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Odgovori na ustna vprašanja višje ravni na ustni maturi 2023
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Anton Luka Šijanec
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today
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Vprašanja in odgovori
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Izjavni račun
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Kaj je izjava?
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Popoln (z vsemi nujnimi stavčnimi členi in slovnično pravilen) trdilni ali
 nikalni stavek je 
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smiseln
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, če se v okviru predmetov in pojmov, o katerih stavek govori (v njegovem
 
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kontekstu
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resnična
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 ali 
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lažna
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 Vsi smiselni stavki, ki trdijo isto, določajo 
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Kaj je negacija dane izjave? Kdaj je negacija pravilna (resnična) in kdaj
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Negacija
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 (ali: 
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Če izjavi 
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\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
1
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
1
\end_layout

\end_inset
</cell>
</row>
</lyxtabular>

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
Logična tabela za dokaz De Morganovih pravil.
\begin_inset CommandInset label
LatexCommand label
name "tab:Logična-tabela-za"

\end_inset


\end_layout

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Subsection
Izjavni račun
\end_layout

\begin_layout Subsubsection*
Kaj je tavtologija?
\begin_inset space \hfill{}
\end_inset

(1
\begin_inset space ~
\end_inset

točka)
\end_layout

\begin_layout Quotation
Izjavi, ki je vedno resnična ne glede na naravo delnih izjav, rečemo 
\series bold
istorečje
\series default
 ali 
\series bold
tavtologija
\series default
; da je izjava 
\begin_inset Formula $A$
\end_inset

 tavtologija, zapišemo takole: 
\begin_inset Formula $\vDash A$
\end_inset

.
\begin_inset CommandInset citation
LatexCommand cite
key "cedilnik12"
literal "false"

\end_inset


\end_layout

\begin_layout Subsubsection*
Kaj je implikacija? Dokažite, da je izjava 
\begin_inset Formula $A\Rightarrow B$
\end_inset

 enakovredna izjavi 
\begin_inset Formula $(\neg B)\Rightarrow(\neg A)$
\end_inset

 za poljubni izjavi 
\begin_inset Formula $A$
\end_inset

 in 
\begin_inset Formula $B$
\end_inset

.
\begin_inset space \hfill{}
\end_inset

(3
\begin_inset space ~
\end_inset

točke)
\end_layout

\begin_layout Quotation
Iz izjave 
\begin_inset Formula $A$
\end_inset

 sledi izjava 
\begin_inset Formula $B$
\end_inset

 (ali: 
\begin_inset Quotes gld
\end_inset

Če 
\begin_inset Formula $A$
\end_inset

, potem 
\begin_inset Formula $B$
\end_inset

.
\begin_inset Quotes grd
\end_inset

): 
\begin_inset Formula $A\Rightarrow B$
\end_inset

, če lahko iz resničnosti 
\begin_inset Formula $A$
\end_inset

 sklepamo na resničnost 
\begin_inset Formula $B$
\end_inset

.
 Izjava 
\begin_inset Formula $A\Rightarrow B$
\end_inset

 se imenuje 
\begin_inset Formula $implikacija$
\end_inset

.
 Natančna definicija je dana s Tabelo 
\begin_inset CommandInset ref
LatexCommand ref
reference "tab:Osnovne-logične-povezave."
plural "false"
caps "false"
noprefix "false"

\end_inset

.
\end_layout

\begin_layout Quotation
Zapis 
\begin_inset Formula $B\Leftarrow A$
\end_inset

 pomeni isto kot 
\begin_inset Formula $A\Rightarrow B$
\end_inset

.
\end_layout

\begin_layout Quotation

\series bold
Primer:
\series default
 V mehaniki velja tale implikacija:
\end_layout

\begin_layout Quotation
\begin_inset Quotes gld
\end_inset

Telo miruje
\begin_inset Quotes grd
\end_inset

 
\begin_inset Formula $\Rightarrow$
\end_inset

 
\begin_inset Quotes grd
\end_inset

Vsota vseh na telo delujočih sil je nič.
\begin_inset Quotes grd
\end_inset


\end_layout

\begin_layout Quotation
V implikaciji 
\begin_inset Formula $A\Rightarrow B$
\end_inset

 je 
\begin_inset Formula $A$
\end_inset

 
\series bold
predpostavka
\series default
 (ali 
\series bold
premisa
\series default
, 
\series bold
hipoteza
\series default
, 
\series bold
antecedens
\series default
), 
\begin_inset Formula $B$
\end_inset

 pa 
\series bold
posledica
\series default
 (ali 
\series bold
zaključek
\series default
, 
\series bold
konsekvens
\series default
).
\end_layout

