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Rigorous proof

  1. Sep 27, 2009 #1
    I was asked to write a rigorous proof for the following theorem:

    0x = 0 ,for all x.

    Is the following rigorous proof correct??

    1) 0x = 0x+0.....................................................by using the axiom:for all ,a : a+0=a

    2) x+(-x) = 0....................................................by using the axiom: for all ,a: a+(-a) = 0

    3) 0x = 0x +(x+(-x))..........................................by substituting (2) into (1)

    4) 0x+(x+(-x)) = (0x+x)+(-x)...............................by using the axiom:for all a,b,c:a+(b+c)=(a+b)+c

    5) 0x = (0x+x)+(-x)............................................by substituting (4) into (3)

    6) 0x+x = x+0x..................................................by using the axiom:for all a,b:a+b=b+a

    7) 0x = (x+0x)+(-x)............................................by substituting (6) into (5)

    8) 1x = x..........................................................by using the axiom:for all,a:1a = a

    9) 0x = (1x+0x)+(-x)..........................................by substituting (8) into (7)

    10) 1x+0x = (1+0)x............................................by using the axiom: for all a,b,c:(a+b)c= ac+bc

    11) 0x = (1+0)x+(-x)..........................................by substituting (10) into (9)

    12) 1+0 = 1.....................................................by using the axiom:for all,a:a+0=a

    13) 0x = 1x+(-x)...............................................by substituting (12) into (11)

    14) 1x = x........................................................by using the axiom:for all,a:1a = a

    15) 0x = x+(-x)................................................by substituting (14) into (13)

    16) x+(-x) = 0..................................................by using the axiom:for all,a:a+(-a) = 0

    17) 0x = 0.......................................................by substituting (16) into (15)

    Thanx ,any help will be wellcomed
     
  2. jcsd
  3. Sep 27, 2009 #2

    EnumaElish

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    I see redundant steps. How about:

    Let n be an integer.
    0x = n
    0x - x = n - x
    (0 - 1)x = n - x
    -x = n - x
    x - x = n
    What can you say about n?
     
  4. Sep 27, 2009 #3
    Thanks ,a good way to find how much is 0x.

    B,t.w, x is a real No
    There is an even shorter proof :

    0x =0 <===> 0x +x =0+x <===> x(0+1) = x <===> x=x.

    But in a rigorous proof we must show the axioms involved .

    Where are the redundant steps??

    Thanks again
     
  5. Sep 27, 2009 #4

    EnumaElish

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    You are correct, I should have written "n is real."

    As for redundancy, I think you can start with 6; because 0x + x = x + 0x as a postulate. I am not saying you are wrong, but I do not see why you cannot start with 6.
     
    Last edited: Sep 27, 2009
  6. Sep 27, 2009 #5

    statdad

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    "But in a rigorous proof we must show the axioms involved."

    No, in a rigorous proof you must make sure details are explicit. Stating the axioms is not a requirement.

    Is there a reason for this discussion?
     
  7. Sep 27, 2009 #6
    yes i agree after that statement there is no reason for further discussion
     
  8. Sep 27, 2009 #7

    EnumaElish

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    evagelos, I think statdad is referencing to widely accepted standards of proof in general math, statistics, and related fields.

    If your instructor/professor has explicitly asked you to state each axiom, then those specific instructions take precedence over general ones.
     
  9. Sep 27, 2009 #8

    statdad

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    EnumaElish is correct; my response was a little terse, my apologies, but I simply did not want to get involved in another long "gotcha" post involving what is and what is not a "rigorous proof".

    Never email in haste - never post on a forum in haste: words I need to live by.
     
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