Metamath Proof Explorer


Theorem addsq2nreurex

Description: For each complex number C , there is no unique complex number a added to the square of another complex number b resulting in the given complex number C . (Contributed by AV, 2-Jul-2023)

Ref Expression
Assertion addsq2nreurex C¬∃!aba+b2=C

Proof

Step Hyp Ref Expression
1 peano2cnm CC1
2 id CC
3 4cn 4
4 3 a1i C4
5 2 4 subcld CC4
6 1cnd C1
7 1re 1
8 1lt4 1<4
9 7 8 ltneii 14
10 9 a1i C14
11 2 6 4 10 subneintrd CC1C4
12 oveq1 b=1b2=12
13 12 oveq2d b=1C-1+b2=C-1+12
14 13 eqeq1d b=1C-1+b2=CC-1+12=C
15 14 adantl Cb=1C-1+b2=CC-1+12=C
16 sq1 12=1
17 16 oveq2i C-1+12=C-1+1
18 npcan1 CC-1+1=C
19 17 18 eqtrid CC-1+12=C
20 6 15 19 rspcedvd CbC-1+b2=C
21 2cnd C2
22 oveq1 b=2b2=22
23 22 oveq2d b=2C-4+b2=C-4+22
24 23 eqeq1d b=2C-4+b2=CC-4+22=C
25 24 adantl Cb=2C-4+b2=CC-4+22=C
26 2cn 2
27 26 sqcli 22
28 27 a1i C22
29 2 4 28 subadd23d CC-4+22=C+22-4
30 sq2 22=4
31 30 a1i C22=4
32 28 31 subeq0bd C224=0
33 27 3 subcli 224
34 addid0 C224C+22-4=C224=0
35 33 34 mpan2 CC+22-4=C224=0
36 32 35 mpbird CC+22-4=C
37 29 36 eqtrd CC-4+22=C
38 21 25 37 rspcedvd CbC-4+b2=C
39 oveq1 a=C1a+b2=C-1+b2
40 39 eqeq1d a=C1a+b2=CC-1+b2=C
41 40 rexbidv a=C1ba+b2=CbC-1+b2=C
42 oveq1 a=C4a+b2=C-4+b2
43 42 eqeq1d a=C4a+b2=CC-4+b2=C
44 43 rexbidv a=C4ba+b2=CbC-4+b2=C
45 41 44 2nreu C1C4C1C4bC-1+b2=CbC-4+b2=C¬∃!aba+b2=C
46 45 imp C1C4C1C4bC-1+b2=CbC-4+b2=C¬∃!aba+b2=C
47 1 5 11 20 38 46 syl32anc C¬∃!aba+b2=C