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Proving that 1 > 0

  1. Jun 23, 2011 #1
    Hi

    I am trying to prove that 1 > 0. I am going to assume a contradiction .

    Assume [tex] 1 \leqslant 0[/tex]

    First consider 1=0, Let [itex]a \in \mathbb{R}[/itex] be arbitrary. So

    [tex] a=1.a = 0.a = 0 [/tex] using field axioms. Actually 0.a=0 is not field axiom , but I have
    proved it separately .

    that means [tex] \forall \;\; a\in \mathbb{R} \Rightarrow a=0 [/tex]

    since all numbers are equal to 0 , no inverse exists in the set R. So R is not a field. Hence
    the contradiction. So

    [tex] 1 \neq 0 [/tex]

    that means we have to consider the second option

    [tex] 1 < 0 [/tex]

    Consider [itex] a > 0 \backepsilon a \in R [/itex]

    Now I have already proved another theorem.

    If q,b,c are in R and q < b , c < 0 then qc > bc

    letting c =1 and q = 0 , b= a we can say that

    (0)(1) > (a)(1) which means 0 > a or a < 0 but this is contradiction since
    we assumed that a> 0 . So our assumption that 1 < 0 , is wrong.

    Hence 1 > 0

    is my proof right ?
     
  2. jcsd
  3. Jun 23, 2011 #2

    micromass

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    Hi issacnewton! :smile:

    It might help if you list the axioms you use. Some of the comments I make will be about how I've seen the field axioms, but you might have different axioms.

    This is not a contradiction. The field axioms say that every nonzero element in R must have an inverse. But taking R={0} satisfies this: every nonzero element has an inverse, because there are no nonzero elements!!

    In fact, most field axioms take [itex]1\neq 0[/itex] as an axiom. So there's nothing to prove here. However, your axioms might do things differently...

    Who says such an a exists?

    The rest of the proof looks right.
     
  4. Jun 23, 2011 #3
    Hi Micro

    Your comments about the first part are right. Only non zero elements of R have inverse.
    I will try to modify that part. The axioms I am using doesn't say that [itex] 1 \neq 0 [/itex]
    as part of the any axiom. So I will need to come up with some contradiction.

    For the second part, I am using some [itex]a \in \mathbb{R} [/itex] which is greater than 0. Is that wrong ?
     
  5. Jun 23, 2011 #4

    micromass

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    Your axioms somehow need to rule out the possibility that R={0}. So try to see why this can't be possible.

    That isn't wrong, and I'm being extremely pedantic here. But who says that there exists an a which is greater than 0? That is, can't it happen that all elements in [itex]\mathbb{R}[/itex] are <0?? (it can't, but why not?)
     
  6. Jun 23, 2011 #5
    Here's another attempt for the first part

    Let [itex] a,b \in \mathbb{R} [/itex] and [itex] a < b [/itex]

    then [tex] 1.a < 1.b [/tex] since 1 is multiplicative identity

    [tex] 0.a < 0.b [/tex] since 1=0

    [tex] 0 < 0 [/tex] a.0 = 0 (I proved this elsewhere)

    this is a contradiction. Hence

    [tex] 1 \neq 0 [/tex]

    does it look right now ?
     
  7. Jun 23, 2011 #6

    micromass

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    Hmm, you're going to find me annoying. But how do you know there exists numbers in [itex]\mathbb{R}[/itex] such that a<b?

    May I know what field axioms you're starting from?
     
  8. Jun 23, 2011 #7
    All elements of [itex]\mathbb{R}[/itex] can't be < 0 since we need an identity element for the operation of addition. One of the axioms says that there exists a zero element(
    identity element) for the operation of addition.
     
  9. Jun 23, 2011 #8

    micromass

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    OK, but a=0 won't be good in your proof. You'll need to find an element >0...
     
