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  1. #1
    Sea Torques
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    Help with math proof.

    Let a and b be integers. Assume 3|(a² + b²). Prove 3|a and 3|b.
    I need some help here. I'm not sure how to even begin. I've been trying to use different forms of the implication, but the contrapositive has me working with a not equal sign and I don't know how to prove something using an inequality; the contradiction has me working with the assumption, which I don't know how to take apart. I'm not asking for a full proof, just a nod in the right direction would suffice.

  2. #2
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    Your proof statement's not showing up.

  3. #3
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    yeah, actually have to take a picture of it or whatever

    that or i'm proving and ,

  4. #4

    Let's play, "guess what he's being asked to prove" based only on the information in the OP. My money is on m=gcd(a,b) and show m is a linear combination of a and b.

  5. #5
    okay guy I guess
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    if i is a variable, as opposed to the imaginary number, and if u is a different variable how do I simplify:
    2i<6u
    ?

  6. #6
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    If A is some real number greater than zero and f: R→R is a function such that |f(x)| ≤ A for all x in R, then prove that lim x → 0 Axf(x) = 0.

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    If A is a ray defined as A(t) = P + t(Q - P), 0 <= t <= 1, and B is a triangle defined by the points C, D and E, prove that A intersects B when ((P - Q) . ((D - C) x (E - C))) > 0 and provide a general method to calculate the intersection point.

  8. #8
    Ridill
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    Quote Originally Posted by Qalbert View Post
    if i is a variable, as opposed to the imaginary number, and if u is a different variable how do I simplify:
    2i<6u
    ?
    icwutudidthar

  9. #9
    Sea Torques
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    Pfft Mathematica. Here it is again:

    Let a and b be integers. Assume 3|(a² + b²). Prove 3|a and 3|b.

  10. #10
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    This statement is the converse of a statement that I have already proved true. Can I then take the inverse of the original statement (which would be three doesn't divide a or three doesn't divide b implies three doesn't divide...), then prove the inverse true by contradiction, and since the inverse is true that means the converse is true, which implies that the converse, my proposition here, is true?

  11. #11
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    Quote Originally Posted by Avé View Post
    Pfft Mathematica. Here it is again:

    Let a and b be integers. Assume 3|(a² + b²). Prove 3|a and 3|b.
    What's | mean?

  12. #12
    Sea Torques
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    Divides. In other words, Assume 3m = (a² + b²). Prove 3n=a and 3p=b.

  13. #13

    I hope this isn't too much of a nod, but assuming you have done some modular arithmetic, take a look at the possibilities for a and b mod 3 and what they would mean for a^2 + b^2.

  14. #14
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    Quote Originally Posted by Avé View Post
    This statement is the converse of a statement that I have already proved true. Can I then take the inverse of the original statement (which would be three doesn't divide a or three doesn't divide b implies three doesn't divide...), then prove the inverse true by contradiction, and since the inverse is true that means the converse is true, which implies that the converse, my proposition here, is true?
    Yep you can do this. Logically, it would look like this ( '^' denotes 'and,' 'v' denotes 'or'):

    1) x => y ^ z (where x is 3 l (a^2 + b^2), y is 3 l a and z is 3 l b)
    2) ~(y ^ z) => ~x (where ~ means not) by transposition
    3) ~(y ^ z) => ~y v ~z by DeMorgans Law
    4) ~y v ~z => ~x

    So if you prove that 3 X a v 3 X b => 3 X a^2 + b^2 (where 'X' means 'does not divide') then you will have proven 3 l a^2 + b^2 => 3 l a ^ 3 l b.

    So yea, logically what you said is correct so I don't see why you can't prove it that way.

  15. #15

    Quote Originally Posted by Avé View Post

    Let a and b be integers. Assume 3|(a² + b²). Prove 3|a and 3|b.

    I need some help here. I'm not sure how to even begin. I've been trying to use different forms of the implication, but the contrapositive has me working with a not equal sign and I don't know how to prove something using an inequality; the contradiction has me working with the assumption, which I don't know how to take apart. I'm not asking for a full proof, just a nod in the right direction would suffice.
    I can try and put you in the right direction.

    First, this is what you should know:

    (you're probably not going to use all of these for this proof, but some are interesting to know)

    1. If a|b and a|c then a|(b + c)
    2. If a|b and a|c then a|(b − c)
    3. If a|b then a|(bc)
    4 . If a|b and b|c then a|c
    5. If a|b and b|a then a = ±b
    6. If a|b and a and b are both > 0 then a ≤ b
    7. If m != 0 and a|b then (am)|(bm)

    Note: by "!=" I mean "is not equal to" and by "X" I mean "does not divide"

    I would start like this:
    (Keep in mind that there are over 9000 ways of solving this)

    Assume aXb and aXc.
    After some crunching, you will end up with:
    Therefore 3X(a² + b²), which is a contradiction. So a|b and a|c.

    OR … another way of doing this (a little harder) is to start by assuming 3X(a² + b²) and proceed in a similar way as above.

    Hope this helped!

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