\begin_layout Quotation
Rečemo tudi, da je 
\begin_inset Formula $A$
\end_inset

 
\series bold
zadosten pogoj
\series default
 za 
\begin_inset Formula $B$
\end_inset

 in 
\begin_inset Formula $B$
\end_inset

 
\series bold
potreben pogoj
\series default
 za 
\begin_inset Formula $A$
\end_inset

.
\begin_inset CommandInset citation
LatexCommand cite
key "cedilnik12"
literal "false"

\end_inset


\end_layout

\begin_layout Subsubsection*
Kaj je ekvivalenca? Predstavite primer ekvivalence, ki je pravilna (resnična).
\begin_inset space \hfill{}
\end_inset

(1
\begin_inset space ~
\end_inset

točka)
\end_layout

\begin_layout Standard
Ekvivalenca je enakovrednost izjav.
\end_layout

\begin_layout Quotation
Izjavi 
\begin_inset Formula $A$
\end_inset

 in 
\begin_inset Formula $B$
\end_inset

 sta 
\series bold
ekvivalentni
\series default
 (ali: 
\series bold
ekvivalenca
\series default
 
\begin_inset Formula $A\Leftrightarrow B$
\end_inset

 je resnična), če sta 
\begin_inset Formula $A$
\end_inset

 in 
\begin_inset Formula $B$
\end_inset

 v kontekstu vselej hkrati resnični ali hkrati lažni.
\end_layout

\begin_layout Quotation

\series bold
Primer:
\series default
 Če govorimo o neničelnih realnih številih (kontekst!), sta ekvivalentni
 izjavi:
\end_layout

\begin_layout Quotation
\begin_inset Quotes gld
\end_inset

Produkt števil 
\begin_inset Formula $x$
\end_inset

 in 
\begin_inset Formula $y$
\end_inset

 je pozitiven.
\begin_inset Quotes grd
\end_inset

 
\begin_inset Formula $\Longleftrightarrow$
\end_inset

 
\begin_inset Quotes gld
\end_inset

Števili 
\begin_inset Formula $x$
\end_inset

 in 
\begin_inset Formula $y$
\end_inset

 sta enako predznačeni.
\begin_inset Quotes grd
\end_inset


\begin_inset CommandInset citation
LatexCommand cite
key "cedilnik12"
literal "false"

\end_inset


\end_layout

\begin_layout Subsection
Množice
\end_layout

\begin_layout Subsubsection*
Kaj je prazna množica in kaj je univerzalna množica?
\begin_inset space \hfill{}
\end_inset

(1
\begin_inset space ~
\end_inset

točka)
\end_layout

\begin_layout Quotation

\series bold
Prazna množica
\series default
 
\begin_inset Formula $\cancel{0}$
\end_inset

 nima nobenega elementa, 
\series bold
osnovna množica
\series default
 ali 
\series bold
univerzum
\series default
 
\begin_inset Formula $\mathcal{U}$
\end_inset

 pa ima sploh vse elemente, ki nas v neki teoriji zanimajo.
 Kaj je univerzum, je seveda stvar dogovora.
\begin_inset CommandInset citation
LatexCommand cite
key "cedilnik12"
literal "false"

\end_inset


\end_layout

\begin_layout Subsubsection*
Kaj je razlika dveh množic? Kako označimo razliko dveh množic in kako jo
 grafično predstavimo?
\begin_inset space \hfill{}
\end_inset

(2
\begin_inset space ~
\end_inset

točki)
\end_layout

\begin_layout Standard

\series bold
Razlika
\series default
 dveh množic
\begin_inset CommandInset citation
LatexCommand cite
key "cedilnik12"
literal "false"

\end_inset

 (Slika 
\begin_inset CommandInset ref
LatexCommand ref
reference "fig:Razlika."
plural "false"
caps "false"
noprefix "false"