  10. Jun 23, 2011 #9
    here are the axioms I am using
    1) for all [itex] a,b \in \mathbb{R} [/itex] we have [itex] a+b , a.b \in \mathbb{R}[/itex]

    2)[tex] \forall a,b \in \mathbb{R} , a+b=b+a \;\; a.b=b.a [/tex]

    3)[tex] \forall a,b,c \in \mathbb{R} ,\; (a+b)+c=a+(b+c) \;\; (a.b).c=a.(b.c) [/tex]

    4)there exists a zero element in [itex]\mathbb{R}[/itex] , denoted by 0, such that
    [itex]a+0=a \;\; \forall a \in \mathbb{R} [/itex]

    5)[itex] \forall a \in \mathbb{R}[/itex] there exists an element -a in
    [itex]\mathbb{R}[/itex] , such that [itex]a+(-a)=0[/itex]

    6) there exists an element in [itex]\mathbb{R}[/itex] , which we denote by 1, such that
    [itex]a.1=a \;\; \forall a\in \mathbb{R}[/itex]

    7)[itex]\forall a \in \mathbb{R}[/itex] with [itex] a\neq 0[/itex] there exists an element
    in [itex]\mathbb{R}[/itex] denoted by [itex]\frac{1}{a}[/itex] or [itex]a^{-1}[/itex]
    such that

    [tex] a.a^{-1}=1[/tex]

    8)[itex]\forall a,b,c \in \mathbb{R}[/itex] we have

    [tex]a.(b+c)=(a.b)+(a.c)[/tex]
     
  11. Jun 23, 2011 #10

    Hurkyl

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    Okay. Then your first challenge is to prove the following statement is not true:
    [tex] \forall a,b \in \mathbb{R}: a = b[/tex]​
    (hint: this challenge is actually impossible to complete)

    (also, those axioms don't say anything about > anyways, so you couldn't possibly use them to prove anything about orderings)
     
  12. Jun 23, 2011 #11

    micromass

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    I'm afraid that this list of axioms is incomplete then, since R={0} satisfies all these axioms. There should at least by a remark that says that 1 is distinct from 0, or even something that says that R has more than one element.

    If you check wiki's entry for a field you'll see one of the following axioms:

    This last sentence is essential.
     
  13. Jun 23, 2011 #12
    Sorry to interrupt, but strictly speaking you also need the order axioms:

    1) Trichotomy property: for all a, b in R, only one of the following holds: a < b, b < a, or a = b.

    2) Transitive property: for all a, b, c in R, if a < b and b < c, then a < c.

    3) Additive property: for all a, b, c in R, if a < b, then a + c < b + c.

    4) Multiplicative property: for all a, b, c in R, if a < b and 0 < c then ac < bc; if a < b and c < 0 then bc < ac.

    EDIT: As Hurkyl and micromass have pointed out, the additive and multiplicative identities must be distinct in any field; 0 ≠ 1.
     
  14. Jun 23, 2011 #13
    Hi Hurkyl

    I think you are also saying that [itex] \mathbb{R}\neq \{0\}[/itex] . My starting point that 1=0 lead me to conclude that [itex] \mathbb{R}= \{0\}[/itex] . Now you guys are saying that this can't be true. I don't see which of the axioms are violated. Or do I have
    to think about the completeness axiom ?

    Also some comments about the latex thing. The latex output here doesn't look so nice. Is
    there some maintenance going on ?
     
  15. Jun 23, 2011 #14
    Unit, oh yeah,

    I forgot about the order axioms. They are there.
     
  16. Jun 23, 2011 #15
    I think the axiom no 7 listed by implicitly says that [itex] 1\neq 0[/itex]

    Since we have [itex] 1.1^{-1}=1[/itex] that means , inverse of 1 exists and so
    it must be true that [itex] 1\neq 0[/itex] . am I right ?
     
  17. Jun 23, 2011 #16
    How do you know 1.1 is in R?

    The only numbers you know are in R are 1, 0, and the additive inverse of 1. Try cases, noting that 1 times 1 is 1.
     
  18. Jun 23, 2011 #17

    micromass

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    None of the axioms are violated, that's the point. There's an axiom missing that says that [itex]1\neq 0[/itex].
     
  19. Jun 23, 2011 #18
    closure property. Since [itex]1 \in \mathbb{R}[/itex]

    [tex]1.1 \in \mathbb{R}[/tex]
     
  20. Jun 23, 2011 #19

    Fredrik

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    It doesn't. However, you proved in post #1 that if 1=0, the field is the trivial field {0}.

    Most authors include the axiom 1≠0. Some don't. I would choose not to include it, and instead do what you did: Prove that the only field that doesn't satisfy this condition is trivial.
     
  21. Jun 23, 2011 #20
    Micromass, as I pointed out , doesn't that follow implicitly from axiom 7. go 2-3 posts back.

    thanks
     
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