Gli Infiniti Valori Derivanti dalla Frazione 5/3 - Due Formule - Sette Dimostrazioni - Tanti Esempi - 27 Slide - Divisione di Numeri Naturali - Divisioni Parziali - Infiniti Quozienti - Infiniti Resti #Matematica #Aritmetica #Frazioni #Numeratore #Denominatore #Divisioni #Dividendo #Divisore #Quoziente #Resto #Divisioni_Parziali #Divisioni_Totali #Sottrazioni #Numeri_Naturali #Numeri_Razionali #Infinito #Math #Maths #Mathematics #Arithmetic #Fractions #Numerator #Denominator #Divisions #Dividend #Divider #Quotient #Remainder #Partial_Divisions #Total_Divisions #Subtractions #Natural_Numbers #Rational_Numbers #Infinity https://www.youtube.com/user/enzoexposito http://www.enzoexposito.it/mobile/matematica.html http://www.enzoexposito.it/mobile/aritmetica.html https://www.slideshare.net/EnzoExposito1 https://www.linkedin.com/in/enzo-exposyto-aa970530/detail/recent-activity/shares/ https://www.facebook.com/enzoexposyto https://twitter.com/enzoexposyto https://www.tumblr.com/blog/enzoexposyto https://www.pinterest.it/enzoexposyto/_saved/

1. DIVISIONI PARZIALI
i VALORI di 5
3
FORMULA A)
a cura di Enzo Exposyto
A)
5
3
= 1,
n − 1
6 5 +
5
3 ⋅ 1
n
0
∀n ∈ N+
= {1,2,3,...}

2. DIVISIONI PARZIALI
i VALORI di 5
3
FORMULA B)
a cura di Enzo Exposyto
B)
5
3
= 1,
n
6 +
2
3 ⋅ 1
n
0
∀n ∈ N = {0,1,2,3,...}

3. A) i VALORI di 5
a cura di Enzo Exposyto 3
La SCRITTURA “n-1” SUL 6 INDICA CHE
la CIFRA 6 È RIPETUTA “n-1” VOLTE;
la “n” SOPRA lo 0 INDICA CHE
la CIFRA 0 È RIPETUTA “n” VOLTE.
n È un QUALSIASI NUMERO INTERO,
appartenente all’insieme N+ = {1,2,3,...}.
Ad esempio, con n = 5, la FORMULA A) diventa
4 volte 6 5 volte 0
A)
5
3
= 1,
n − 1
6 5 +
5
3 ⋅ 1
n
0
∀n ∈ N+
= {1,2,3,...}
A)
5
3
= 1,66665 +
5
3 ⋅ 100000 n = 5

4. B) i VALORI di 5
a cura di Enzo Exposyto 3
Le DUE “n” -SUL 6 e SOPRA lo 0- INDICANO CHE
la CIFRA 6 È RIPETUTA “n” VOLTE,
la CIFRA 0 È RIPETUTA “n” VOLTE.
n è un QUALSIASI NUMERO INTERO,
appartenente all’insieme N = {0,1,2,3,...}.
Ad esempio, con n = 5, la FORMULA B) diventa
5 volte 6 5 volte 0
B)
5
3
= 1,
n
6 +
2
3 ⋅ 1
n
0
B)
5
3
= 1,
5
6 +
2
3 ⋅ 1
5
0
= 1,66666 +
2
3 ⋅ 100000
∀n ∈ N = {0,1,2,3,...}
n = 5

5. FORMULA A)
1^ DIMOSTRAZIONE
A)
5
3
= 1,
n − 1
6 5 +
5
3 ⋅ 1
n
0
∀n ∈ N+
= {1,2,3,...}

6. Formula A) - 1^ dimostrazione
Quindi
5
3
= 5 ⋅
1
3
= 5 ⋅ 0,
n
3 +
5
3 ⋅ 1
n
0
= 5 ⋅ (0,
n
3 +
1
3 ⋅ 1
n
0
)
= 1,
n − 1
6 5 +
5
3 ⋅ 1
n
0
5
3
= 5 ⋅
1
3
= 1,
n − 1
6 5 +
5
3 ⋅ 1
n
0
∀n ∈ N = {0,1,2,3,...}
∀n ∈ N+
= {1,2,3,...}
∀n ∈ N+
= {1,2,3,...}