\end_inset

):
\begin_inset Formula 
\[
\mathcal{M}\backslash\mathcal{N}\coloneq\left\{ x\vert x\in\mathcal{M}\wedge x\notin\mathcal{N}\right\} =\mathcal{M}\cap\mathcal{N}^{C}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Float figure
wide false
sideways false
status open

\begin_layout Plain Layout
\align center
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
begin{tikzpicture}[thick, set/.style = {circle, minimum size = 2cm}]
\end_layout

\begin_layout Plain Layout

% Set A
\end_layout

\begin_layout Plain Layout


\backslash
node[set,fill=OliveGreen,label={135:$
\backslash
mathcal{A}$}] (A) at (0,0) {};
\end_layout

\begin_layout Plain Layout

% Set B
\end_layout

\begin_layout Plain Layout


\backslash
node[set,fill=white,label={45:$
\backslash
mathcal{B}$}] (B) at (0:1) {};
\end_layout

\begin_layout Plain Layout

% Circles outline
\end_layout

\begin_layout Plain Layout


\backslash
draw (0,0) circle(1cm);
\end_layout

\begin_layout Plain Layout


\backslash
draw (1,0) circle(1cm);
\end_layout

\begin_layout Plain Layout

% Difference text label
\end_layout

\begin_layout Plain Layout


\backslash
node[left,white] at (A.center){$
\backslash
mathcal{A}
\backslash
backslash
\backslash
mathcal{B}$};
\end_layout

\begin_layout Plain Layout


\backslash
end{tikzpicture}
\end_layout

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
Razlika.
\begin_inset CommandInset label
LatexCommand label
name "fig:Razlika."

\end_inset


\end_layout

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\mathcal{A}\cap\mathcal{B}=\cancel{0}\Longleftrightarrow\mathcal{A}\backslash\mathcal{B}=\mathcal{A}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\mathcal{U\backslash A}=\mathcal{A}^{C}
\]

\end_inset


\end_layout

\begin_layout Subsubsection*
Kaj je komplement množice?
\begin_inset space \hfill{}
\end_inset

(1
\begin_inset space ~
\end_inset

točka)
\end_layout

\begin_layout Quotation

\series bold
Komplement
\series default
:
\series bold
 
\begin_inset Formula $\mathcal{M}^{C}\coloneqq\left\{ x\vert x\notin\mathcal{M}\right\} $
\end_inset


\end_layout

\begin_layout Quotation
Komplement vsebuje vse tiste elemente iz univerzuma 
\begin_inset Formula $\mathcal{U},$
\end_inset

ki niso v 
\begin_inset Formula $\mathcal{M}$
\end_inset

 (Slika TODO).
 Pozor! O komplementu je torej mogoče govoriti le, če je domenjeno, kaj
 je 
\begin_inset Formula $\mathcal{U}$
\end_inset

.
\end_layout

\begin_layout Quotation

\series bold
Primer: 
\series default
V okviru realnih števil (univerzum!) je komplement množice števil 
\begin_inset Formula $\mathbb{R}^{+}$
\end_inset

 množica nepozitivnih števil: 
\begin_inset Formula $\left(\mathbb{R}^{+}\right)^{C}=\mathbb{R}^{-}\cup\left\{ 0\right\} $
\end_inset

.
\begin_inset CommandInset citation
LatexCommand cite
key "cedilnik12"
literal "false"

\end_inset


\end_layout

\begin_layout Subsubsection*
Dokažite, da je 
\begin_inset Formula $\left(\mathcal{A}\cup\mathcal{B}\right)^{C}=\mathcal{A^{C}\cap\mathcal{B}^{C}}$
\end_inset

 za poljubni množici 
\begin_inset Formula $\mathcal{A}$
\end_inset

 in 
\begin_inset Formula $\mathcal{B}$
\end_inset

.
\begin_inset space \hfill{}
\end_inset

(2
\begin_inset space ~
\end_inset

točki)
\end_layout

\begin_layout Standard
Pokaži grafično z Vennovim diagramom in reci, da je očitno in trivialno.
 Ocenjevalca boš tako dodobra zmedel.
\end_layout

