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 ∈ ℂ ↦ ( √ ‘ ( 𝑥 · ( ∗ ‘ 𝑥 ) ) ) )

Detailed syntax breakdown

Step Hyp Ref Expression
0 cabs abs
1 vx 𝑥
2 cc
3 csqrt
4 1 cv 𝑥
5 cmul ·
6 ccj
7 4 6 cfv ( ∗ ‘ 𝑥 )
8 4 7 5 co ( 𝑥 · ( ∗ ‘ 𝑥 ) )
9 8 3 cfv ( √ ‘ ( 𝑥 · ( ∗ ‘ 𝑥 ) ) )
10 1 2 9 cmpt ( 𝑥 ∈ ℂ ↦ ( √ ‘ ( 𝑥 · ( ∗ ‘ 𝑥 ) ) ) )
11 0 10 wceq abs = ( 𝑥 ∈ ℂ ↦ ( √ ‘ ( 𝑥 · ( ∗ ‘ 𝑥 ) ) ) )