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Math Help - Need help proving a couple theorems

  1. #1
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    Need help proving a couple theorems

    Prove the following theorems.

    1) If x is any number, then 0*x = 0.

    2) If x is any number, there do not exist two numbers y and y' such that x*y = 1 and x*y' = 1.

    Use any of the following axioms to help prove the above theorems:
    1) If each of x and y is a number, then x*y is a number.
    2) If each of x and y is a number, then x*y = y*x.
    3) If each of x, y, and z is a number, then (x*y)*z = x*(y*z).
    4) There is a number U such that if x is any number, then U*x = x.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by noles2188 View Post
    Prove the following theorems.

    1) If x is any number, then 0*x = 0.
    this is actually an interesting one. you usually see this as an axiom. not always, but a lot of the time.

    0 \cdot x = 0 \cdot (U \cdot x) ....axiom (4)

    = 0 \cdot \underbrace{(U + U + \cdots + U)}_{x \text{ times}}

    = 0 \cdot U + 0 \cdot U + \cdots + 0 \cdot U.....distributive property

    = 0 + 0 + \cdots + 0 .......axiom (4)

    = 0

    2) If x is any number, there do not exist two numbers y and y' such that x*y = 1 and x*y' = 1.
    i suppose you mean distinct numbers y and y', of course.

    assume to the contrary that there does exist distinct numbers y and y' such that

    x*y = 1.........(a) and,
    x*y' = 1 ........(b)

    subtracting (b) from (a) we get

    x*y - x*y' = 0

    => x(y - y') = 0

    thus, x = 0 or y = y'

    in either case, we have a contradiction. (do you see why? in particular, why does x = 0 yield a contradiction?)
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