Metamath Proof Explorer


Theorem logbgcd1irr

Description: The logarithm of an integer greater than 1 to an integer base greater than 1 is an irrational number if the argument and the base are relatively prime. For example, ( 2 logb 9 ) e. ( RR \ QQ ) (see 2logb9irr ). (Contributed by AV, 29-Dec-2022)

Ref Expression
Assertion logbgcd1irr X 2 B 2 X gcd B = 1 log B X

Proof

Step Hyp Ref Expression
1 eluz2nn B 2 B
2 1 nnrpd B 2 B +
3 2 3ad2ant2 X 2 B 2 X gcd B = 1 B +
4 eluz2nn X 2 X
5 4 nnrpd X 2 X +
6 5 3ad2ant1 X 2 B 2 X gcd B = 1 X +
7 eluz2b3 B 2 B B 1
8 7 simprbi B 2 B 1
9 8 3ad2ant2 X 2 B 2 X gcd B = 1 B 1
10 3 6 9 3jca X 2 B 2 X gcd B = 1 B + X + B 1
11 relogbcl B + X + B 1 log B X
12 10 11 syl X 2 B 2 X gcd B = 1 log B X
13 eluz2gt1 X 2 1 < X
14 13 adantr X 2 B 2 1 < X
15 4 adantr X 2 B 2 X
16 15 nnrpd X 2 B 2 X +
17 1 adantl X 2 B 2 B
18 17 nnrpd X 2 B 2 B +
19 eluz2gt1 B 2 1 < B
20 19 adantl X 2 B 2 1 < B
21 logbgt0b X + B + 1 < B 0 < log B X 1 < X
22 16 18 20 21 syl12anc X 2 B 2 0 < log B X 1 < X
23 14 22 mpbird X 2 B 2 0 < log B X
24 23 anim1ci X 2 B 2 log B X log B X 0 < log B X
25 elpq log B X 0 < log B X m n log B X = m n
26 24 25 syl X 2 B 2 log B X m n log B X = m n
27 26 ex X 2 B 2 log B X m n log B X = m n
28 oveq2 m n = log B X B m n = B log B X
29 28 eqcoms log B X = m n B m n = B log B X
30 eluzelcn B 2 B
31 30 adantl X 2 B 2 B
32 nnne0 B B 0
33 1 32 syl B 2 B 0
34 33 8 nelprd B 2 ¬ B 0 1
35 34 adantl X 2 B 2 ¬ B 0 1
36 31 35 eldifd X 2 B 2 B 0 1
37 eluzelcn X 2 X
38 37 adantr X 2 B 2 X
39 nnne0 X X 0
40 nelsn X 0 ¬ X 0
41 4 39 40 3syl X 2 ¬ X 0
42 41 adantr X 2 B 2 ¬ X 0
43 38 42 eldifd X 2 B 2 X 0
44 cxplogb B 0 1 X 0 B log B X = X
45 36 43 44 syl2anc X 2 B 2 B log B X = X
46 45 adantr X 2 B 2 m n B log B X = X
47 29 46 sylan9eqr X 2 B 2 m n log B X = m n B m n = X
48 47 ex X 2 B 2 m n log B X = m n B m n = X
49 oveq1 B m n = X B m n n = X n
50 31 adantr X 2 B 2 m n B
51 nncn m m
52 51 adantr m n m
53 nncn n n
54 53 adantl m n n
55 nnne0 n n 0
56 55 adantl m n n 0
57 52 54 56 3jca m n m n n 0
58 divcl m n n 0 m n
59 57 58 syl m n m n
60 59 adantl X 2 B 2 m n m n
61 nnnn0 n n 0
62 61 adantl m n n 0
63 62 adantl X 2 B 2 m n n 0
64 50 60 63 3jca X 2 B 2 m n B m n n 0
65 cxpmul2 B m n n 0 B m n n = B m n n
66 65 eqcomd B m n n 0 B m n n = B m n n
67 64 66 syl X 2 B 2 m n B m n n = B m n n
