The document presents a mathematical exploration of the division of the value 1 by 9, demonstrating that 1/9 equals 0.111... using various proofs, including induction. It details formulas involving the repetition of digits and uses integer variables to illustrate these concepts. The proofs establish a formula for integer values of n, reinforcing the conclusions derived from multiple cases.

1. DIVISIONI PARZIALI
i VALORI di 1
9
a cura di Enzo Exposyto
A)
1
9
= 0,
n
1 +
1
9 ⋅ 1
n
0
∀n ∈ N = {0,1,2,3,...}

2. i VALORI di 1
a cura di Enzo Exposyto 9
Le DUE “n” -SULL’1 e SOPRA lo 0- INDICANO CHE
la CIFRA 1 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 A) diventa
5 volte 1 5 volte 0
A)
1
9
= 0,
n
1 +
1
9 ⋅ 1
n
0
A)
1
9
= 0,
5
1 +
1
9 ⋅ 1
5
0
= 0,11111 +
1
9 ⋅ 100000
∀n ∈ N = {0,1,2,3,...}
n = 5

3. FORMULA A)
1^ DIMOSTRAZIONE
a cura di Enzo Exposyto
A)
1
9
= 0,
n
1 +
1
9 ⋅ 1
n
0
∀n ∈ N = {0,1,2,3,...}

4. FORMULA A) - 1^ dimostrazione
QUINDI
1
9
=
1
3
⋅
1
3
=
1
3
⋅ 0,
n
3 +
1
3 ⋅ 3 ⋅ 1
n
0
=
1
3
⋅ {0,
n
3 +
1
3 ⋅ 1
n
0
}
= 0,
n
1 +
1
9 ⋅ 1
n
0
A)
1
9
=
1
3
⋅
1
3
= 0,
n
1 +
1
9 ⋅ 1
n
0
∀n ∈ N = {0,1,2,3,...}
∀n ∈ N = {0,1,2,3,...}

5. FORMULA A)
2^ DIMOSTRAZIONE
(MODUS PONES e PRINCIPIO d’INDUZIONE)
a cura di Enzo Exposyto
A)
1
9
= 0,
n
1 +
1
9 ⋅ 1
n
0
∀n ∈ N = {0,1,2,3,...}

6. Formula A) - 2^ dimostrazione sintetica
La formula A) è dimostrata visto che
P(0) = 0,
0
1 +
1
9 ⋅ 1
0
0
= 0 +
1
9 ⋅ 1
=
1
9
P(n) = 0,
n
1 +
1
9 ⋅ 1
n
0
=
9 ⋅ 1
n
0 ⋅ 0,
n
1 + 1
9 ⋅ 1
n
0
=
n
9 + 1
9 ⋅ 1
n
0
=
1
n
0
9 ⋅ 1
n
0
=
1
9
P(1) = 0,
1
1 +
1
9 ⋅ 1
1
0
= 0,1 +
1
9 ⋅ 10
=
9 ⋅ 10 ⋅ 0,1 + 1
9 ⋅ 10
=
9 + 1
9 ⋅ 10
=
10
90
=
1
9
A) 0,
n
1 +
1
9 ⋅ 1
n
0
=
1
9
∀n ∈ N = {0,1,2,3,...}
P(0) =
1
9
P(1) =
1
9
P(n) =
1
9
P(n + 1) =
1
9
P(n) =
1
9
→ P(n + 1) =
1
9
P(0) =
1
9
∧ P(1) =
1
9
∀n ∈ N = {0,1,2,3,...}
P(n + 1) = 0,
n + 1
1 +
1
9 ⋅ 1
n + 1
0
=
9 ⋅ 1
n + 1
0 ⋅ 0,
n + 1
1 + 1
9 ⋅ 1
n + 1
0
=
n + 1
9 + 1
9 ⋅ 1
n + 1
0
=
1
n + 1
0
9 ⋅ 1
n + 1
0
=
1
9

