Metamath Proof Explorer


Theorem sqrtcvallem4

Description: Equivalent to saying that the square of the real component of the square root of a complex number is a non-negative real number. Lemma for sqrtcval . See resqrtval . (Contributed by RP, 11-May-2024)

Ref Expression
Assertion sqrtcvallem4
|- ( A e. CC -> 0 <_ ( ( ( abs ` A ) + ( Re ` A ) ) / 2 ) )

Proof

Step Hyp Ref Expression
1 abscl
 |-  ( A e. CC -> ( abs ` A ) e. RR )
2 recl
 |-  ( A e. CC -> ( Re ` A ) e. RR )
3 1 2 readdcld
 |-  ( A e. CC -> ( ( abs ` A ) + ( Re ` A ) ) e. RR )
4 2rp
 |-  2 e. RR+
5 4 a1i
 |-  ( A e. CC -> 2 e. RR+ )
6 negcl
 |-  ( A e. CC -> -u A e. CC )
7 6 releabsd
 |-  ( A e. CC -> ( Re ` -u A ) <_ ( abs ` -u A ) )
8 6 abscld
 |-  ( A e. CC -> ( abs ` -u A ) e. RR )
9 6 recld
 |-  ( A e. CC -> ( Re ` -u A ) e. RR )
10 8 9 subge0d
 |-  ( A e. CC -> ( 0 <_ ( ( abs ` -u A ) - ( Re ` -u A ) ) <-> ( Re ` -u A ) <_ ( abs ` -u A ) ) )
11 7 10 mpbird
 |-  ( A e. CC -> 0 <_ ( ( abs ` -u A ) - ( Re ` -u A ) ) )
12 absneg
 |-  ( A e. CC -> ( abs ` -u A ) = ( abs ` A ) )
13 reneg
 |-  ( A e. CC -> ( Re ` -u A ) = -u ( Re ` A ) )
14 12 13 oveq12d
 |-  ( A e. CC -> ( ( abs ` -u A ) - ( Re ` -u A ) ) = ( ( abs ` A ) - -u ( Re ` A ) ) )
15 1 recnd
 |-  ( A e. CC -> ( abs ` A ) e. CC )
16 2 recnd
 |-  ( A e. CC -> ( Re ` A ) e. CC )
17 15 16 subnegd
 |-  ( A e. CC -> ( ( abs ` A ) - -u ( Re ` A ) ) = ( ( abs ` A ) + ( Re ` A ) ) )
18 14 17 eqtrd
 |-  ( A e. CC -> ( ( abs ` -u A ) - ( Re ` -u A ) ) = ( ( abs ` A ) + ( Re ` A ) ) )
19 11 18 breqtrd
 |-  ( A e. CC -> 0 <_ ( ( abs ` A ) + ( Re ` A ) ) )
20 3 5 19 divge0d
 |-  ( A e. CC -> 0 <_ ( ( ( abs ` A ) + ( Re ` A ) ) / 2 ) )