Metamath Proof Explorer


Theorem sumsqeq0

Description: Two real numbers are equal to 0 iff their Euclidean norm is. (Contributed by NM, 29-Apr-2005) (Revised by Stefan O'Rear, 5-Oct-2014) (Proof shortened by Mario Carneiro, 28-May-2016)

Ref Expression
Assertion sumsqeq0 ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( ( 𝐴 = 0 ∧ 𝐵 = 0 ) ↔ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = 0 ) )

Proof

Step Hyp Ref Expression
1 resqcl ( 𝐴 ∈ ℝ → ( 𝐴 ↑ 2 ) ∈ ℝ )
2 sqge0 ( 𝐴 ∈ ℝ → 0 ≤ ( 𝐴 ↑ 2 ) )
3 1 2 jca ( 𝐴 ∈ ℝ → ( ( 𝐴 ↑ 2 ) ∈ ℝ ∧ 0 ≤ ( 𝐴 ↑ 2 ) ) )
4 resqcl ( 𝐵 ∈ ℝ → ( 𝐵 ↑ 2 ) ∈ ℝ )
5 sqge0 ( 𝐵 ∈ ℝ → 0 ≤ ( 𝐵 ↑ 2 ) )
6 4 5 jca ( 𝐵 ∈ ℝ → ( ( 𝐵 ↑ 2 ) ∈ ℝ ∧ 0 ≤ ( 𝐵 ↑ 2 ) ) )
7 add20 ( ( ( ( 𝐴 ↑ 2 ) ∈ ℝ ∧ 0 ≤ ( 𝐴 ↑ 2 ) ) ∧ ( ( 𝐵 ↑ 2 ) ∈ ℝ ∧ 0 ≤ ( 𝐵 ↑ 2 ) ) ) → ( ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = 0 ↔ ( ( 𝐴 ↑ 2 ) = 0 ∧ ( 𝐵 ↑ 2 ) = 0 ) ) )
8 3 6 7 syl2an ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = 0 ↔ ( ( 𝐴 ↑ 2 ) = 0 ∧ ( 𝐵 ↑ 2 ) = 0 ) ) )
9 recn ( 𝐴 ∈ ℝ → 𝐴 ∈ ℂ )
10 sqeq0 ( 𝐴 ∈ ℂ → ( ( 𝐴 ↑ 2 ) = 0 ↔ 𝐴 = 0 ) )
11 9 10 syl ( 𝐴 ∈ ℝ → ( ( 𝐴 ↑ 2 ) = 0 ↔ 𝐴 = 0 ) )
12 recn ( 𝐵 ∈ ℝ → 𝐵 ∈ ℂ )
13 sqeq0 ( 𝐵 ∈ ℂ → ( ( 𝐵 ↑ 2 ) = 0 ↔ 𝐵 = 0 ) )
14 12 13 syl ( 𝐵 ∈ ℝ → ( ( 𝐵 ↑ 2 ) = 0 ↔ 𝐵 = 0 ) )
15 11 14 bi2anan9 ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( ( ( 𝐴 ↑ 2 ) = 0 ∧ ( 𝐵 ↑ 2 ) = 0 ) ↔ ( 𝐴 = 0 ∧ 𝐵 = 0 ) ) )
16 8 15 bitr2d ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( ( 𝐴 = 0 ∧ 𝐵 = 0 ) ↔ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = 0 ) )