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i am proving this statement about strict isotoneness. i will try on my own and you will tell me whether i am okay or not :)

$A$ is subset of $\mathbb{R}$
$f$ is strict isotone $ \Longleftrightarrow \forall x,y\in A (x \neq y \Longrightarrow \frac{f(y)-f(x)}{y-x} > 0)$

i say: if $f$ is strict monotone, then for any x,y with $x<y$ it is $f(x)<f(y)$. so in the right part, $\frac{f(y)-f(x)}{y-x} > 0$ is then true because $f(y)-f(x)>0$ and $y-x>0$ are true, since $x<y$ and f is isotone.

i feel, i amnot showing enough proof here. how to develop the proof technic?

thanks

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Have you noticed how I keep removing [set-theory] tags from your questions? Why do you think I'd do something like that? –  Asaf Karagila Dec 1 '12 at 19:17
    
why? it is also a set theory, isnot it? –  doniyor Dec 1 '12 at 19:20
4  
@doniyor: in that case, every question on this site should have the [set-theory] tag. [set-theory] is about ZFC and such. –  akkkk Dec 1 '12 at 19:30
1  
Not every thing involving "sets" is set theory. In this aspect everything involving numbers is number theory. –  Asaf Karagila Dec 1 '12 at 21:35

1 Answer 1

up vote 3 down vote accepted
  • For the forward implication, assuming $f$ is strict monotone to deduce $$ \forall x,y\in A (x \neq y \longrightarrow \dfrac{f(y)-f(x)}{y-x} > 0),$$ you haven't dealt with what the case is when $x>y$.

    $x\neq y$ includes two cases you need to consider: $x<y$ or $x>y$.

    It should follow the same pattern: except for $x > y$, you'll have $f(x) > f(y)$ so $f(y)-f(x) < 0$ (by definition of strict monotonicity), and $y - x < 0$, and negative divided by negative is positive.

  • Also, you need to prove the "backward* direction of the double implication: You need to prove $$\forall x,y\in A (x \neq y \Longrightarrow \dfrac{f(y)-f(x)}{y-x} > 0) \longrightarrow f \text{ is strict monotone.}\;$$ (1) Assume $\forall x,y\in A \left(x \neq y \Longrightarrow \dfrac{f(y)-f(x)}{y-x} > 0\right)$.

    (2) $x \neq y$ means either $x>y$ or $y>x$.

    (3) If $x>y$ implies $\dfrac{f(y)-f(x)}{y-x} > 0$ and $y>x$ implies $\dfrac{f(y)-f(x)}{y-x}> 0$, then ...

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okay. thanks man, i will try then now –  doniyor Dec 1 '12 at 19:53
    
yes, you are right. great! –  doniyor Dec 1 '12 at 20:06
    
:), clean and simple explanation. i am getting better by learning here than in my lectures.. –  doniyor Dec 1 '12 at 20:36

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