Metamath Proof Explorer


Definition df-abs

Description: Define the function for the absolute value (modulus) of a complex number. See abscli for its closure and absval or absval2i for its value. For example, ( abs-u 2 ) = 2 ( ex-abs ). (Contributed by NM, 27-Jul-1999)

Ref Expression
Assertion df-abs
|- abs = ( x e. CC |-> ( sqrt ` ( x x. ( * ` x ) ) ) )

Detailed syntax breakdown

Step Hyp Ref Expression
0 cabs
 |-  abs
1 vx
 |-  x
2 cc
 |-  CC
3 csqrt
 |-  sqrt
4 1 cv
 |-  x
5 cmul
 |-  x.
6 ccj
 |-  *
7 4 6 cfv
 |-  ( * ` x )
8 4 7 5 co
 |-  ( x x. ( * ` x ) )
9 8 3 cfv
 |-  ( sqrt ` ( x x. ( * ` x ) ) )
10 1 2 9 cmpt
 |-  ( x e. CC |-> ( sqrt ` ( x x. ( * ` x ) ) ) )
11 0 10 wceq
 |-  abs = ( x e. CC |-> ( sqrt ` ( x x. ( * ` x ) ) ) )