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+#LyX 2.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 544
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass article
+\begin_preamble
+\usepackage[dvipsnames]{xcolor}
+\usepackage{siunitx}
+\usepackage{pgfplots}
+\usepackage{listings}
+\usepackage{multicol}
+\sisetup{output-decimal-marker = {,}, quotient-mode=fraction, output-exponent-marker=\ensuremath{\mathrm{3}}}
+\usepackage{tikz}
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children false
+\language slovene
+\language_package default
+\inputencoding utf8
+\fontencoding global
+\font_roman "default" "default"
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+\font_math "auto" "auto"
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+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\float_placement H
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification false
+\use_refstyle 1
+\use_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\leftmargin 1cm
+\topmargin 0cm
+\rightmargin 1cm
+\bottommargin 2cm
+\headheight 1cm
+\headsep 1cm
+\footskip 1cm
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+\tocdepth 3
+\paragraph_separation indent
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+\output_changes false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\end_header
+
+\begin_body
+
+\begin_layout Title
+Odgovori na ustna vprašanja višje ravni na ustni maturi 2023
+\end_layout
+
+\begin_layout Author
+
+\noun on
+Anton Luka Šijanec
+\end_layout
+
+\begin_layout Date
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+today
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+newcommand
+\backslash
+euler{e}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset toc
+LatexCommand tableofcontents
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Vprašanja in odgovori
+\end_layout
+
+\begin_layout Subsection
+Izjavni račun
+\end_layout
+
+\begin_layout Subsubsection*
+Kaj je izjava?
+\begin_inset space \hfill{}
+\end_inset
+
+(1
+\begin_inset space ~
+\end_inset
+
+točka)
+\end_layout
+
+\begin_layout Quotation
+Popoln (z vsemi nujnimi stavčnimi členi in slovnično pravilen) trdilni ali
+ nikalni stavek je 
+\series bold
+smiseln
+\series default
+, če se v okviru predmetov in pojmov, o katerih stavek govori (v njegovem
+ 
+\series bold
+kontekstu
+\series default
+), vsaj načelno lahko lahko odločimo, ali je njegova vsebina 
+\series bold
+resnična
+\series default
+ ali 
+\series bold
+lažna
+\series default
+.
+ Vsi smiselni stavki, ki trdijo isto, določajo 
+\series bold
+izjavo
+\series default
+.
+\begin_inset CommandInset citation
+LatexCommand cite
+key "cedilnik12"
+literal "false"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection*
+Kaj je negacija dane izjave? Kdaj je negacija pravilna (resnična) in kdaj
+ nepravilna (neresnična)?
+\begin_inset space \hfill{}
+\end_inset
+
+(1
+\begin_inset space ~
+\end_inset
+
+točka)
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Negacija
+\series default
+ 
+\begin_inset Formula $\neg A$
+\end_inset
+
+ (ali: 
+\begin_inset Formula $\overline{A}$
+\end_inset
+
+) izjave 
+\begin_inset Formula $A$
+\end_inset
+
+ je izjava, ki je resnična natanko tedaj, ko je 
+\begin_inset Formula $A$
+\end_inset
+
+ lažna (Tabela 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "tab:Negacija."
+plural "false"
+caps "false"
+noprefix "false"
+
+\end_inset
+
+).
+\begin_inset CommandInset citation
+LatexCommand cite
+key "cedilnik12"
+literal "false"
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+\end_inset
+
+
+\end_layout
+
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+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\neg A$
+\end_inset
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+
+\end_layout
+
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+\begin_layout Plain Layout
+\align center
+\begin_inset Caption Standard
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+\begin_layout Plain Layout
+Negacija.
+\begin_inset CommandInset label
+LatexCommand label
+name "tab:Negacija."
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+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection*
+Kaj je konjunkcija izjav?
