1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Proving that 1 > 0

  1. Jun 23, 2011 #1

    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
    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
    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
    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
    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

  11. Jun 23, 2011 #10


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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
    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
    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook