Metamath Proof Explorer


Theorem sqrtmsq

Description: Square root of square. (Contributed by NM, 2-Aug-1999) (Revised by Mario Carneiro, 29-May-2016)

Ref Expression
Assertion sqrtmsq ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) → ( √ ‘ ( 𝐴 · 𝐴 ) ) = 𝐴 )

Proof

Step Hyp Ref Expression
1 simpl ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) → 𝐴 ∈ ℝ )
2 1 recnd ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) → 𝐴 ∈ ℂ )
3 2 sqvald ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) → ( 𝐴 ↑ 2 ) = ( 𝐴 · 𝐴 ) )
4 3 fveq2d ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) → ( √ ‘ ( 𝐴 ↑ 2 ) ) = ( √ ‘ ( 𝐴 · 𝐴 ) ) )
5 sqrtsq ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) → ( √ ‘ ( 𝐴 ↑ 2 ) ) = 𝐴 )
6 4 5 eqtr3d ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) → ( √ ‘ ( 𝐴 · 𝐴 ) ) = 𝐴 )