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misham
post Apr 16 2008, 09:44 AM
Post #21





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evo jos jednog zadatka koji me muci, pa ako imate vremena pomagajte:

odrediti sve cele brojeve za koje vazi x^2(x^2+y)=y^{(z+1)}
trivijalno resenje je (0,0,z E Z );
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tijana
post Apr 16 2008, 10:27 PM
Post #22





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QUOTE(misham @ Apr 16 2008, 10:44 AM)
evo jos jednog zadatka koji me muci, pa ako imate vremena pomagajte:

odrediti sve cele brojeve za koje vazi x^2(x^2+y)=y^{(z+1)}
trivijalno resenje je (0,0,z E Z );
*




pa mislim da nisi u pravu proveri malo to trivijalno resenje...ne vazi ako je z=0...posto je 0^0=1 a ne 0


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Nostradamus
post Apr 16 2008, 11:14 PM
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A bas i ne znam koliko je 0^{-1}

This post has been edited by Nostradamus: Apr 16 2008, 11:15 PM


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(a,b) \sim (c,d) \Leftrightarrow (a - c) \in \mathbb{Z} \wedge (b - d) \in \mathbb{Z}

\mathbb{T}^2\equiv\mathbb{R}^2/\sim
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misham
post Apr 17 2008, 07:30 AM
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upravu ste
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tijana
post Apr 17 2008, 10:37 AM
Post #25





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QUOTE(Nostradamus @ Apr 17 2008, 12:14 AM)
A bas i ne znam koliko je 0^{-1}
*


0^{-1}=1/0 a to je ustvari beskonacno...


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Wolfbrother
post Apr 17 2008, 12:11 PM
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Beskonacno je ako broj tezi 0, ali u ovom slucaju nije definisano rekao bih.


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Nostradamus
post Apr 17 2008, 12:45 PM
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QUOTE(tijana @ Apr 17 2008, 11:37 AM)
0^{-1}=1/0 a to je ustvari beskonacno...
*



Au, pa sta vas to sada uce u skoli.

valjda 7 aksioma skupa R. Za svaki element x iz R bez nule, postoji y iz R takav da je x*y=1.

Cak i ako posmatras f-ju 1/x ona u nuli ide u beskonacnost samo sa desne strane, sa leve u minus beskonacno.

A naravno u samoj nuli nije definisana.

This post has been edited by Nostradamus: Apr 17 2008, 12:49 PM


--------------------
(a,b) \sim (c,d) \Leftrightarrow (a - c) \in \mathbb{Z} \wedge (b - d) \in \mathbb{Z}

\mathbb{T}^2\equiv\mathbb{R}^2/\sim
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tijana
post Apr 17 2008, 06:56 PM
Post #28





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pa neki kazu da nije definisana a neki da tezi beskonacno...a i u svakom slucaju ne sme da se deli sa 0. ali opet zadatak nije definisan...posto to sto je on rekao nije trivijalno resenje...


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