7. Formula A) - 2^ dimostrazione estesa
Qui, sarà dimostrata la forma
A) 0,
n
1 +
1
9 ⋅ 1
n
0
=
1
9
∀n ∈ N = {0,1,2,3,...}
A)
1
9
= 0,
n
1 +
1
9 ⋅ 1
n
0
∀n ∈ N = {0,1,2,3,...}

8. 2a) Dimostrazione della Base
Quindi
Quindi
P(0) =
1
9
P(1) =
1
9
P(0) = 0,
0
1 +
1
9 ⋅ 1
0
0
= 0 +
1
9 ⋅ 1
=
1
9
P(1) = 0,
1
1 +
1
9 ⋅ 1
1
0
= 0,1 +
1
9 ⋅ 10
=
9 ⋅ 10 ⋅ 0,1 + 1
9 ⋅ 10
=
9 + 1
9 ⋅ 10
=
10
90
=
1
9

9. 2b) Dimostrazione del Passo Induttivo
P(n) =
1
9
∀n ∈ N = {0,1,2,3,...}
P(n + 1) =
1
9
∀n ∈ N = {0,1,2,3,...}
P(n) = 0,
n
1 +
1
9 ⋅ 1
n
0
=
9 ⋅ 1
n
0 ⋅ 0,
n
1 + 1
9 ⋅ 1
n
0
=
n
9 + 1
9 ⋅ 1
n
0
=
1
n
0
9 ⋅ 1
n
0
=
1
9
P(n + 1) = 0,
n + 1
1 +
1
9 ⋅ 1
n + 1
0
=
9 ⋅ 1
n + 1
0 ⋅ 0,
n + 1
1 + 1
9 ⋅ 1
n + 1
0
=
n + 1
9 + 1
9 ⋅ 1
n + 1
0
=
1
n + 1
0
9 ⋅ 1
n + 1
0
=
1
9

10. Formula A) - Conclusioni da 2a) e 2b)
Poiché
e
ne deriva che
P(0) =
1
9
∧ P(1) =
1
9
P(n) =
1
9
→ P(n + 1) =
1
9
∀n ∈ N = {0,1,2,3,...}
A) 0,
n
1 +
1
9 ⋅ 1
n
0
=
1
9
∀n ∈ N = {0,1,2,3,...}

11. DIVISIONI PARZIALI
i VALORI di 1
9
a cura di Enzo Exposyto
∀n ∈ N+
= {1,2,3,...}
B)
1
9
= 0,
2n
1 +
1
9 ⋅ 1
2n
0

12. i VALORI di 1
a cura di Enzo Exposyto 9
Le “2n” -SULL’1 e SOPRA lo 0- INDICANO CHE
la CIFRA 1 è RIPETUTA “2n” VOLTE,
la CIFRA 0 è RIPETUTA “2n” VOLTE.
n è un QUALSIASI NUMERO INTERO,
appartenente all’insieme N+ = {1,2,3,...}.
Ad esempio, con n = 3, la FORMULA B) diventa
2*3 volte 1 2*3 volte 0
B)
1
9
= 0,
2n
1 +
1
9 ⋅ 1
2n
0
∀n ∈ N+
= {1,2,3,...}
B)
1
9
= 0,
2 ⋅ 3
1 +
1
9 ⋅ 1
2 ⋅ 3
0
= 0,111111 +
1
9 ⋅ 1000000
n = 3

13. FORMULA B)
1^ DIMOSTRAZIONE
a cura di Enzo Exposyto
B)
1
9
= 0,
2n
1 +
1
9 ⋅ 1
2n
0
∀n ∈ N+
= {1,2,3,...}

14. FORMULA B) - 1^ dimostrazione
continua ...
1
9
= {
1
3
}2
= {0,
n
3 +
1
3 ⋅ 1
n
0
}2
= 0,
n
3 ⋅ 0,
n
3 + 2 ⋅ 0,
n
3 ⋅
1
3 ⋅ 1
n
0
+
1
3 ⋅ 1
n
0
⋅
1
3 ⋅ 1
n
0
∀n ∈ N = {0,1,2,3,...}
= {0,
n
3 ⋅ 0,
n
3 = 0,
n − 1
1 0
n − 1
8 9} ∀n ∈ N+
= {1,2,3,...}

15. continua ...
1
9
= 0,
n − 1
1 0
n − 1
8 9 + 2 ⋅
0,
n
3
3
⋅
1
1
n
0
+
1
9 ⋅ 1
n
0 ⋅ 1
n
0
= 0,
n − 1
1 0
n − 1
8 9 + 2 ⋅ 0,
n
1 ⋅
1
1
n
0
+
1
9 ⋅ 1
n
0
n
0
= 0,
n − 1
1 0
n − 1
8 9 +
0,
n
2
1
n
0
+
1
9 ⋅ 1
2n
0
= 0,
n − 1
1 0
n − 1
8 9 + 0,
n
0
n
2 +
1
9 ⋅ 1
2n
0
= {0,
n − 1
1 0
n − 1
8 9 + 0,
n
0
n
2 = 0,
n
1
n
1 = 0,
2n
1 }

16. Quindi
Notiamo che la formula B) è stata ricavata, a un certo punto della
dimostrazione, sotto la condizione: n elemento di N+ = {1,2,3,...}. Tuttavia,
vedremo che ESSA È VALIDA ANCHE con n = 0.
1
9
= 0,
2n
1 +
1
9 ⋅ 1
2n
0
B)
1
9
= {
1
3
}2
= 0,
2n
1 +
1
9 ⋅ 1
2n
0
∀n ∈ N+
= {1,2,3,...}

17. FORMULA B)
2^ DIMOSTRAZIONE
(MODUS PONES e PRINCIPIO d’INDUZIONE)
a cura di Enzo Exposyto
B)
1
9
= 0,
2n
1 +
1
9 ⋅ 1
2n
0
∀n ∈ N+
= {1,2,3,...}

18. Formula B) - 2^ dimostrazione sintetica - Numero di cifre PARI - Caso n = 0
La formula B) è dimostrata visto che, pur omettendo il caso n=0, si ha
B) 0,
2n
1 +
1
9 ⋅ 1
2n
0
=
1
9
∀n ∈ N+
= {1,2,3,...}
P(2 ⋅ 0) = 0,
2 ⋅ 0
1 +
1
9 ⋅ 1
2 ⋅ 0
0
= 0 +
1
9 ⋅ 1
=
1
9
P(2 ⋅ 0) =
1
9
P(2 ⋅ 1) = 0,
2 ⋅ 1
1 +
1
9 ⋅ 1
2 ⋅ 1
0
= 0,
2
1 +
1
9 ⋅ 1
2
0
= 0,11 +
1
9 ⋅ 100
=
9 ⋅ 100 ⋅ 0,11 + 1
9 ⋅ 100
=
99 + 1
9 ⋅ 100
=
100
900
=
1
9
P(2 ⋅ 1) =
1
9
P(2n) = 0,
2n
1 +
1
9 ⋅ 1
2n
0
=
9 ⋅ 1
2n
0 ⋅ 0,
2n
1 + 1
9 ⋅ 1
2n
0
=
2n
9 + 1
9 ⋅ 1
2n
0
=
1
2n
0
9 ⋅ 1
2n
0
=
1
9
P(2n) =
1
9
P[2(n + 1)] = 0,
2(n + 1)
1 +
1
9 ⋅ 1
2(n + 1)
0
=
9 ⋅ 1
2(n + 1)
0 ⋅ 0,
2(n + 1)
1 + 1
9 ⋅ 1
2(n + 1)
0
=
2(n + 1)
9 + 1
9 ⋅ 1
2(n + 1)
0
=
1
2(n + 1)
0
9 ⋅ 1
2(n + 1)
0
=
1
9
P[2(n + 1)] =
1
9
P(2 ⋅ 1) =
1
9
∀n ∈ N+
= {1,2,3,...}
P(2n) =
1
9
→ P[2(n + 1)] =
1
9

19. Formula B) - 2^ dimostrazione sintetica - Numero di cifre DISPARI, salvo 2n
La formula B) è dimostrata ancora, visto che, pur senza il caso n=0, si ha
P(2n + 1) = 0,
2n + 1
1 +
1
9 ⋅ 1
2n + 1
0
=
9 ⋅ 1
2n + 1
0 ⋅ 0,
2n + 1
1 + 1
9 ⋅ 1
2n + 1
0
=
2n + 1
9 + 1
9 ⋅ 1
2n + 1
0
=
1
2n + 1
0
9 ⋅ 1
2n + 1
0
=
1
9
P(2 ⋅ 0 + 1) = 0,
2 ⋅ 0 + 1
1 +
1
9 ⋅ 1
2 ⋅ 0 + 1
0
= 0,
1
1 +
1
9 ⋅ 1
1
0
= 0,1 +
1
9 ⋅ 10
=
9 ⋅ 10 ⋅ 0,1 + 1
9 ⋅ 10
=
9 + 1
9 ⋅ 10
=
10
90
=
1
9
P(2 ⋅ 1 + 1) = 0,
2 ⋅ 1 + 1
1 +
1
9 ⋅ 1
2 ⋅ 1 + 1
0
= 0,
3
1 +
1
9 ⋅ 1
3
0
= 0,111 +
1
9 ⋅ 1000
=
9 ⋅ 1000 ⋅ 0,111 + 1
9 ⋅ 1000
=
999 + 1
9 ⋅ 1000
=
1000
9000
=
1
9
P(2n) = 0,
2n
1 +
1
9 ⋅ 1
2n
0
=
9 ⋅ 1
2n
0 ⋅ 0,
2n
1 + 1
9 ⋅ 1
2n
0
=
2n
9 + 1
9 ⋅ 1
2n
0
=
1
2n
0
9 ⋅ 1
2n
0
=
1
9
P(2 ⋅ 1 + 1) =
1
9
P(2n) =
1
9
→ P(2n + 1) =
1
9
∀n ∈ N+
= {1,2,3,...}
B) 0,
2n
1 +
1
9 ⋅ 1
2n
0
=
1
9
∀n ∈ N+
= {1,2,3,...}

20. FORMULA C)
(CONTIENE NUMERO con 2 CIFRE RIPETUTE)
a cura di Enzo Exposyto
C)
1
9
=
9 ⋅
n − 1
1 0
n − 1
8 9 + 6 ⋅
n
3 + 1
9 ⋅ 1
2n
0
∀n ∈ N+
= {1,2,3,...}

21. FORMULA C)
La CIFRA 1 è RIPETUTA “n-1” VOLTE,
la CIFRA 8 è RIPETUTA “n-1” VOLTE,
la CIFRA 3 è RIPETUTA “n” VOLTE,
la CIFRA 0 è RIPETUTA “2n” VOLTE,
con n QUALSIASI appartenente all’insieme N+ = {1,2,3,...}.
Ad esempio, con n = 3, la FORMULA C) diventa
2 volte 1 e 8 3 volte 3
2*3 volte 0
n = 3
C)
1
9
=
9 ⋅
n − 1
1 0
n − 1
8 9 + 6 ⋅
n
3 + 1
9 ⋅ 1
2n
0
∀n ∈ N+
= {1,2,3,...}
C)
1
9
=
9 ⋅
3 − 1
1 0
3 − 1
8 9 + 6 ⋅
3
3 + 1
9 ⋅ 1
2 ⋅ 3
0
=
9 ⋅ 110889 + 6 ⋅ 333 + 1
9 ⋅ 1000000