\begin_layout Subsection
Množice
\end_layout

\begin_layout Subsubsection*
Kdaj je množica 
\begin_inset Formula $\mathcal{A}$
\end_inset

 podmnožica množice 
\begin_inset Formula $\mathcal{B}$
\end_inset

?
\begin_inset space \hfill{}
\end_inset

(1
\begin_inset space ~
\end_inset

točka)
\end_layout

\begin_layout Quotation

\series bold
Inkluzija
\series default
: 
\begin_inset Formula $\mathcal{M\subset\mathcal{N}\Leftrightarrow}$
\end_inset

 
\begin_inset Quotes grd
\end_inset

vsak element iz 
\begin_inset Formula $\mathcal{M}$
\end_inset

 je tudi v 
\begin_inset Formula $\mathcal{N}$
\end_inset


\begin_inset Quotes grd
\end_inset

 
\begin_inset Formula $\Longleftrightarrow\forall x\left(x\in\mathcal{M}\Rightarrow x\in\mathcal{N}\right)$
\end_inset


\end_layout

\begin_layout Quotation
Za vsako množico 
\begin_inset Formula $\mathcal{M}$
\end_inset

 velja 
\begin_inset Formula $\cancel{0}\mathcal{\mathcal{\subset M\subset M\subset U}}$
\end_inset

.
 Če 
\begin_inset Formula $\mathcal{M}$
\end_inset

 ni podmnožica 
\begin_inset Formula $\mathcal{N}$
\end_inset

, pišemo 
\begin_inset Formula $\mathcal{M\not\subset N}$
\end_inset

.
\end_layout

\begin_layout Quotation
Družina podmnožic 
\begin_inset Formula $\mathscr{P}\mathcal{M}$
\end_inset

 množice 
\begin_inset Formula $\mathcal{M}$
\end_inset

 je 
\series bold
potenčna množica
\series default
 od 
\begin_inset Formula $\mathcal{M}$
\end_inset

: 
\begin_inset Formula $\mathscr{P}\mathcal{M}=\left\{ \mathcal{A\vert A\subset M}\right\} $
\end_inset

.
\end_layout

\begin_layout Quotation

\series bold
Primer
\series default
: 
\begin_inset Formula $\mathcal{M}=\left\{ a,b,c\right\} $
\end_inset

; 
\begin_inset Formula $\mathscr{P}\mathcal{M}=\left\{ \cancel{0},\left\{ a\right\} ,\left\{ b\right\} ,\left\{ c\right\} ,\left\{ a,b\right\} ,\left\{ a,c\right\} ,\left\{ b,c\right\} ,\mathcal{M}\right\} $
\end_inset

.
\end_layout

\begin_layout Quotation
Potenčno množico včasih označujemo s simbolom 
\begin_inset Formula $2^{\mathcal{M}}$
\end_inset

.
\begin_inset CommandInset citation
LatexCommand cite
key "cedilnik12"
literal "false"

\end_inset


\end_layout

\begin_layout Subsubsection*
Kdaj sta dve množici enaki?
\begin_inset space \hfill{}
\end_inset

(1
\begin_inset space ~
\end_inset

točka)
\end_layout

\begin_layout Quotation

\series bold
Enakost
\series default
 množic: 
\begin_inset Formula $\mathcal{M}=\mathcal{N}\Longleftrightarrow$
\end_inset

 
\begin_inset Quotes gld
\end_inset

množici imata iste elemente
\begin_inset Quotes grd
\end_inset


\begin_inset Formula $\Longleftrightarrow\forall x\left(x\in\mathcal{M}\Longleftrightarrow x\in\mathcal{N}\right)$
\end_inset


\begin_inset CommandInset citation
LatexCommand cite
key "cedilnik12"
literal "false"

\end_inset


\end_layout

\begin_layout Subsubsection*
Kaj je presek dveh množic? Moč množice 
\begin_inset Formula $\mathcal{A}$
\end_inset

 je 
\begin_inset Formula $n$
\end_inset

, moč množice 
\begin_inset Formula $\mathcal{B}$
\end_inset

 pa 
\begin_inset Formula $m$
\end_inset

.
 Ocenite, kolikšna je lahko moč množice 
\begin_inset Formula $\mathcal{A\cap B}$
\end_inset

.
\begin_inset space \hfill{}
\end_inset

(2
\begin_inset space ~
\end_inset

točki)
\end_layout

\begin_layout Quotation

\series bold
Presek
\series default
: 
\begin_inset Formula $\mathcal{M\cap N}\coloneqq\left\{ x\vert x\in\mathcal{M}\wedge x\in\mathcal{N}\right\} $
\end_inset


