Grundlagen der Algebra und der elementaren Zahlentheorie

UNIVERSITÄT KOBLENZ LANDAU, CAMPUS LANDAU
INSTITUT FÜR MATHEMATIK
Dr. Dominik Faas
Grundlagen der Algebra und der elementaren Zahlentheorie
Sommersemester 2011
Lösungen zu den Übungsaufgaben zur Vorlesung vom 26.05.2011
Aufgabe Ü12
(a) Gesucht ist jeweils die Lösungsmenge L der angegebenen Diophantischen Gleichung:
(i) 27 · x − 8 · y = 2
Also: a = 27, b = −8, c = 2
Erweiterter Euklidischer Algorithmus:
1
(1) 27 = 3 · 8 + 3
(2) 8 = 2 · 3 + 2
(3) 3 = 1 · 2 + 1
(2 = 2 · 1)
(3)
=
(2)
=
=
⇒
(1)
=
=
3−1·2
3 − 1 · (8 − 2 · 3)
3·3−1·8
3 · (27 − 3 · 8) − 1 · 8
3 · 27 − 10 · 8
Wegen ggT(a, b) = 1 ist die Gleichung lösbar. Eine Lösung findet man nun:
s.o
⇒
1 = 3 · a + 10 · b
·2
⇒
c = 6 · a + 20 · b
Also ist (6, 20) ∈ L.
Nach 3.15 (b) ist nun
L = {(6 + t · (−8), 20 − t · 27); t ∈ Z}
= {. . . , (30, 101), (22, 74), (14, 47), (6, 20), (−2, −7), (−10, −34), (−18, −61), . . .}
(ii) 0 · x + 15 · y = −105
Also: a = 0, b = 15, c = −105
Es ist
:15
0 · x + 15 · y = −105 ⇔ 15 · y = −105 ⇔
y = −7
Also ist
L = {(x, 7); x ∈ Z} = {. . . , (−3, 7), (−2, 7), (−1, 7), (0, 7), (1, 7), (2, 7), (3, 7), . . .}
(iii) 1665 · x + 592 · y = 167
Also: a = 1665, b = 592, c = 167
Euklidischer Algorithmus:
1665
592
481
(111
=
=
=
=
2 · 592 + 481
1 · 481 + 111
4 · 111 + 37
3 · 37)
Also ist ggT(a, b) = 37. Es gilt 37 - 167, das heißt ggT(a, b) - c. Somit ist L = ∅.
(iv) −48 · x + 30 · y = 54
Also: a = −48, b = 30, c = 54
Erweiterter Euklidischer Algorithmus:
(1)
(2)
(3)
48
30
18
(12
=
=
=
=
6
1 · 30 + 18
1 · 18 + 12
1 · 12 + 6
2 · 6)
(3)
=
(2)
=
=
⇒
(1)
=
=
18 − 1 · 12
18 − 1 · (30 − 1 · 18)
2 · 18 − 1 · 30
2 · (48 − 1 · 30) − 1 · 30
2 · 48 − 3 · 30
Also ist ggT(a, b) = 6. Es gilt 6 | 54, das heißt ggT(a, b) | c. Somit ist die
Gleichung lösbar. Eine Lösung findet man nun:
s.o
⇒
·9
6 = (−2) · a + (−3) · b
⇒
c = (−18) · a + (−27) · b
Also ist (−18, −27) ∈ L.
Nach 3.15 (c) ist nun
30
−48
L =
(−18 + t · , −27 − t ·
); t ∈ Z
6
6
= {(−18 + 5t, −27 + 8t); t ∈ Z}
= {. . . , (−23, −35), (−18, −27), (−13, −19), (−8, −11), (−3, −3), (2, 5), (7, 13), . . .}
(iv) (alternativ) Es gilt ggT(a, b) = 6 und damit ggT(a, b) | c. Wir können die Gleicuung also zunächst durch ggT(a, b) teilen.
−48 · x + 30 · y = 54
:6
⇒
−8 · x + 5 · y = 9
Die auftretenden Koeffizienten −8, 5 sind nun teilerfremd (wie nach 3.4 (d) erwartet).
Erweiterter Euklidischer Algorithmus:
(1)
(2)
(3)
8
5
3
(2
=
=
=
=
1
1·5+3
1·3+2
1·2+1
2 · 1)
(3)
=
(2)
=
=
⇒
(1)
=
=
3−1·2
3 − 1 · (5 − 1 · 3)
2·3−1·5
2 · (8 − 1 · 5) − 1 · 5
2·8−3·5
Eine Lösung findet man nun:
s.o
⇒
1 = (−2) · (−8) + (−3) · 5
·9
⇒
9 = (−18) · (−8) + (−27) · 5
Also ist (−18, −27) ∈ L.
Nach 3.15 (b) ist nun L = {(−18 + t · 5, −27 − t · (−8); t ∈ Z} wie oben.
