Page 2 of 2 FirstFirst 12
Results 16 to 26 of 26
Like Tree2Thanks

Math Help - equivalence

  1. #16
    Member
    Joined
    Jul 2011
    Posts
    121
    Thanks
    1

    Re: equivalence

    Quote Originally Posted by Deveno View Post
    We want to show (3) --> (1). Therefore, we assume (3) to be true, and derive (1) conditionally.

    .
    YOU have not shown : [(ac>bc & c>0) => a>b] =>1>0

    But you have shown : {[(ac>bc & c>0) => a>b]&c>0} =>1>0

    There is a big difference between the two
    Follow Math Help Forum on Facebook and Google+

  2. #17
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,880
    Thanks
    742

    Re: equivalence

    Quote Originally Posted by psolaki View Post
    YOU have not shown : [(ac>bc & c>0) => a>b] =>1>0

    But you have shown : {[(ac>bc & c>0) => a>b]&c>0} =>1>0

    There is a big difference between the two
    Deveno seems correct in everything he wrote. I don't understand your argument against it. Here is an expanded proof (the second part is not necessary, but maybe it will better explain things).

    Assume (ac>bc \wedge c>0)\Rightarrow a>b. Choose any c>0. Then c = 1\cdot c > 0\cdot c = 0. This gives a=1,b=0, so according to the implication we assumed, 1>0. Next, choose any c\le 0. Now, the implication is not relevant since we only care when c>0. So, we don't have to use every possible value for c. We are showing that given that implication, we can arrive at the conclusion that 1>0. We can have c\le 0, a=1, b=0, and the implication is still true. It is simply not relevant.
    Follow Math Help Forum on Facebook and Google+

  3. #18
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,392
    Thanks
    759

    Re: equivalence

    Quote Originally Posted by psolaki View Post
    YOU have not shown : [(ac>bc & c>0) => a>b] =>1>0

    But you have shown : {[(ac>bc & c>0) => a>b]&c>0} =>1>0

    There is a big difference between the two
    Good luck in your uphill battle to teach me how to think. While as prone to error as the next person, in this particular instance, I am quite certain I am correct.

    I am always quite amused when someone invokes Modus Ponens as if it were the gods' gift to mankind. When a mathematician says: "If A holds, then B holds" she is not interested in what happens if A does not hold. In fact, whether or not A is even true isn't relevant. But if it (A) is, then B must be true, or we've got a spanner in the works.

    There is no need to add an extra "c > 0" statement. We are allowed to assume that in the statement of (3). If, in fact, c is not greater than 0, we're not talking about (3) anymore, but some other statement in which, frankly, I have no interest.

    What is, indeed, your point? Do you wish to further your understanding of real numbers, or do you just want to engage in futile bickering? You asked for help with a problem on a math help forum. I gave you what seemed a relevant response. If you want to argue about whether or not "I'm right", I'm not interested. I'm not here to show off, just to help. If you don't really want help, fine. Not my problem.
    Follow Math Help Forum on Facebook and Google+

  4. #19
    Member
    Joined
    Jul 2011
    Posts
    121
    Thanks
    1

    Re: equivalence

    Quote Originally Posted by SlipEternal View Post
    Deveno seems correct in everything he wrote. I don't understand your argument against it. Here is an expanded proof (the second part is not necessary, but maybe it will better explain things).

    Assume (ac>bc \wedge c>0)\Rightarrow a>b. Choose any c>0. Then c = 1\cdot c > 0\cdot c = 0. This gives a=1,b=0, so according to the implication we assumed, 1>0. Next, choose any c\le 0. Now, the implication is not relevant since we only care when c>0. So, we don't have to use every possible value for c. We are showing that given that implication, we can arrive at the conclusion that 1>0. We can have c\le 0, a=1, b=0, and the implication is still true. It is simply not relevant.
    `

    If you choose c>0 ,then you Have to choose c\leq 0 this dictated by the trichotomy law.

    So you have to show what happens when c\leq 0 otherwise you violate the symmetry of the proof

    Anyway ,i think your proof is not the same with that Deveno wrote
    Follow Math Help Forum on Facebook and Google+

  5. #20
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,880
    Thanks
    742

    Re: equivalence

    Quote Originally Posted by psolaki View Post
    `

    If you choose c>0 ,then you Have to choose c\leq 0 this dictated by the trichotomy law.

    So you have to show what happens when c\leq 0 otherwise you violate the symmetry of the proof

    Anyway ,i think your proof is not the same with that Deveno wrote
    Another way to prove (3) implies (1) is a proof by contrapositive. Show that not (1) implies not (3). Not (1) is 1 \le 0, which is false by the axioms of the real numbers. Since a false statement implies anything, we don't even need to check the second part of the implication. It is automatically true.
    Follow Math Help Forum on Facebook and Google+

  6. #21
    Member
    Joined
    Oct 2009
    From
    Detroit
    Posts
    158
    Thanks
    5

    Re: equivalence

    You guys are getting all confused. psolaki's using a propositional logic system. The only thing he's doing wrong is not recognizing that 1>0 is a tautology, it follows from any set of premises including the empty set, and so you can just assert it freely.
    Follow Math Help Forum on Facebook and Google+