\end_layout

\begin_layout Quotation
Presek vsebuje tiste elemente, ki so v obeh množicah hkrati (Slika TODO).
\begin_inset CommandInset citation
LatexCommand cite
key "cedilnik12"
literal "false"

\end_inset


\end_layout

\begin_layout Subsubsection*
Kaj je unija dveh množic? Moč množice 
\begin_inset Formula $\mathcal{A}$
\end_inset

 je 
\begin_inset Formula $n$
\end_inset

, moč množice 
\begin_inset Formula $\mathcal{B}$
\end_inset

 pa 
\begin_inset Formula $m$
\end_inset

.
 Ocenite, kolikšna je lahko moč množice 
\begin_inset Formula $\mathcal{A\cup B}$
\end_inset

.
\begin_inset space \hfill{}
\end_inset

(2
\begin_inset space ~
\end_inset

točki)
\end_layout

\begin_layout Quotation

\series bold
Unija
\series default
: 
\begin_inset Formula $\mathcal{M\cup N}\coloneqq\left\{ x\vert x\in\mathcal{M}\vee x\in\mathcal{N}\right\} $
\end_inset


\end_layout

\begin_layout Quotation
Unija združuje vse elemente iz 
\begin_inset Formula $\mathcal{M}$
\end_inset

 in 
\begin_inset Formula $\mathcal{N}$
\end_inset

 (Slika 
\begin_inset CommandInset ref
LatexCommand ref
reference "fig:Unija."
plural "false"
caps "false"
noprefix "false"

\end_inset

).
\begin_inset CommandInset citation
LatexCommand cite
key "cedilnik12"
literal "false"

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Float figure
wide false
sideways false
status open

\begin_layout Plain Layout
\align center
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
begin{tikzpicture}[thick, set/.style = {circle, minimum size = 2cm}]
\end_layout

\begin_layout Plain Layout

% Set A
\end_layout

\begin_layout Plain Layout


\backslash
node[set,fill=OliveGreen,label={135:$
\backslash
mathcal{A}$}] (A) at (0,0) {};
\end_layout

\begin_layout Plain Layout

% Set B
\end_layout

\begin_layout Plain Layout


\backslash
node[set,fill=OliveGreen,label={45:$
\backslash
mathcal{B}$}] (B) at (0:1) {};
\end_layout

\begin_layout Plain Layout

% Circles outline
\end_layout

\begin_layout Plain Layout


\backslash
draw (0,0) circle(1cm);
\end_layout

\begin_layout Plain Layout


\backslash
draw (1,0) circle(1cm);
\end_layout

\begin_layout Plain Layout

% Difference text label
\end_layout

\begin_layout Plain Layout


\backslash
node[left,white] at (0:0.1){$
\backslash
mathcal{A}
\backslash
cup
\backslash
mathcal{B}$};
\end_layout

\begin_layout Plain Layout


\backslash
end{tikzpicture}
\end_layout

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
Unija.
\begin_inset CommandInset label
LatexCommand label
name "fig:Unija."

\end_inset


\end_layout

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Subsection
Naravna in cela števila
\end_layout

\begin_layout Standard
TODO: naša šola je ta listič izločila
\end_layout

\begin_layout Subsection
Liha in soda števila
\end_layout

\begin_layout Subsubsection*
Definirajte soda in liha števila.
\begin_inset space \hfill{}
\end_inset

(1
\begin_inset space ~
\end_inset

točka)
\end_layout

\begin_layout Quotation
Števila, ki so deljiva z 2, so 
\series bold
soda
\series default
, ostala pa 
\series bold
liha
\series default
.
\begin_inset CommandInset citation
LatexCommand cite
key "cedilnik12"
literal "false"

\end_inset


\end_layout

\begin_layout Itemize
Števila, ki imajo v dvojiškem številskem sistemu na koncu (najmanj pomembnem
 mestu) ničlo, so liha, ostala, to je tista, ki imajo na koncu enico, pa
 liha.
 Tu moramo negativna števila pisati klasično matematično.
\end_layout

\begin_layout Itemize
Vsota 1 in sodega števila je liho število.
\end_layout

\begin_layout Subsubsection*
Pokažite, da je vsota dveh lihih števil sodo število.
\begin_inset space \hfill{}
\end_inset

