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 ⊢ ℂ
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 ) ) ) )