From 4f8031a27285553746ac1c043e1f9b7bf885a7b0 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Anton=20Luka=20=C5=A0ijanec?= Date: Sun, 2 Jul 2023 22:01:05 +0200 Subject: eden izmed zadnjih commitov --- mat/ustno.lyx | 3112 +++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 3112 insertions(+) create mode 100644 mat/ustno.lyx (limited to 'mat') diff --git a/mat/ustno.lyx b/mat/ustno.lyx new file mode 100644 index 0000000..d0d444c --- /dev/null +++ b/mat/ustno.lyx @@ -0,0 +1,3112 @@ +#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" +\font_sans "default" "default" +\font_typewriter "default" "default" +\font_math "auto" "auto" +\font_default_family default +\use_non_tex_fonts false +\font_sc false +\font_osf false +\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 +\secnumdepth 3 +\tocdepth 3 +\paragraph_separation indent +\paragraph_indentation default +\is_math_indent 0 +\math_numbering_side default +\quotes_style german +\dynamic_quotes 0 +\papercolumns 1 +\papersides 1 +\paperpagestyle default +\tracking_changes false +\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" + +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Float table +wide false +sideways false +status open + +\begin_layout Plain Layout +\begin_inset Tabular + + + + + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $A$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $\neg A$ +\end_inset + + +\end_layout + +\end_inset + + + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + + + +\end_inset + + +\end_layout + +\begin_layout Plain Layout +\align center +\begin_inset Caption Standard + +\begin_layout Plain Layout +Negacija. +\begin_inset CommandInset label +LatexCommand label +name "tab:Negacija." + +\end_inset + + +\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" + +\end_inset + + +\end_layout + +\begin_layout Quotation +\begin_inset Float table +wide false +sideways false +status open + +\begin_layout Plain Layout +\align center +\begin_inset Tabular + + + + + + + + + + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $A$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $B$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $A\Leftrightarrow B$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $A\Rightarrow B$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $A\wedge B$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $A\vee B$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $A\veebar B$ +\end_inset + + +\end_layout + +\end_inset + + + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + + + +\end_inset + + +\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 + +\begin_layout Standard +\begin_inset Float table +wide false +sideways false +status open + +\begin_layout Plain Layout +\align center +\begin_inset Tabular + + + + + + + + + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $A$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $B$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $\neg\left(A\wedge B\right)$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $\neg A\vee\neg B$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $\neg\left(A\vee B\right)$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $\neg A\wedge\neg B$ +\end_inset + + +\end_layout + +\end_inset + + + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +0 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + + + +\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