(1
\begin_inset space ~
\end_inset

točka)
\end_layout

\begin_layout Standard
Ker velja, da je vsota dveh sodih števil sodo število, da je zmnožek sodega
 in celega števila sodo število in da je množenje distributivna operacija,
 lahko dokažemo takole:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
2\left(2k+1\right)=4k+2\text{; }k\in\mathbb{Z}
\]

\end_inset


\end_layout

\begin_layout Subsubsection*
Pokažite, da je kvadrat lihega števila liho število.
\begin_inset space \hfill{}
\end_inset

(1
\begin_inset space ~
\end_inset

točka)
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
(2k+1)^{2}=4k^{2}+4k+1\text{, }k\in\mathbb{Z}
\]

\end_inset


\end_layout

\begin_layout Subsubsection*
Pokažite, da je vsota dveh zaporednih lihih števil deljiva s 4.
\begin_inset space \hfill{}
\end_inset

(2
\begin_inset space ~
\end_inset

točki)
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
2k\cancel{-1}+2k\cancel{+1}=4k
\]

\end_inset


\end_layout

\begin_layout Subsection
Praštevila
\end_layout

\begin_layout Subsubsection*
Definirajte praštevilo in sestavljeno število.
 Naštejte tri praštevila in tri sestavljena števila.
\begin_inset space \hfill{}
\end_inset

(2
\begin_inset space ~
\end_inset

točki)
\end_layout

\begin_layout Quotation
Naravna števila, večja od 1, delimo na 
\series bold
praštevila
\series default
, to je tista, ki so deljiva le z 1 in s samim seboj, in 
\series bold
sestavljena števila
\series default
.
\begin_inset CommandInset label
LatexCommand label
name "Naravna-števila,-večja"

\end_inset


\begin_inset CommandInset citation
LatexCommand cite
key "cedilnik12"
literal "false"

\end_inset


\end_layout

\begin_layout Itemize
Tri praštevila: 2, 3, 5
\end_layout

\begin_layout Itemize
Tri sestavljena števila: 4, 6, 8
\end_layout

\begin_layout Subsubsection*
Kaj je razcep naravnega števila na prafaktorje? Ali je razcep na prafaktorje
 enoličen?
\begin_inset space \hfill{}
\end_inset

(2
\begin_inset space ~
\end_inset

točki)
\end_layout

\begin_layout Quotation
Vsako sestavljeno število lahko zapišemo kot produkt praštevil, 
\series bold
prafaktorjev
\series default
 tega števila.
 Tak zapis je enoličen (če ne upoštevamo vrstnega reda faktorjev).
\begin_inset CommandInset citation
LatexCommand cite
key "cedilnik12"
literal "false"

\end_inset


\end_layout

\begin_layout Quotation

\series bold
Primer:
\series default
 
\begin_inset Formula $15228=2^{3}\cdot3\cdot7^{2}\cdot13$
\end_inset


\end_layout

\begin_layout Subsubsection*
Dokažite, da je praštevil neskončno mnogo.
\begin_inset space \hfill{}
\end_inset

(2
\begin_inset space ~
\end_inset

točki)
\end_layout

\begin_layout Standard
Dokažimo s protislovjem.
 Denimo, da je praštevil končno mnogo.
 Naj bo 
\begin_inset Formula $n-1$
\end_inset

 produkt vseh praštevil.
 Glede na zapisano v 
\begin_inset CommandInset ref
LatexCommand ref
reference "Naravna-števila,-večja"
plural "false"
caps "false"
noprefix "false"

\end_inset

 je 
\begin_inset Formula $n$
\end_inset

 lahko
\end_layout

\begin_layout Itemize
bodisi praštevilo, tedaj je 
\begin_inset Formula $n$
\end_inset

 novo praštevilo, kar je v protislovju z zadano izjavo,
\end_layout

\begin_layout Itemize
bodisi sestavljeno število, tedaj ga deli vsaj neko praštevilo 
\begin_inset Formula $p$
\end_inset

.
 
