Are the three inequalities;
1) 1>0
2) a>0 => (1/a)>0
3) ac>bc & c>0 => a>b
equivalent??
I can prove that (1) implies (2) and that (2) implies (3) ,but i cannot prove that (3) implies (1)
Well 1 > 0 is true no matter what.
The second inequation is false for all values of a, as reciprocating REVERSES the direction of the inequality sign (so > becomes < ).
The third is fine because you can divide both sides by c, and since c is positive that doesn't change the inequality sign.
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I would like to see a reference from ANY book stating that : for any real,a : a>o => (1/a)>0 is false
You confuse the above inequality with the following inequality:
0<a<b => 0<(1/b)< (1/a).
Note :even in this inequality (1/a) and (1/b) remain bigger than zero.
But my problem is ,if the three inqualities are equivalent.
The problem with trying to help is that we have no clue what other premises you are permitted to use in proofs.
For example
$a > 0 \implies (1/a) > 0.$ Given.
$a * (1/a) = 1.$ By definition of (1/a).
$x > 0\ and\ y > z \implies xy > xz.$ Is that rule about multiplication by a positive allowed, but 1 > 0 is not allowed?
$\therefore a * (1/a) > a * 0.$
$But\ a * 0 = 0.$ Is that allowed?
$THUS,\ 1 > 0.$
I have no idea whether that proof is valid under the premises that you are permitted to use. My basic problem is that if we must prove 1 > 0, what else must we prove.
In any case, using some premises that seem rather basic, I can prove 1 from 2 and 2 from 1 and from 2 to 3. I too have trouble with reversing from 3.
The medium is the real Nos ,so any property or theorem of the real Nos can be used.
Now to prove the equivalence we have to prove the following circle:
(1)=>(2)=>(3)=>(1),or any other proving circle for that matter will do.
As i said ican prove (1)=>(2)=>(3). But i cannot prove (3)=>(1)
For example in proving (1)=> (2) we have:
Assume (1)
Let a>0 & ~((1/a)>0). That implies (a>0)&( (1/a)<0 v (1/a)=0)) =>(a>0 & (1/a)<0) v (a>0& (1/a)=0)=>(a>0&(1/a).a=1 & (1/a)<0) v (a>0 & (1/a).a=1 & (1/a)=0)............................................. ...................A
Since $\displaystyle a>0\Longrightarrow a\neq 0$
Now A implies ( (1/a).a<0 & (1/a).a=1) v ( 0.a=1) => (1<0) v (1=0) => ~(1>0) v ~(1>0) => ~(1>0) and since 1>0 we have a contradiction
Hence a>0 => (1/a)>0
In the same style we can prove : (2) => (3).
The problem is in proving : (3) => (1)
Suppose $c > 0$ is given.
Taking $a = 1, b = 0$ in (3), we have: $1c = c > 0 = 0c$, so we may conclude $1 > 0$, which is (1). So (3) implies (1).
(note: by definition, we have $1x = x$, for all real numbers $x$. To see that $0x = 0$, for all real numbers $x$, note that:
$0x + 0x = (0+0)x = 0x$, so that:
$0x = 0x + 0 = 0x + (0x + (-0x)) = (0x + 0x) + (-0x) = 0x + (-0x) = 0$).
O.K (3) is: ac>bc & c>0 => a>b.
Now if you put a=1 ,b=0 ,as Deveno suggested (3) becomes.
(1.c>0.c & c>0 ) => 1>0.
Now to show 1>0 ,YOU MUST SHOW : 1.c>0.c & c>0,so that you can use M.Ponens and get 1>0
Deveno did not show that ,he simply assumed it.
He assumed : c>0
But c>0 => 1.c>0.c ,because 1.c=c and 0.c =0
Hence by M.Ponens we have : 1.c>0.c.
Thus , 1.c>0.c & c>0
But by using (3) we have : 1.c>0.c &c>0 => 1>0.
And by using M.Ponens again we have : 1>0.
So far we have ASSUMED , c>0 and ended up with 1>0 ..
But ,to be exact ,we also have assumed : ac>bc &c>0 => a>b
So we have proved : [(ac>bc& c>0) =>a>b]& c>0 => 1>0 and not 1>0.
IF you assume ,p and end with q ,you have proved : p=>q and not q
We want to show (3) --> (1). Therefore, we assume (3) to be true, and derive (1) conditionally.