22. FORMULA C)
DIMOSTRAZIONE
a cura di Enzo Exposyto
∀n ∈ N+
= {1,2,3,...}
C)
1
9
=
9 ⋅
n − 1
1 0
n − 1
8 9 + 6 ⋅
n
3 + 1
9 ⋅ 1
2n
0

23. FORMULA C) - Dimostrazione
continua ...
1
9
= {
1
3
}2
= {0,
n
3 +
1
3 ⋅ 1
n
0
}2 ∀n ∈ N = {0,1,2,3,...}
= 0,
n
3 ⋅ 0,
n
3 + 2 ⋅ 0,
n
3 ⋅
1
3 ⋅ 1
n
0
+
1
3 ⋅ 1
n
0
⋅
1
3 ⋅ 1
n
0
= 0,
n
3 ⋅ 0,
n
3 + 2 ⋅ 0,
n
3 ⋅
1
3 ⋅ 1
n
0
+
1
9 ⋅ 1
n
0 ⋅ 1
n
0

24. continua ...
= {
n
3 ⋅
n
3 =
n − 1
1 0
n − 1
8 9} ∀n ∈ N+
= {1,2,3,...}
1
9
=
9 ⋅ 1
n
0 ⋅ 1
n
0 ⋅ 0,
n
3 ⋅ 0,
n
3 + 3 ⋅ 1
n
0 ⋅ 2 ⋅ 0,
n
3 + 1
9 ⋅ 1
n
0 ⋅ 1
n
0
=
9 ⋅
n
3 ⋅
n
3 + 6 ⋅
n
3 + 1
9 ⋅ 1
2n
0
=
9 ⋅ 1
n
0 ⋅ 0,
n
3 ⋅ 1
n
0 ⋅ 0,
n
3 + 6 ⋅ 1
n
0 ⋅ 0,
n
3 + 1
9 ⋅ 1
n
0 ⋅ 1
n
0

25. Quindi
Notiamo che anche la formula C) è stata ricavata, a un certo punto della
dimostrazione, sotto la condizione: n elemento di N+ = {1,2,3,...}.
1
9
=
9 ⋅
n − 1
1 0
n − 1
8 9 + 6 ⋅
n
3 + 1
9 ⋅ 1
2n
0
∀n ∈ N+
= {1,2,3,...}
C)
1
9
=
9 ⋅
n − 1
1 0
n − 1
8 9 + 6 ⋅
n
3 + 1
9 ⋅ 1
2n
0

26. FORMULA C)
2 ESEMPI
(CONTENGONO NUMERI con 2 CIFRE RIPETUTE)
a cura di Enzo Exposyto
C)
1
9
=
9 ⋅
n − 1
1 0
n − 1
8 9 + 6 ⋅
n
3 + 1
9 ⋅ 1
2n
0
∀n ∈ N+
= {1,2,3,...}

27. 9 ⋅
3 − 1
1 0
3 − 1
8 9 + 6 ⋅
3
3 + 1
9 ⋅ 1
2 ⋅ 3
0
=
9 ⋅ 110.889 + 6 ⋅ 333 + 1
9 ⋅ 1.000.000
=
3 − 1
9 8
3 − 1
0 1 + 1
3 − 1
9 8 + 1
9 ⋅ 1.000.000
=
2 ⋅ 3
9 + 1
9 ⋅ 1.000.000
n = 3
n = 5
9 ⋅
5 − 1
1 0
5 − 1
8 9 + 6 ⋅
5
3 + 1
9 ⋅ 1
2 ⋅ 5
0
=
9 ⋅ 1.111.088.889 + 6 ⋅ 33.333 + 1
9 ⋅ 10.000.000.000
=
5 − 1
9 8
5 − 1
0 1 + 1
5 − 1
9 8 + 1
9 ⋅ 10.000.000.000
=
2 ⋅ 5
9 + 1
90.000.000.000
=
9.999.999.999 + 1
90.000.000.000
=
10.000.000.000
90.000.000.000
=
1
9
=
999.999 + 1
9.000.000
=
1.000.000
9.000.000
=
1
9

