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Math Help - 1>0 proof

  1. #1
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    1>0 proof

    I am trying to write this proof for class. I am trying to prove by contradiction so I try to prove that 0 is greater than or equal to 1.
    "Case 1" gives 1>0 which is a contradiction.
    "Case 2" is 1=0, which I can't disprove. Anyone have a proof for that? I've tried a=1 and playing with the algebra (this seems like the right track, though I'm not sure), but I can't get anywhere from there.
    I have a test tomorrow, so either a proof that 1 doesn't equal 0 or a big hint would be greatly appreciated.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by reilyj View Post
    I am trying to write this proof for class. I am trying to prove by contradiction so I try to prove that 0 is greater than or equal to 1.
    "Case 1" gives 1>0 which is a contradiction.
    "Case 2" is 1=0, which I can't disprove. Anyone have a proof for that? I've tried a=1 and playing with the algebra (this seems like the right track, though I'm not sure), but I can't get anywhere from there.
    I have a test tomorrow, so either a proof that 1 doesn't equal 0 or a big hint would be greatly appreciated.
    to prove something like this, you need to rely on axioms. what axioms are you using?

    you can use the axiomatic definitions of 1 and 0 respectively to prove they are not equal

    and if 1 < 0 that would mean 1 is negative. what do the axioms regarding "1" have to say against this?
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  3. #3
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    i can use the associative laws, the communitative laws, and the distributive law.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by reilyj View Post
    i can use the associative laws, the communitative laws, and the distributive law.
    those won't help too much here.

    usually the axioms regarding 1 and 0 are:

    1*x = x for all x

    x + 0 = x for all x

    respectively
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  5. #5
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    yeah I can use those
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by reilyj View Post
    yeah I can use those
    ok, so use those.

    disproving 1 < 0 is easy.

    assume 1 is negative. then 1*1 = 1 by the afore mentioned axiom. but that means 1^2 is negative, clearly a contradiction, as a negative number times a negative one should yield a positive number.

    now, try to use the axioms to disprove 1 = 0
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  7. #7
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    cool, thanks for the help buddy I got it. Sorry about the double post. catch you later.
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  8. #8
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    i seen this question a few times, you can do a contradiction...
    Suppose 1<=0, case 1. 1=o, then 0<=0, which holds, now consider 1not=0, if you are using Kirdwood books is theorem 1-7e) that states anot+=0 then a^2>0, so 1^2<=0, contradiction to the theorem....
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  9. #9
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    And both of those proofs involve axioms reilyj did not mention:

    If a> 0 and b> 0 then ab> 0

    If a> 0 and b> 0 then a+ b> 0

    If a and b are two numbers then one and only one of these must hold:
    i) a= b
    ii) a> b
    iii) b> a
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