7. FORMULA A)
2^ DIMOSTRAZIONE
A)
5
3
= 1,
n − 1
6 5 +
5
3 ⋅ 1
n
0
∀n ∈ N+
= {1,2,3,...}

8. Formula A) - 2^ dimostrazione
Quindi
5
3
=
1
3
+
4
3
= (0,
n
3 +
1
3 ⋅ 1
n
0
) + (1,
n − 1
3 2 +
4
3 ⋅ 1
n
0
) ∀n ∈ N+
= {1,2,3,...}
= 1,
n − 1
6 5 +
5
3 ⋅ 1
n
0
5
3
=
1
3
+
4
3
= 1,
n − 1
6 5 +
5
3 ⋅ 1
n
0
∀n ∈ N+
= {1,2,3,...}
= 0,
n
3 + 1,
n − 1
3 2 +
1
3 ⋅ 1
n
0
+
4
3 ⋅ 1
n
0

9. FORMULA A)
3^ DIMOSTRAZIONE con PREMESSA
A)
5
3
= 1,
n − 1
6 5 +
5
3 ⋅ 1
n
0
∀n ∈ N+
= {1,2,3,...}

10. Formula A)
PREMESSA per la 3^ DIMOSTRAZIONE
Quindi
2
3
= 2 ⋅
1
3
= 2 ⋅ (0,
n
3 +
1
3 ⋅ 1
n
0
)
= 2 ⋅ 0,
n
3 +
2
3 ⋅ 1
n
0
= 0,
n
6 +
2
3 ⋅ 1
n
0
∀n ∈ N = {0,1,2,3,...}
2
3
= 2 ⋅
1
3
= 0,
n
6 +
2
3 ⋅ 1
n
0
∀n ∈ N = {0,1,2,3,...}

11. Formula A) - 3^ dimostrazione
Quindi
5
3
=
2
3
+
3
3
= (0,
n
6 +
2
3 ⋅ 1
n
0
) + (0,
n
9 +
1
1
n
0
) ∀n ∈ N = {0,1,2,3,...}
∀n ∈ N+
= {1,2,3,...}
5
3
=
2
3
+
3
3
= 1,
n − 1
6 5 +
5
3 ⋅ 1
n
0
∀n ∈ N+
= {1,2,3,...}
= 0,
n
6 + 0,
n
9 +
2
3 ⋅ 1
n
0
+
3
3 ⋅ 1
n
0
= 1,
n − 1
6 5 +
5
3 ⋅ 1
n
0

12. FORMULA A)
4^ DIMOSTRAZIONE
A)
5
3
= 1,
n − 1
6 5 +
5
3 ⋅ 1
n
0
∀n ∈ N+
= {1,2,3,...}

13. 4^ dimostrazione sintetica della formula A)
La formula è dimostrata visto che
P(1) = 1,
1 − 1
6 5 +
5
3 ⋅ 1
1
0
= 1,
0
65 +
5
3 ⋅ 1
1
0
= 1,5 +
5
3 ⋅ 10
=
3 ⋅ 10 ⋅ 1,5 + 5
3 ⋅ 10
=
45 + 5
3 ⋅ 10
=
50
30
=
5
3
1,
n − 1
6 5 +
5
3 ⋅ 1
n
0
=
5
3
∀n ∈ N+
= {1,2,3,...}
P(1) =
5
3
P(n) =
5
3
P(n + 1) =
5
3
P(n) =
5
3
→ P(n + 1) =
5
3
P(1) =
5
3
∀n ∈ N+
= {1,2,3,...}
P(n) = 1,
n − 1
6 5 +
5
3 ⋅ 1
n
0
=
3 ⋅ 1
n
0 ⋅ 1,
n − 1
6 5 + 5
3 ⋅ 1
n
0
=
3 ⋅ 1
n − 1
6 5 + 5
3 ⋅ 1
n
0
=
4
n − 1
9 5 + 5
3 ⋅ 1
n
0
=
5
n
0
3 ⋅ 1
n
0
=
5 ⋅ 1
n
0
3 ⋅ 1
n
0
=
5
3
P(n + 1) = 1,
n + 1 − 1
6 5 +
5
3 ⋅ 1
n + 1
0
=
3 ⋅ 1
n + 1
0 ⋅ 1,
n
65 + 5
3 ⋅ 1
n + 1
0
=
3 ⋅ 1
n
65 + 5
3 ⋅ 1
n + 1
0
=
4
n
95 + 5
3 ⋅ 1
n + 1
0
=
5
n + 1
0
3 ⋅ 1
n + 1
0
=
5 ⋅ 1
n + 1
0
3 ⋅ 1
n + 1
0
=
5
3

