Difference between revisions of "TU Wien:Algebra und Diskrete Mathematik UE (diverse)/Übungen SS19/Beispiel 271"

From VoWi
Jump to navigation Jump to search
Line 2: Line 2:
 
Man bestimme <math>G_1 \cap G_2</math> und <math>G_1 \cup G_2</math>:  
 
Man bestimme <math>G_1 \cap G_2</math> und <math>G_1 \cup G_2</math>:  
  
<math>G_1: V(G_1) = \{1, 2, ..., 8\}, E(G_1) = \{\langle x, y \rangle | x teilt <, x < y\}</math>  
+
<math>G_1: V(G_1) = \{1, 2, ..., 8\}, E(G_1) = \{\langle x, y \rangle | x teilt y, x < y\}</math>  
  
 
<math>G_2: V(G_2) = \{1, 2, ..., 5\}, E(G_2) = \{\langle x, y  \rangle | x < y <= x + 3\}</math>
 
<math>G_2: V(G_2) = \{1, 2, ..., 5\}, E(G_2) = \{\langle x, y  \rangle | x < y <= x + 3\}</math>

Revision as of 13:35, 24 February 2013

Angabe

Man bestimme G_1 \cap G_2 und G_1 \cup G_2:

G_1: V(G_1) = \{1, 2, ..., 8\}, E(G_1) = \{\langle x, y \rangle | x teilt y, x < y\}

G_2: V(G_2) = \{1, 2, ..., 5\}, E(G_2) = \{\langle x, y  \rangle | x < y <= x + 3\}

Lösungsvorschlag

für V(G_1) \cap V(G_2) = \{1, 2, 3, 4, 5\}

für E(G_1) \cap E(G_2) = \{(1,2), (1,3), (1, 4), (2,4)\}


für V(G_1) \cup V(_G2) = \{1, 2, 3, 4, 5, 6, 7, 8\}

für E(G_1) \cup E(G_2) = \{(1,2), (1,3), (1,4), (1, 5), (1, 6), (1, 7), (1,8), (2, 4), (2,5), (2, 6), (2, 8), (3,4), (3,5), (3, 6)\}