+\begin_inset space \hfill{}
+\end_inset
+
+(1
+\begin_inset space ~
+\end_inset
+
+točka)
+\end_layout
+
+\begin_layout Quotation
+Če izjavi 
+\begin_inset Formula $A$
+\end_inset
+
+ in 
+\begin_inset Formula $B$
+\end_inset
+
+ povežemo z veznikom 
+\begin_inset Quotes gld
+\end_inset
+
+in
+\begin_inset Quotes grd
+\end_inset
+
+, dobimo 
+\series bold
+konjunkcijo
+\series default
+ 
+\begin_inset Formula $A\wedge B$
+\end_inset
+
+ (ali tudi 
+\begin_inset Formula $A\&B$
+\end_inset
+
+).
+ Konjunkcija je resnična le tedaj, kadar sta oba člena resnični izjavi (Tabela
+ 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "tab:Osnovne-logične-povezave."
+plural "false"
+caps "false"
+noprefix "false"
+
+\end_inset
+
+).
+\begin_inset CommandInset citation
+LatexCommand cite
+key "cedilnik12"
+literal "false"
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+
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+\begin_inset Float table
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+\begin_inset Text
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+\begin_inset Formula $A\Rightarrow B$
+\end_inset
+
+
+\end_layout
+
+\end_inset
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+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $A\wedge B$
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+
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+
+\end_inset
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+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $A\vee B$
+\end_inset
+
+
+\end_layout
+
+\end_inset
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+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $A\veebar B$
+\end_inset
+
+
+\end_layout
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+\end_layout
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+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
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+\begin_layout Plain Layout
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+\end_layout
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+\end_inset
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+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
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+\end_layout
+
+\end_inset
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+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
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+\begin_inset Text
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+\begin_layout Plain Layout
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+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+1
+\end_layout
+
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+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+0
+\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
+0
+\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" 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" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+0
+\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
+0
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+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
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+
+\begin_layout Plain Layout
+0
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+</lyxtabular>
+
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+
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Osnovne logične povezave.
+\begin_inset CommandInset label
+LatexCommand label
+name "tab:Osnovne-logične-povezave."
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection*
+Kaj je disjunkcija izjav? Dokažite, da je izjava 
+\begin_inset Formula $\neg\left(A\wedge B\right)$
+\end_inset
+
+ enakovredna izjavi 
+\begin_inset Formula $\neg\left(A\right)\vee\neg\left(B\right)$
+\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 Standard
+De Morganovi
+\begin_inset CommandInset citation
+LatexCommand cite
+key "cedilnik12"
+literal "false"
+
+\end_inset
+
+ pravili 
+\begin_inset Formula $\neg\left(A\wedge B\right)\sim\neg A\vee\neg B$
+\end_inset
+
+ in 
+\begin_inset Formula $\neg\left(A\vee B\right)\sim\neg A\wedge\neg B$
+\end_inset
+
+ najlažje dokažemo z logično tabelo (Tabela 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "tab:Logična-tabela-za"
+plural "false"
+caps "false"
+noprefix "false"
+
+\end_inset
+
+).
+\end_layout
+
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+\begin_inset Float table
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+\align center
+\begin_inset Tabular
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+
+\begin_layout Plain Layout
+\begin_inset Formula $A$
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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
+\begin_inset Formula $B$
+\end_inset
+
+
+\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
+\begin_inset Formula $\neg\left(A\wedge B\right)$
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+
+
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+
+\end_inset
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+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\neg A\vee\neg B$
+\end_inset
+
+
+\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
+\begin_inset Formula $\neg\left(A\vee B\right)$
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+
+\end_layout
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+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
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+\begin_layout Plain Layout
+\begin_inset Formula $\neg A\wedge\neg B$
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+\begin_inset Text
+
+\begin_layout Plain Layout
+0
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="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" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+0
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="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" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+1
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+0
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+0
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+0
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="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" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+1
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="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" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+0
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+0
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+0
+\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
+0
+\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" 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" 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