Metamath Proof Explorer


Theorem 01sqrex

Description: Existence of a square root for reals in the interval ( 0 , 1 ] . (Contributed by Mario Carneiro, 10-Jul-2013)

Ref Expression
Assertion 01sqrex ( ( 𝐴 ∈ ℝ+𝐴 ≤ 1 ) → ∃ 𝑥 ∈ ℝ+ ( 𝑥 ≤ 1 ∧ ( 𝑥 ↑ 2 ) = 𝐴 ) )

Proof

Step Hyp Ref Expression
1 eqid { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } = { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 }
2 eqid sup ( { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } , ℝ , < ) = sup ( { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } , ℝ , < )
3 1 2 sqrlem4 ( ( 𝐴 ∈ ℝ+𝐴 ≤ 1 ) → ( sup ( { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } , ℝ , < ) ∈ ℝ+ ∧ sup ( { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } , ℝ , < ) ≤ 1 ) )
4 eqid { 𝑧 ∣ ∃ 𝑤 ∈ { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } ∃ 𝑥 ∈ { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } 𝑧 = ( 𝑤 · 𝑥 ) } = { 𝑧 ∣ ∃ 𝑤 ∈ { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } ∃ 𝑥 ∈ { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } 𝑧 = ( 𝑤 · 𝑥 ) }
5 1 2 4 sqrlem7 ( ( 𝐴 ∈ ℝ+𝐴 ≤ 1 ) → ( sup ( { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } , ℝ , < ) ↑ 2 ) = 𝐴 )
6 breq1 ( 𝑥 = sup ( { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } , ℝ , < ) → ( 𝑥 ≤ 1 ↔ sup ( { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } , ℝ , < ) ≤ 1 ) )
7 oveq1 ( 𝑥 = sup ( { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } , ℝ , < ) → ( 𝑥 ↑ 2 ) = ( sup ( { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } , ℝ , < ) ↑ 2 ) )
8 7 eqeq1d ( 𝑥 = sup ( { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } , ℝ , < ) → ( ( 𝑥 ↑ 2 ) = 𝐴 ↔ ( sup ( { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } , ℝ , < ) ↑ 2 ) = 𝐴 ) )
9 6 8 anbi12d ( 𝑥 = sup ( { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } , ℝ , < ) → ( ( 𝑥 ≤ 1 ∧ ( 𝑥 ↑ 2 ) = 𝐴 ) ↔ ( sup ( { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } , ℝ , < ) ≤ 1 ∧ ( sup ( { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } , ℝ , < ) ↑ 2 ) = 𝐴 ) ) )
10 9 rspcev ( ( sup ( { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } , ℝ , < ) ∈ ℝ+ ∧ ( sup ( { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } , ℝ , < ) ≤ 1 ∧ ( sup ( { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } , ℝ , < ) ↑ 2 ) = 𝐴 ) ) → ∃ 𝑥 ∈ ℝ+ ( 𝑥 ≤ 1 ∧ ( 𝑥 ↑ 2 ) = 𝐴 ) )
11 10 anassrs ( ( ( sup ( { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } , ℝ , < ) ∈ ℝ+ ∧ sup ( { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } , ℝ , < ) ≤ 1 ) ∧ ( sup ( { 𝑦 ∈ ℝ+ ∣ ( 𝑦 ↑ 2 ) ≤ 𝐴 } , ℝ , < ) ↑ 2 ) = 𝐴 ) → ∃ 𝑥 ∈ ℝ+ ( 𝑥 ≤ 1 ∧ ( 𝑥 ↑ 2 ) = 𝐴 ) )
12 3 5 11 syl2anc ( ( 𝐴 ∈ ℝ+𝐴 ≤ 1 ) → ∃ 𝑥 ∈ ℝ+ ( 𝑥 ≤ 1 ∧ ( 𝑥 ↑ 2 ) = 𝐴 ) )