  7. #22
    Member
    Joined
    Jul 2011
    Posts
    121
    Thanks
    1

    Re: equivalence

    Quote Originally Posted by SlipEternal View Post
    Another way to prove (3) implies (1) is a proof by contrapositive. Show that not (1) implies not (3). Not (1) is 1 \le 0, which is false by the axioms of the real numbers. Since a false statement implies anything, we don't even need to check the second part of the implication. It is automatically true.
    In a similar way then we can prove:

    a\neq 0\Longrightarrow 1>0

    Using contrapositive we must show that : 1\leq 0 implies a=0 and since 1\leq 0is false and false can imply anything we can say a=0 is true
    Follow Math Help Forum on Facebook and Google+

  8. #23
    Member
    Joined
    Jul 2011
    Posts
    121
    Thanks
    1

    Re: equivalence

    Quote Originally Posted by Deveno View Post
    Good luck in your uphill battle to teach me how to think. While as prone to error as the next person, in this particular instance, I am quite certain I am correct.

    I am always quite amused when someone invokes Modus Ponens as if it were the gods' gift to mankind. When a mathematician says: "If A holds, then B holds" she is not interested in what happens if A does not hold. In fact, whether or not A is even true isn't relevant. But if it (A) is, then B must be true, or we've got a spanner in the works.

    There is no need to add an extra "c > 0" statement. We are allowed to assume that in the statement of (3). If, in fact, c is not greater than 0, we're not talking about (3) anymore, but some other statement in which, frankly, I have no interest.

    What is, indeed, your point? Do you wish to further your understanding of real numbers, or do you just want to engage in futile bickering? You asked for help with a problem on a math help forum. I gave you what seemed a relevant response. If you want to argue about whether or not "I'm right", I'm not interested. I'm not here to show off, just to help. If you don't really want help, fine. Not my problem.
    Your help is allways inspring me and of great help.
    Follow Math Help Forum on Facebook and Google+

  9. #24
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,880
    Thanks
    742

    Re: equivalence

    Quote Originally Posted by psolaki View Post
    In a similar way then we can prove:

    a\neq 0\Longrightarrow 1>0

    Using contrapositive we must show that : 1\leq 0 implies a=0 and since 1\leq 0is false and false can imply anything we can say a=0 is true
    Not quite. You cannot say a=0 is true. A false antecedent does not imply that the sequent is true. The implication is a true statement whether the sequent is true OR false.
    Follow Math Help Forum on Facebook and Google+

  10. #25
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,392
    Thanks
    759

    Re: equivalence

    Quote Originally Posted by bkbowser View Post
    You guys are getting all confused. psolaki's using a propositional logic system. The only thing he's doing wrong is not recognizing that 1>0 is a tautology, it follows from any set of premises including the empty set, and so you can just assert it freely.
    Depends on the scope of the problem (what is meant by 1, what is meant by > and what is meant by 0). As an example, in a finite field there is no notion of > that is compatible with the field operations, and 1 > 0 doesn't even make sense.

    So what we can prove in showing, or not being able to show, the equivalence of these statements, depends on what is assumed beforehand. Do we have a field? Which field? An ordered field? Any order, or some particular one? A real-closed field? The real numbers? Which axioms are specified (there are a number of equivalent formulations, and what we can freely use in a proof depends on which formulation we are using)?

    From a logical standpoint, the variables in the statement of the OP should be quantified, and the constants 1 and 0, defined. The general idea I used in my original post is called "instantiation" that is:

    $(\forall x \in S: P(x)) \implies [(c \in S) \implies P(c)]$

    (this is also called universal quantifier elimination).

    The difficulty in showing that the set of positive reals is non-empty (or indeed, that the set of real numbers itself is non-empty) is a valid objection: HOWEVER, if one is using the "positive cone" (or the trichotomy rule, which is an equivalent formulation) concept to define the order on $\Bbb R$, this is moot:

    we have 3 subsets of $\Bbb R$ that form a partition: $P - \{0\},\{0\}$ and $-P - \{0\}$. 1 lies in one of these 3 sets, and since 1 is a square (of itself, by definition of multiplicative identity), it lies in $P -\{0\}$.

    (I might add at this point there is a deep connection between the set of squares, and the ordering in a real-closed field, of which the real numbers are ONE example).

    I find it very likely that the OP is actually working from an axiomatic definition of the real numbers, it would be helpful to know "which" axiomatic definition. Again, I find it amusing he gets finicky about which impllication of an implication we are actually proving, and neglects the even more important omission of what sets we are quantifying over.
    Follow Math Help Forum on Facebook and Google+

  11. #26
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,880
    Thanks
    742

    Re: equivalence

    @Deveno, in post #6, the OP says that any property or theorem of the real numbers may be used. I assumed that included the property that 1>0, making its statement a tautology.
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: November 18th 2013, 10:09 AM
  2. Replies: 2
    Last Post: November 17th 2013, 04:12 PM
  3. Replies: 10
    Last Post: January 14th 2010, 12:28 PM
  4. Equivalence relation and order of each equivalence class
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 30th 2009, 09:03 AM
  5. Equivalence relation and Equivalence classes?
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: January 7th 2009, 03:39 AM

Search Tags


/mathhelpforum @mathhelpforum