Now (3) itself is a statement of the form: (p&q)-->r. Thus in taking (3) to be true, we are taking it to be of the form: T-->T.
I have NOT shown that , in fact, 1 > 0. I have shown that if (3) is true, for any $a,b \in \Bbb R$, and $c \in \Bbb R^+$, then in particular it is true for some particular pair $a,b$, namely $a = 1,b = 0$.
If you want to QUIBBLE, the only "hole" is that the set $\Bbb R^+$ may be empty. This is actually somewhat problemmatic, as one has to define exactly what is MEANT by $x > y$, for two real numbers $x,y$.
The standard way of doing this, is to use the density of the rational numbers in the reals, and induce the ordering on $\Bbb R$ from that in $\Bbb Q$. That is, we say, for two real numbers $x,y$, that $x < y$, if for every rational number $p < x$, and every rational number $y < q$, we have $p < q$. To an extent, this "begs the question": what does it mean, in the rationals, for $p < q$ to be true? (Note that this specifically addresses the fact that the ordering of the rationals in Archimedean, and the the supremum of a set of rational numbers is a real number (the "least upper bound" property of the reals)).
Here, we resort to the ordering on the integers: we say that $p = \dfrac{a}{b} < q = \dfrac{c}{d}$, if $ad < bc$. Again, we must ask: how do we define the ordering on the integers?
We say that for two integers: $m > n$, if $m - n = m + (-n)$ is a (non-zero) natural number. Alternatively, writing an integer as a an equivalence of pairs of natural numbers $[(a,b)]$ ($a$ is the "positive part" and $b$ is the "negative part"), we say that $[(a,b)]$ is positive (> 0 = [(0,0)]) if $a > b$, as natural numbers. So, again, we have "pushed the problem" to yet another ordered set, the natural numbers.
For two natural numbers, $k,k'$ we say that $k > k'$ if there exists a natural number $n$ such that $k$ is the $n$-th successor of $k'$.
This means that the integer $k$, that is the equivalence class of $(k,0)$, for a non-zero natural number $k$ is positive (greater than 0), whence it follows that the rational number $\dfrac{k}{1}$ is likewise positive (greater than $0 = \dfrac{0}{1}$), and thus that real number $k$ is also greater than 0. This shows that we have (at least) an infinite choice of real numbers $c > 0$ we might choose from (any real form of a natural number).
Of course, 1 is one of those numbers, so it appears that we have inadvertently used "circular reasoning" (used (1) to assert the existence of $c$).
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So, can we justify that there exists at least ONE positive real number (without appealing to $1$)? I claim we can, if we can use THESE axioms (in addition to the field axioms):
A1) $a > 0,b > 0 \implies ab > 0$, for any $a,b \in \Bbb R$.
A2) $a > 0$ or $-a > 0$, or $a = 0$, and exactly one of these holds.
Claim: if $a \neq 0$, then $a^2 > 0$.
Since $\Bbb R$ is a field, if $a^2 = 0$, then either $a = 0$, or...$a = 0$ (fields are integral domains, and posses no zero divisors). Since $a \neq 0$, it follws $a^2 \neq 0$, and thus by (A2) either $a^2 > 0$, or $-a^2 > 0$.
Now, also by (A2), since $a \neq 0$, we have either $a > 0$ or $a < 0$. If $a > 0$, then by (A1), we have $a^2 > 0$.
So suppose $- a > 0$.
Then $(-a)^2 > 0$, by (A1).
However, $(-a)^2 = [(-1)(a)]^2 = (-1)(a)(-1)(a) = (-1)(-1)(a)(a) = (-1)(-1)a^2 = -(-1)a^2 = 1a^2 = a^2$, so in all cases, $a^2 > 0$.
If $\displaystyle 1>0$, then True. If True, then $\displaystyle 1>0$. These statements are both true. Hence, $\displaystyle 1>0$ is equivalent to True. So, you just need to show (3) implies True. Then, True implies $\displaystyle 1>0$. Suppose (3) is false. Then the statement (3) implies True is true. Suppose (3) is true. Then the statement (3) implies True is true. Hence, (3) implies True. True implies 1>0. Hence, (3) implies 1>0?
By the way, my point was, $\displaystyle 1>0$ is a basic axiom of integers and continues to hold for the integers embedded within the real numbers. Hence, any implication of the form $\displaystyle P \Rightarrow 1>0$ is true regardless of the statement $\displaystyle P$.