Description: Lemma for sqrt2irr . This is the core of the proof: if A / B = sqrt ( 2 ) , then A and B are even, so A / 2 and B / 2 are smaller representatives, which is absurd by the method of infinite descent (here implemented by strong induction). This is Metamath 100 proof #1. (Contributed by NM, 20-Aug-2001) (Revised by Mario Carneiro, 12-Sep-2015) (Proof shortened by JV, 4-Jan-2022)
Ref | Expression | ||
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Hypotheses | sqrt2irrlem.1 | |
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sqrt2irrlem.2 | |
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sqrt2irrlem.3 | |
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Assertion | sqrt2irrlem | |
Step | Hyp | Ref | Expression |
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1 | sqrt2irrlem.1 | |
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2 | sqrt2irrlem.2 | |
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3 | sqrt2irrlem.3 | |
|
4 | 2cnd | |
|
5 | 4 | sqsqrtd | |
6 | 3 | oveq1d | |
7 | 5 6 | eqtr3d | |
8 | 1 | zcnd | |
9 | 2 | nncnd | |
10 | 2 | nnne0d | |
11 | 8 9 10 | sqdivd | |
12 | 7 11 | eqtrd | |
13 | 12 | oveq1d | |
14 | 8 | sqcld | |
15 | 2 | nnsqcld | |
16 | 15 | nncnd | |
17 | 15 | nnne0d | |
18 | 14 16 17 | divcan1d | |
19 | 13 18 | eqtrd | |
20 | 19 | oveq1d | |
21 | 2ne0 | |
|
22 | 21 | a1i | |
23 | 16 4 22 | divcan3d | |
24 | 20 23 | eqtr3d | |
25 | 24 15 | eqeltrd | |
26 | 25 | nnzd | |
27 | zesq | |
|
28 | 1 27 | syl | |
29 | 26 28 | mpbird | |
30 | 4 | sqvald | |
31 | 30 | oveq2d | |
32 | 8 4 22 | sqdivd | |
33 | 14 4 4 22 22 | divdiv1d | |
34 | 31 32 33 | 3eqtr4d | |
35 | 24 | oveq1d | |
36 | 34 35 | eqtrd | |
37 | zsqcl | |
|
38 | 29 37 | syl | |
39 | 36 38 | eqeltrrd | |
40 | 15 | nnrpd | |
41 | 40 | rphalfcld | |
42 | 41 | rpgt0d | |
43 | elnnz | |
|
44 | 39 42 43 | sylanbrc | |
45 | nnesq | |
|
46 | 2 45 | syl | |
47 | 44 46 | mpbird | |
48 | 29 47 | jca | |