14. 4^ dimostrazione estesa della formula A)
Qui, sarà dimostrata la forma
∀n ∈ N+
= {1,2,3,...}
∀n ∈ N+
= {1,2,3,...}
A)
5
3
= 1,
n − 1
6 5 +
5
3 ⋅ 1
n
0
A) 1,
n − 1
6 5 +
5
3 ⋅ 1
n
0
=
5
3

15. 4a) Dimostrazione della Base
Quindi
P(1) =
5
3
P(1) = 1,
1 − 1
6 5 +
5
3 ⋅ 1
1
0
= 1,
0
65 +
5
3 ⋅ 1
1
0
= 1,5 +
5
3 ⋅ 10
=
3 ⋅ 10 ⋅ 1,5 + 5
3 ⋅ 10
=
45 + 5
3 ⋅ 10
=
50
30
=
5
3

16. 4b) Dimostrazione del Passo Induttivo
P(n) =
5
3
∀n ∈ N+
= {1,2,3,...}
P(n + 1) =
5
3
∀n ∈ N+
= {1,2,3,...}
P(n) = 1,
n − 1
6 5 +
5
3 ⋅ 1
n
0
=
3 ⋅ 1
n
0 ⋅ 1,
n − 1
6 5 + 5
3 ⋅ 1
n
0
=
3 ⋅ 1
n − 1
6 5 + 5
3 ⋅ 1
n
0
=
4
n − 1
9 5 + 5
3 ⋅ 1
n
0
=
5
n
0
3 ⋅ 1
n
0
=
5 ⋅ 1
n
0
3 ⋅ 1
n
0
=
5
3
P(n + 1) = 1,
n + 1 − 1
6 5 +
5
3 ⋅ 1
n + 1
0
=
3 ⋅ 1
n + 1
0 ⋅ 1,
n
65 + 5
3 ⋅ 1
n + 1
0
=
3 ⋅ 1
n
65 + 5
3 ⋅ 1
n + 1
0
=
4
n
95 + 5
3 ⋅ 1
n + 1
0
=
5
n + 1
0
3 ⋅ 1
n + 1
0
=
5 ⋅ 1
n + 1
0
3 ⋅ 1
n + 1
0
=
5
3

17. Conclusioni da 4a) e 4b)
Poiché
e
ne deriva che
P(1) =
5
3
P(n) =
5
3
→ P(n + 1) =
5
3
∀n ∈ N+
= {1,2,3,...}
∀n ∈ N+
= {1,2,3,...}
A) 1,
n − 1
6 5 +
5
3 ⋅ 1
n
0
=
5
3

18. FORMULA B)
1^ DIMOSTRAZIONE
B)
5
3
= 1,
n
6 +
2
3 ⋅ 1
n
0
∀n ∈ N = {0,1,2,3,...}

19. Formula B) - 1^ Dimostrazione
Quindi
5
3
= 1 +
2
3
= 1 + (0,
n
6 +
2
3 ⋅ 1
n
0
) ∀n ∈ N = {0,1,2,3,...}
= 1 + 0,
n
6 +
2
3 ⋅ 1
n
0
= 1,
n
6 +
2
3 ⋅ 1
n
0
∀n ∈ N = {0,1,2,3,...}
5
3
= 1 +
2
3
= 1,
n
6 +
2
3 ⋅ 1
n
0
∀n ∈ N = {0,1,2,3,...}

20. FORMULA B)
2^ DIMOSTRAZIONE
B)
5
3
= 1,
n
6 +
2
3 ⋅ 1
n
0
∀n ∈ N = {0,1,2,3,...}

21. Formula B) - 2^ Dimostrazione
Quindi
5
3
=
4
3
+
1
3
= (1,
n
3 +
1
3 ⋅ 1
n
0
) + (0,
n
3 +
1
3 ⋅ 1
n
0
) ∀n ∈ N = {0,1,2,3,...}
= 1,
n
6 +
2
3 ⋅ 1
n
0
∀n ∈ N = {0,1,2,3,...}
5
3
=
4
3
+
1
3
= 1,
n
6 +
2
3 ⋅ 1
n
0
∀n ∈ N = {0,1,2,3,...}
= 1,
n
3 + 0,
n
3 +
1
3 ⋅ 1
n
0
+
1
3 ⋅ 1
n
0

