1. Re: equivalence

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

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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

2. Re: equivalence

Originally Posted by psolaki
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.

3. Re: equivalence

Originally Posted by psolaki
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.

4. Re: equivalence

Originally Posted by SlipEternal
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

5. Re: equivalence

Originally Posted by psolaki

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.

6. 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.

7. Re: equivalence

Originally Posted by SlipEternal
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 $1\leq 0$is false and false can imply anything we can say a=0 is true

8. Re: equivalence

Originally Posted by Deveno
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.

9. Re: equivalence

Originally Posted by psolaki
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 $1\leq 0$is 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.

10. Re: equivalence

Originally Posted by bkbowser
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.

11. 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.

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