Metamath Proof Explorer

Theorem sqrtsq2

Description: Relationship between square root and squares. (Contributed by NM, 31-Jul-1999) (Revised by Mario Carneiro, 29-May-2016)

Ref Expression
Assertion sqrtsq2 ( ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) ∧ ( 𝐵 ∈ ℝ ∧ 0 ≤ 𝐵 ) ) → ( ( √ ‘ 𝐴 ) = 𝐵 ↔ 𝐴 = ( 𝐵 ↑ 2 ) ) )

Proof

Step Hyp Ref Expression
1 resqrtcl ⊢ ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) → ( √ ‘ 𝐴 ) ∈ ℝ )
2 sqrtge0 ⊢ ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) → 0 ≤ ( √ ‘ 𝐴 ) )
3 1 2 jca ⊢ ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) → ( ( √ ‘ 𝐴 ) ∈ ℝ ∧ 0 ≤ ( √ ‘ 𝐴 ) ) )
4 sq11 ⊢ ( ( ( ( √ ‘ 𝐴 ) ∈ ℝ ∧ 0 ≤ ( √ ‘ 𝐴 ) ) ∧ ( 𝐵 ∈ ℝ ∧ 0 ≤ 𝐵 ) ) → ( ( ( √ ‘ 𝐴 ) ↑ 2 ) = ( 𝐵 ↑ 2 ) ↔ ( √ ‘ 𝐴 ) = 𝐵 ) )
5 3 4 sylan ⊢ ( ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) ∧ ( 𝐵 ∈ ℝ ∧ 0 ≤ 𝐵 ) ) → ( ( ( √ ‘ 𝐴 ) ↑ 2 ) = ( 𝐵 ↑ 2 ) ↔ ( √ ‘ 𝐴 ) = 𝐵 ) )
6 resqrtth ⊢ ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) → ( ( √ ‘ 𝐴 ) ↑ 2 ) = 𝐴 )
7 6 adantr ⊢ ( ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) ∧ ( 𝐵 ∈ ℝ ∧ 0 ≤ 𝐵 ) ) → ( ( √ ‘ 𝐴 ) ↑ 2 ) = 𝐴 )
8 7 eqeq1d ⊢ ( ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) ∧ ( 𝐵 ∈ ℝ ∧ 0 ≤ 𝐵 ) ) → ( ( ( √ ‘ 𝐴 ) ↑ 2 ) = ( 𝐵 ↑ 2 ) ↔ 𝐴 = ( 𝐵 ↑ 2 ) ) )
9 5 8 bitr3d ⊢ ( ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) ∧ ( 𝐵 ∈ ℝ ∧ 0 ≤ 𝐵 ) ) → ( ( √ ‘ 𝐴 ) = 𝐵 ↔ 𝐴 = ( 𝐵 ↑ 2 ) ) )