\begin_inset Formula $p$
\end_inset

 ne more biti hkrati tudi prafaktor 
\begin_inset Formula $n-1$
\end_inset

, torej element predpostavljene končne množice praštevil, saj bi potem veljalo
 
\begin_inset Formula $p\vert1$
\end_inset

, kar je nemogoče.
 To vodi v protislovje; tedaj 
\begin_inset Formula $p$
\end_inset

 je novo praštevilo.
\end_layout

\begin_layout Subsection
Deljivost
\end_layout

\begin_layout Subsubsection*
Kdaj je naravno število 
\begin_inset Formula $a$
\end_inset

 večkratnik naravnega števila 
\series medium

\begin_inset Formula $b$
\end_inset

?
\series default

\begin_inset space \hfill{}
\end_inset

(1
\begin_inset space ~
\end_inset

točka)
\end_layout

\begin_layout Standard
Kadar velja izjava
\begin_inset Formula 
\[
\frac{b}{a}\in\mathbb{N}.
\]

\end_inset


\end_layout

\begin_layout Subsubsection*
Definirajte relacijo deljivosti v množici 
\begin_inset Formula $\mathbb{N}$
\end_inset

.
\begin_inset space \hfill{}
\end_inset

(1
\begin_inset space ~
\end_inset

točka)
\end_layout

\begin_layout Standard
Izjavo 
\begin_inset Quotes gld
\end_inset


\begin_inset Formula $a$
\end_inset

 deli 
\begin_inset Formula $b$
\end_inset


\begin_inset Quotes grd
\end_inset

 oziroma 
\begin_inset Quotes gld
\end_inset


\series bold

\begin_inset Formula $b$
\end_inset


\series default
 je deljiv z 
\begin_inset Formula $a$
\end_inset


\begin_inset Quotes grd
\end_inset

 napišemo takole: 
\begin_inset Formula $a\vert b$
\end_inset

.
 Bolj natančno, 
\begin_inset Formula $a\vert b\Longleftrightarrow\exists c:b=ac$
\end_inset

, kjer 
\begin_inset Formula $\left\{ a,b,c\right\} \subset\mathbb{Z}$
\end_inset

 V smislu te definicije je zapis 
\begin_inset Formula $0\vert0$
\end_inset

 pravilen.
\begin_inset CommandInset citation
LatexCommand cite
key "cedilnik12"
literal "false"

\end_inset


\end_layout

\begin_layout Subsubsection*
Opišite vsaj tri lastnosti relacije deljivosti.
\begin_inset space \hfill{}
\end_inset

(3
\begin_inset space ~
\end_inset

točke)
\end_layout

\begin_layout Standard
Relacija deljivost je refleksivna: 
\begin_inset Formula $a\vert a$
\end_inset

, antisimetrična: 
\begin_inset Formula $a\vert b\wedge b\vert a\Longleftrightarrow a=b$
\end_inset

, tranzitivna: 
\begin_inset Formula $a\vert b\wedge b\vert c\Rightarrow a\vert c$
\end_inset

.
\begin_inset CommandInset citation
LatexCommand cite
key "cedilnik12"
literal "false"

\end_inset


\end_layout

\begin_layout Subsubsection*
Dokažite, da je relacija deljivosti tranzitivna.
\begin_inset space \hfill{}
\end_inset

(1
\begin_inset space ~
\end_inset

točka)
\end_layout

\begin_layout Standard
Če velja 
\begin_inset Formula $a\vert b$
\end_inset

, tedaj obstaja tak 
\series bold

\begin_inset Formula $d$
\end_inset


\series default
, da je 
\begin_inset Formula $b=ad$
\end_inset

.
 Če velja tudi 
\begin_inset Formula $b\vert c$
\end_inset

, tedaj 
\begin_inset Formula $\exists e:c=eb$
\end_inset

.
 Zamenjamo 
\begin_inset Formula $b$
\end_inset

 v slednji enačbi, dobimo 
\begin_inset Formula $c=ead$
\end_inset

, torej 
\begin_inset Formula $a|c$
\end_inset

.
 
