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Math Help - Another proof

  1. #1
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    Another proof

    Assume x,y e N. If y != 0 (not equal) , then x + y > x.

    Again, this is obvious, but Im having trouble giving a solid mathematical proof.
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    Quote Originally Posted by jzellt View Post
    Assume x,y e N. If y != 0 (not equal) , then x + y > x.

    Again, this is obvious, but Im having trouble giving a solid mathematical proof.
    is this for a set theory class? i just want to know how rigorous your proofs have to be. here are some simple approaches, they probably wouldn't hold up in a set theory class, but they'd be ok in most other classes:

    (1) let x be fixed and do induction on y

    (2) a proof by contradiction. show that x + y \le x is false if y is not zero.

    (3) since y is a natural number that is not zero, it must be greater than 0 (since zero is the least element). so we have y > 0. now add x to both sides.
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    Quote Originally Posted by jzellt View Post
    Assume x,y e N. If y != 0 (not equal) , then x + y > x.

    Again, this is obvious, but Im having trouble giving a solid mathematical proof.
    You prove it by induction. The base case is y=1, it is true because x < x+1 since x\in x \cup \{ x\}.
    If x<x+y then x<(x+y)+1 = x+(y+1).
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  4. #4
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    Quote Originally Posted by jzellt View Post
    Assume x,y e N. If y != 0 (not equal) , then x + y > x.

    Again, this is obvious, but Im having trouble giving a solid mathematical proof.

    For that you need the following 2 axioms and a definition:

    1) Axiom: for all x,yεN ...x+y'=(x+y)'...........where y' is the successor of y,hence ,if y=5 then y' =5'=6....and in symbols:

    ........ \forall x\forall y[ x+y'=(x+y)'].................................................. ..........................................1


    2) Axiom: for all xεN,....0+x=x............and in symbols.....


    .......... \forall x( 0+x=x).................................................. ........................................2


    3) Definition: for all r,s εN ......r<s <===> there exists uεN and (r+u)'=s ,and in symbols:

    ....... \forall r\forall s[r<s\Longleftrightarrow\exists u( (r+u)'=s)].................................................. .................................................. ..3


    Now for the proof:

    NOTE we want to prove x<x+y.According to the definition (2) we must prove that there exists z such (x+z)'=x+y.


    So,since  y\neq 0 ,0<y and by definition (2) there exists z such that (0+z)'=y ,or z'=y by using axiom (2)............................................... .................................................. 4


    But ..........x+y=x+y,or x+z'=x+y by using (4),or (x+z)'=x+y by using axiom (1)

    Hence there exists z such that (x+z)'=x+y which implies :


    ..................................x<x+y........... ..................................................
    Last edited by archidi; February 15th 2009 at 06:54 AM. Reason: not finished yet
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