the proof for the (<=) direction
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  4. #4
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    thanks for the quick responce and I can't believe I made such a silly oversite...apparently I am good at making thing harder than they need to be...
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by algebrapro18 View Post
    3) For n>2 prove or dispirove that n^2 -1 is composite.
    note that n^2 - 1 = (n + 1)(n - 1).

    claim: (n + 1)(n - 1) is composite for n > 2

    there are two cases to consider: (1) n is even, and (2) n is odd


    case 1:
    if n is odd, then we can write n = 2y + 1 for some integer y

    thus, (n + 1)(n - 1) = (2y + 2)(2y) = 2(2y^2 + 2y) which is an even integer, and is therefore composite.


    case 2:
    if n is even, then we can write n = 2x for some integer x

    thus, (n + 1)(n - 1) = (2x + 1)(2x - 1) = 4x^2 - 1

    i believe this will always give a composite number, i don't remember how to prove it--i'm not a number theorist. (can you?)

    this completes the proof (if you can prove the last bit that is )
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    so apparently i was making this last question a lot more complicated than it really is.(*)

    just noting that n^2 - 1 = (n + 1)(n - 1) is enough to show that this is composite. because clearly this is an integer that is divisible by both of the integers (n + 1) and (n - 1) and is therefore not prime



    *) TPH helped me to see the error in my complicated ways
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