22. FORMULA B)
3^ DIMOSTRAZIONE
B)
5
3
= 1,
n
6 +
2
3 ⋅ 1
n
0
∀n ∈ N = {0,1,2,3,...}

23. 3^ dimostrazione sintetica della formula B)
La formula è dimostrata visto che
1,
n
6 +
2
3 ⋅ 1
n
0
=
5
3
∀n ∈ N = {0,1,2,3,...}
P(1) =
5
3
P(1) = 1,
1
6 +
2
3 ⋅ 1
1
0
= 1,6 +
2
3 ⋅ 10
=
3 ⋅ 10 ⋅ 1,6 + 2
3 ⋅ 10
=
48 + 2
3 ⋅ 10
=
50
30
=
5
3
P(n) = 1,
n
6 +
2
3 ⋅ 1
n
0
=
3 ⋅ 1
n
0 ⋅ 1,
n
6 + 2
3 ⋅ 1
n
0
=
3 ⋅ 1
n
6 + 2
3 ⋅ 1
n
0
=
4
n − 1
9 8 + 2
3 ⋅ 1
n
0
=
5
n
0
3 ⋅ 1
n
0
=
5 ⋅ 1
n
0
3 ⋅ 1
n
0
=
5
3
P(n + 1) = 1,
n + 1
6 +
2
3 ⋅ 1
n + 1
0
=
3 ⋅ 1
n + 1
0 ⋅ 1,
n + 1
6 + 2
3 ⋅ 1
n + 1
0
=
3 ⋅ 1
n + 1
6 + 2
3 ⋅ 1
n + 1
0
=
4
n
98 + 2
3 ⋅ 1
n + 1
0
=
5
n + 1
0
3 ⋅ 1
n + 1
0
=
5 ⋅ 1
n + 1
0
3 ⋅ 1
n + 1
0
=
5
3
P(n) =
5
3
P(n + 1) =
5
3
P(0) =
5
3
∧ P(1) =
5
3
P(n) =
5
3
→ P(n + 1) =
5
3
∀n ∈ N = {0,1,2,3,...}
P(0) = 1,
0
6 +
2
3 ⋅ 1
0
0
= 1 +
2
3 ⋅ 1
=
3 ⋅ 1 ⋅ 1 + 2
3 ⋅ 1
=
5
3
P(0) =
5
3

24. 3^ dimostrazione estesa della formula B)
Qui, sarà dimostrata la forma
B)
5
3
= 1,
n
6 +
2
3 ⋅ 1
n
0
∀n ∈ N = {0,1,2,3,...}
B) 1,
n
6 +
2
3 ⋅ 1
n
0
=
5
3
∀n ∈ N = {0,1,2,3,...}

25. 3a) Dimostrazione della Base
Quindi
Quindi
P(1) =
5
3
P(0) = 1,
0
6 +
2
3 ⋅ 1
0
0
= 1 +
2
3 ⋅ 1
=
3 ⋅ 1 ⋅ 1 + 2
3 ⋅ 1
=
5
3
P(0) =
5
3
P(1) = 1,
1
6 +
2
3 ⋅ 1
1
0
= 1,6 +
2
3 ⋅ 10
=
3 ⋅ 10 ⋅ 1,6 + 2
3 ⋅ 10
=
48 + 2
3 ⋅ 10
=
50
30
=
5
3

26. 3b) Dimostrazione del Passo Induttivo
P(n) =
5
3
∀n ∈ N = {0,1,2,3,...}
P(n + 1) =
5
3
∀n ∈ N = {0,1,2,3,...}
P(n) = 1,
n
6 +
2
3 ⋅ 1
n
0
=
3 ⋅ 1
n
0 ⋅ 1,
n
6 + 2
3 ⋅ 1
n
0
=
3 ⋅ 1
n
6 + 2
3 ⋅ 1
n
0
=
4
n − 1
9 8 + 2
3 ⋅ 1
n
0
=
5
n
0
3 ⋅ 1
n
0
=
5 ⋅ 1
n
0
3 ⋅ 1
n
0
=
5
3
P(n + 1) = 1,
n + 1
6 +
2
3 ⋅ 1
n + 1
0
=
3 ⋅ 1
n + 1
0 ⋅ 1,
n + 1
6 + 2
3 ⋅ 1
n + 1
0
=
3 ⋅ 1
n + 1
6 + 2
3 ⋅ 1
n + 1
0
=
4
n
98 + 2
3 ⋅ 1
n + 1
0
=
5
n + 1
0
3 ⋅ 1
n + 1
0
=
5 ⋅ 1
n + 1
0
3 ⋅ 1
n + 1
0
=
5
3

27. Conclusioni da 3a) e 3b)
Poiché
e
ne deriva che
P(n) =
5
3
→ P(n + 1) =
5
3
∀n ∈ N = {0,1,2,3,...}
B) 1,
n
6 +
2
3 ⋅ 1
n
0
=
5
3
∀n ∈ N = {0,1,2,3,...}
P(0) =
5
3
∧ P(1) =
5
3