28. dalla C), la FORMULA D)
(CONTIENE NUMERO con 2 CIFRE RIPETUTE)
a cura di Enzo Exposyto
D)
9 ⋅
n − 1
1 0
n − 1
8 9 + 6 ⋅
n
3 + 1
1
2n
0
= 1
∀n ∈ N+
= {1,2,3,...}

29. FORMULA D)
La CIFRA 1 è RIPETUTA “n-1” VOLTE,
la CIFRA 8 è RIPETUTA “n-1” VOLTE,
la CIFRA 3 è RIPETUTA “n” VOLTE,
la CIFRA 0 è RIPETUTA “2n” VOLTE,
con n QUALSIASI appartenente all’insieme N+ = {1,2,3,...}.
Ad esempio, con n = 3, la FORMULA D) diventa
2 volte 1 e 8 3 volte 3
2*3 volte 0
D)
9 ⋅
n − 1
1 0
n − 1
8 9 + 6 ⋅
n
3 + 1
1
2n
0
= 1 ∀n ∈ N+
= {1,2,3,...}
D)
9 ⋅
3 − 1
1 0
3 − 1
8 9 + 6 ⋅
3
3 + 1
1
2 ⋅ 3
0
=
9 ⋅ 110889 + 6 ⋅ 333 + 1
1000000
= 1 n = 3

30. FORMULA D)
DIMOSTRAZIONE
a cura di Enzo Exposyto
∀n ∈ N+
= {1,2,3,...}
D)
9 ⋅
n − 1
1 0
n − 1
8 9 + 6 ⋅
n
3 + 1
1
2n
0
= 1

31. Quindi
1
9
=
9 ⋅
n − 1
1 0
n − 1
8 9 + 6 ⋅
n
3 + 1
9 ⋅ 1
2n
0
1
9
=
1
9
⋅
9 ⋅
n − 1
1 0
n − 1
8 9 + 6 ⋅
n
3 + 1
1
2n
0
9 ⋅
n − 1
1 0
n − 1
8 9 + 6 ⋅
n
3 + 1
1
2n
0
= 1
1
9
=
1
9
⋅ 1
∀n ∈ N+
= {1,2,3,...}
∀n ∈ N+
= {1,2,3,...}

32. FORMULA D)
2 ESEMPI
(CONTENGONO NUMERI con 2 CIFRE RIPETUTE)
a cura di Enzo Exposyto
D)
9 ⋅
n − 1
1 0
n − 1
8 9 + 6 ⋅
n
3 + 1
1
2n
0
= 1
∀n ∈ N+
= {1,2,3,...}

33. n = 3
9 ⋅
3 − 1
1 0
3 − 1
8 9 + 6 ⋅
3
3 + 1
1
2 ⋅ 3
0
=
9 ⋅ 110.889 + 6 ⋅ 333 + 1
1.000.000
=
3 − 1
9 8
3 − 1
0 1 + 1
3 − 1
9 8 + 1
1.000.000
=
2 ⋅ 3
9 + 1
1.000.000
=
999.999 + 1
1.000.000
=
1.000.000
1.000.000
= 1
n = 5
9 ⋅
5 − 1
1 0
5 − 1
8 9 + 6 ⋅
5
3 + 1
1
2 ⋅ 5
0
=
9 ⋅ 1.111.088.889 + 6 ⋅ 33.333 + 1
10.000.000.000
=
5 − 1
9 8
5 − 1
0 1 + 1
5 − 1
9 8 + 1
10.000.000.000
=
10.000.000.000
10.000.000.000
= 1
=
2 ⋅ 5
9 + 1
10.000.000.000
=
9.999.999.999 + 1
10.000.000.000