if k is any natural numb, then n^3>kn^2 for all sufficent large n.

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- Apr 30th 2011, 05:26 AMikurwae89Write this in fopa lang?
if k is any natural numb, then n^3>kn^2 for all sufficent large n.

- Apr 30th 2011, 05:32 AMPlato
- Apr 30th 2011, 06:49 PMikurwae89
first order penos arithmetic

- May 2nd 2011, 03:41 AMemakarov
$\displaystyle \forall k\exists m\forall n.\,n\ge m\to n^3>kn^2$

If >= is not considered a part of the language of PA, then it can be replaced using the following equivalence.

$\displaystyle \forall n\forall m.\,n\ge m\leftrightarrow\exists p.\,n=m+p$ - May 3rd 2011, 07:24 PMikurwae89
can u explain to me how u got that?

This is what i got.

((For all k)( there exists m)( there exists n)(n.n.n > k.m.m.m))

i dont have the upside A and E.

sorry

what is the difference? - May 4th 2011, 12:50 AMemakarovQuote:

((For all k)( there exists m)( there exists n)(n.n.n > k.m.m.m))

**all**sufficent large n," while your formula says "there exists n."

(2) I don't understand why you replaced "n^3 > kn^2" with "n^3 > km^3."

(3) "For all sufficiently large n" means "for all n that exceed a certain lower bound m." That's why I said $\displaystyle \exists m\forall n.\,n>m\to\dots$ (one can also use >= instead of > here). - May 5th 2011, 02:59 AMikurwae89
sorry about my stuff up and thanks for making the for all sufficient large n clear :)

so this is how you would write it.

((For all k)( there exists m)( for all n)(n>=m)--> (n.n.n > k.n.n.))

Thanks a lot btw really helpful :) - May 5th 2011, 04:14 AMemakarovQuote:

so this is how you would write it.

((For all k)( there exists m)( for all n)(n>=m)--> (n.n.n > k.n.n.))