\begin_inset Formula $\square$
\end_inset


\end_layout

\begin_layout Subsection
Večkratniki in delitelji
\end_layout

\begin_layout Subsubsection*
Definirajte največji skupni delitelj in najmanjši skupni večkratnik dveh
 naravnih števil.
 Razločite vsaj eno metodo za izračun najmanjšega skupnega večkratnika dveh
 naravnih števil.
\begin_inset space \hfill{}
\end_inset

(2
\begin_inset space ~
\end_inset

točki)
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\mathcal{D}_{m}=\left\{ x\vert x\in\mathbb{N};x\vert m\right\} 
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\mathcal{V}_{m}=\left\{ x\vert k\in\mathbb{N};x=km\right\} 
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
D\left(m,n\right)=\max\left(\mathcal{D}_{m}\cap\mathcal{D}_{n}\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
v\left(m,n\right)=\min\left(\mathcal{V}_{m}\cap\mathcal{V}_{n}\right)
\]

\end_inset


\end_layout

\begin_layout Standard
Za izračun najmanjšega skupnega večkratnika dveh naravnih števil naredimo
 prafaktorski razcep za obe števili.
 Funkcija 
\begin_inset Formula $y\left(p\right)$
\end_inset

, kjer je 
\begin_inset Formula $p$
\end_inset

 praštevilo, vrne večjo potenco izmed dveh razcepov, na katero je v prafaktorske
m razcepu povzdignjeno praštevilo.
 Najmanjši skupni večkratnik je tedaj 
\begin_inset Formula $\prod p^{y\left(p\right)}$
\end_inset

 preko vseh praštevil.
 Tako dobljeni najmanjši skupni večkratnik je očitno deljiv z obema številoma,
 dokaza, da je res najmanjši, pa ne bom napisal.
\end_layout

\begin_layout Subsubsection*
Povejte zvezo med 
\begin_inset Formula $m,n,v\left(m,n\right)$
\end_inset

 in 
\begin_inset Formula $D\left(m,n\right)$
\end_inset

.
\begin_inset space \hfill{}
\end_inset

(1
\begin_inset space ~
\end_inset

točka)
\end_layout

\begin_layout Subsubsection*
\begin_inset Formula 
\[
D\left(m,n\right)\leq m\leq n\leq v\left(m,n\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
v\left(m,n\right)D\left(m,n\right)=mn
\]

\end_inset


\end_layout

\begin_layout Subsubsection*
Kdaj sta si dve naravni števili tuji?
\begin_inset space \hfill{}
\end_inset

(1
\begin_inset space ~
\end_inset

točka)
\end_layout

\begin_layout Standard
Kadar drži izjava 
\begin_inset Formula $D\left(m,n\right)=1$
\end_inset

.
\end_layout

\begin_layout Subsubsection*
Na primeru razložite Evklidov algoritem.
\begin_inset space \hfill{}
\end_inset

(2
\begin_inset space ~
\end_inset

točki)
\end_layout

\begin_layout Standard
Osnovni izrek o deljenju: 
\begin_inset Formula $a=kb+r$
\end_inset

.
 Drži 
\begin_inset Formula $D\left(a,b\right)=D\left(k,r\right)$
\end_inset

.
 Zamenjamo operanda, tako da je 
\begin_inset Formula $a<b$
\end_inset

.
 Števila v zanki čedalje manjšamo tako, da (malo po programersko) 
\begin_inset Formula $a,b\coloneqq b,a\mod b$
\end_inset

.
\end_layout

\begin_layout Standard

\series bold
Primer:
\series default
 
\begin_inset Formula $D\left(882,666\right)=D\left(666,216\right)=D\left(216,18\right)=18$
\end_inset

.
 Pri zadnji operaciji 
\begin_inset Formula $\mod$
\end_inset

 smo namreč dobili rezultat 0, kar prekine izvajanje algoritma, rezultat
 pa je 
\begin_inset Formula $b$
\end_inset

.
\end_layout

\begin_layout Section
Viri, literatura in dodatno branje
\end_layout

\begin_layout Standard
TODO: popravi vire v slovenščino (namesto and naj bo in)
\end_layout

\begin_layout Standard
\begin_inset CommandInset bibtex
LatexCommand bibtex
btprint "btPrintCited"
bibfiles "IEEEexample,../citati"
options "plain"

\end_inset


\end_layout

\begin_layout Section
Zaključek
\end_layout

\begin_layout Standard
Za možne napake ne odgovarjam, bi bil pa vesel, če mi jih sporočite na 
\begin_inset CommandInset href
LatexCommand href
target "anton@šijanec.eu."
type "mailto:"
literal "true"

\end_inset

 Srečno!
\end_layout

\end_body
\end_document