68 57 adantl X 2 B 2 m n m n n 0
69 divcan1 m n n 0 m n n = m
70 68 69 syl X 2 B 2 m n m n n = m
71 70 oveq2d X 2 B 2 m n B m n n = B m
72 33 adantl X 2 B 2 B 0
73 72 adantr X 2 B 2 m n B 0
74 nnz m m
75 74 adantr m n m
76 75 adantl X 2 B 2 m n m
77 50 73 76 cxpexpzd X 2 B 2 m n B m = B m
78 71 77 eqtrd X 2 B 2 m n B m n n = B m
79 67 78 eqtrd X 2 B 2 m n B m n n = B m
80 79 eqeq1d X 2 B 2 m n B m n n = X n B m = X n
81 simpr m n n
82 rplpwr X B n X gcd B = 1 X n gcd B = 1
83 15 17 81 82 syl2an3an X 2 B 2 m n X gcd B = 1 X n gcd B = 1
84 oveq1 X n = B m X n gcd B = B m gcd B
85 84 eqeq1d X n = B m X n gcd B = 1 B m gcd B = 1
86 85 eqcoms B m = X n X n gcd B = 1 B m gcd B = 1
87 86 adantl X 2 B 2 m n B m = X n X n gcd B = 1 B m gcd B = 1
88 eluzelz B 2 B
89 88 adantl X 2 B 2 B
90 simpl m n m
91 rpexp B B m B m gcd B = 1 B gcd B = 1
92 89 89 90 91 syl2an3an X 2 B 2 m n B m gcd B = 1 B gcd B = 1
93 gcdid B B gcd B = B
94 88 93 syl B 2 B gcd B = B
95 eluzelre B 2 B
96 nnnn0 B B 0
97 nn0ge0 B 0 0 B
98 1 96 97 3syl B 2 0 B
99 95 98 absidd B 2 B = B
100 94 99 eqtrd B 2 B gcd B = B
101 100 eqeq1d B 2 B gcd B = 1 B = 1
102 101 adantl X 2 B 2 B gcd B = 1 B = 1
103 102 adantr X 2 B 2 m n B gcd B = 1 B = 1
104 eqneqall B = 1 B 1 ¬ X gcd B = 1
105 8 104 syl5com B 2 B = 1 ¬ X gcd B = 1
106 105 adantl X 2 B 2 B = 1 ¬ X gcd B = 1
107 106 adantr X 2 B 2 m n B = 1 ¬ X gcd B = 1
108 103 107 sylbid X 2 B 2 m n B gcd B = 1 ¬ X gcd B = 1
109 92 108 sylbid X 2 B 2 m n B m gcd B = 1 ¬ X gcd B = 1
110 109 adantr X 2 B 2 m n B m = X n B m gcd B = 1 ¬ X gcd B = 1
111 87 110 sylbid X 2 B 2 m n B m = X n X n gcd B = 1 ¬ X gcd B = 1
112 111 ex X 2 B 2 m n B m = X n X n gcd B = 1 ¬ X gcd B = 1
113 112 com23 X 2 B 2 m n X n gcd B = 1 B m = X n ¬ X gcd B = 1
114 83 113 syld X 2 B 2 m n X gcd B = 1 B m = X n ¬ X gcd B = 1
115 ax-1 ¬ X gcd B = 1 B m = X n ¬ X gcd B = 1
116 114 115 pm2.61d1 X 2 B 2 m n B m = X n ¬ X gcd B = 1
117 80 116 sylbid X 2 B 2 m n B m n n = X n ¬ X gcd B = 1
118 49 117 syl5 X 2 B 2 m n B m n = X ¬ X gcd B = 1
119 48 118 syld X 2 B 2 m n log B X = m n ¬ X gcd B = 1
120 119 rexlimdvva X 2 B 2 m n log B X = m n ¬ X gcd B = 1
121 27 120 syld X 2 B 2 log B X ¬ X gcd B = 1
122 121 con2d X 2 B 2 X gcd B = 1 ¬ log B X
123 122 3impia X 2 B 2 X gcd B = 1 ¬ log B X
124 12 123 eldifd X 2 B 2 X gcd B = 1 log B X