(b) Es gilt:
−48 · x + 30 · y = 54
⇔
Also ist L(Q) = (x, y) ∈ Q × Q; y =
Achsenabschnitt 95 :
48
54
8
9
·x+
⇔ y = ·x+
30
30
5
5
· x + 95 die Gerade mit Steigung
y=
8
5
8
5
und y-
y
−40
−20
0
20
40
Lösungsmenge
−40
−20
0
20
40
x
Die ganzzahligen Lösungen entsprechen den Punkten der Geraden mit ganzzahligen
Koordinaten (Gitterpunkte):
40
Lösungsmenge
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−40
−20
0
20
40
x
(c) Gesucht sind x, y ∈ N0 mit x · 2 + y · 5 = 53 . Man findet hier leicht eine Lösung
dieser Diophantischen Gleichung durch Ausprobieren, zum Beispiel ist (4, 9) ∈ L.
Wegen ggT(2, 5) = 1 gilt nach 3.15 (b):
L = {(4 + 5t, 9 − 2t); t ∈ Z}
= {. . . , (−6, 13), (−1, 11), (4, 9), (9, 7), (14, 5), (19, 3), (24, 1), (29, −1), (34, −3) . . .}
Tatsächlich realisierbar sind nur die Lösungen (4 + 5t, 9 − 2t) mit 4 + 5t ≥ 0 und
9 − 2t ≥ 0. Das ist genau für t ∈ {0, 1, 2, 3, 4} gegeben. Es gibt also 5 tatsächlich
realisierbare Möglichkeiten, nämlich:
(x, y) ∈ {(4, 9), (9, 7), (14, 5), (19, 3), (24, 1)}
Aufgabe Ü13
(a) In den ersten drei Fällen berechnen wir das kgV mit der PFZ (vergleiche 3.18):
•
10 = 21 · 30 · 51 · 70 · 110 · . . .
15 = 20 · 31 · 51 · 71 · 110 · . . .
kgV(10, 15) = 21 · 31 · 51 · 70 · 110 · . . .
= 30
•
40 = 23 · 30 · 51 · 70 · 110 · . . .
63 = 20 · 32 · 50 · 71 · 110 · . . .
kgV(40, 63) = 23 · 32 · 51 · 71 · 110 · . . .
= 2520
•
900 = 22 · 32 · 52 · 70 · 110 · . . .
56 = 23 · 30 · 50 · 71 · 110 · . . .
675 = 20 · 33 · 52 · 70 · 110 · . . .
kgV(900, 56, 675) = 23 · 33 · 52 · 71 · 110 · . . .
= 37800
• Euklidischer Algorithmus:
12345
6789
5556
1233
624
609
15
9
(6
=
=
=
=
=
=
=
=
=
1 · 6789 + 5556
1 · 5556 + 1233
4 · 1233 + 624
1 · 624 + 609
1 · 609 + 15
40 · 15 + 9
1·9+6
1·6+3
2 · 3)
Also ist ggT(12345, 6789) = 3. Nach 3.19(d) ist
kgV(12345, 6789) =
12345 · 6789
= 27936735
3
(b)
(i) Die Aussage ist wahr.
Beweis: (für a, b > 0) Wir betrachten die PFZ der beiden Zahlen:
a=
∞
Y
pj
ej
und b =
j=0
∞
Y
p j fj
(mit ej , fj ∈ N0 )
j=0
Nach 2.13 (a) und 3.18 gilt:
a2 =
∞
Y
pj 2·ej
und b2 =
j=0
∞
Y
pj 2·fj
⇒
∞
Y
kgV a2 , b2 =
pj max(2·ej ,2·fj )
j=0
j=0
und
kgV (a, b) =
∞
Y
pj max(ej ,fj )
⇒
(kgV (a, b))2 =
j=0
∞
Y
pj 2·max(ej ,fj )
j=0
Offenbar ist aber max (2 · ej , 2 · fj ) = 2 · max (ej , fj ) für alle j ∈ N. Damit folgt
die behauptete Gleichheit.
(ii) Die Aussage ist falsch.
Gegenbeispiel: Für a = 2, b = 2, c = 3 ist ggT(a, b, c) = 1, aber
kgV(a, b, c) = 6 6= 12 = a · b · c
.
(iii) Die Aussage ist wahr.
Beweis: (für a, b, c > 0) Vorausgesetzt sei: ggT(a, b) = ggT(a, c) = ggT(b, c) = 1
Wir betrachten die PFZ der Zahlen:
a=
∞
Y
pj ej
b=
j=0
∞
Y
p j fj
c=
j=0
∞
Y
p j gj
(mit ej , fj , gj ∈ N0 )
j=0
Mit 2.13 (a) und 3.18 folgt:
a·b·c=
∞
Y
pj
(ej +fj +gj )
und
kgV(a, b, c) =
j=0
∞
Y
pj max(ej ,fj ,gj )
j=0
Es muss nun die Gleichheit der Exponenten begründet werden, das heißt:
noch zu zeigen: ej + fj + gj = max (ej , fj , gj )
(für alle j ∈ N)
Wegen ggT(a, b) = 1 ist stets ej = 0 oder fj = 0.
Wegen ggT(a, c) = 1 ist stets ej = 0 oder gj = 0.
Wegen ggT(b, c) = 1 ist stets fj = 0 oder gj = 0.
Zusammen folgt: Für jedes j ∈ N sind immer mindestens zwei der Exponenten
ej , fj , gj gleich 0. Damit ist ihre Summe ej + fj + gj stets gleich dem größten
Summanden max (ej , fj , gj ).
Diese Lösungen finden sie auch unter
http://www.uni-koblenz-landau.de/landau/fb7/mathematik/team/dominik-faas/material/azt