Metamath Proof Explorer

Theorem sqrtle

Description: Square root is monotonic. (Contributed by NM, 17-Mar-2005) (Proof shortened by Mario Carneiro, 29-May-2016)

Ref Expression
Assertion sqrtle ( ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) ∧ ( 𝐵 ∈ ℝ ∧ 0 ≤ 𝐵 ) ) → ( 𝐴 ≤ 𝐵 ↔ ( √ ‘ 𝐴 ) ≤ ( √ ‘ 𝐵 ) ) )

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 resqrtcl ⊢ ( ( 𝐵 ∈ ℝ ∧ 0 ≤ 𝐵 ) → ( √ ‘ 𝐵 ) ∈ ℝ )
5 sqrtge0 ⊢ ( ( 𝐵 ∈ ℝ ∧ 0 ≤ 𝐵 ) → 0 ≤ ( √ ‘ 𝐵 ) )
6 4 5 jca ⊢ ( ( 𝐵 ∈ ℝ ∧ 0 ≤ 𝐵 ) → ( ( √ ‘ 𝐵 ) ∈ ℝ ∧ 0 ≤ ( √ ‘ 𝐵 ) ) )
7 le2sq ⊢ ( ( ( ( √ ‘ 𝐴 ) ∈ ℝ ∧ 0 ≤ ( √ ‘ 𝐴 ) ) ∧ ( ( √ ‘ 𝐵 ) ∈ ℝ ∧ 0 ≤ ( √ ‘ 𝐵 ) ) ) → ( ( √ ‘ 𝐴 ) ≤ ( √ ‘ 𝐵 ) ↔ ( ( √ ‘ 𝐴 ) ↑ 2 ) ≤ ( ( √ ‘ 𝐵 ) ↑ 2 ) ) )
8 3 6 7 syl2an ⊢ ( ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) ∧ ( 𝐵 ∈ ℝ ∧ 0 ≤ 𝐵 ) ) → ( ( √ ‘ 𝐴 ) ≤ ( √ ‘ 𝐵 ) ↔ ( ( √ ‘ 𝐴 ) ↑ 2 ) ≤ ( ( √ ‘ 𝐵 ) ↑ 2 ) ) )
9 resqrtth ⊢ ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) → ( ( √ ‘ 𝐴 ) ↑ 2 ) = 𝐴 )
10 resqrtth ⊢ ( ( 𝐵 ∈ ℝ ∧ 0 ≤ 𝐵 ) → ( ( √ ‘ 𝐵 ) ↑ 2 ) = 𝐵 )
11 9 10 breqan12d ⊢ ( ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) ∧ ( 𝐵 ∈ ℝ ∧ 0 ≤ 𝐵 ) ) → ( ( ( √ ‘ 𝐴 ) ↑ 2 ) ≤ ( ( √ ‘ 𝐵 ) ↑ 2 ) ↔ 𝐴 ≤ 𝐵 ) )
12 8 11 bitr2d ⊢ ( ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) ∧ ( 𝐵 ∈ ℝ ∧ 0 ≤ 𝐵 ) ) → ( 𝐴 ≤ 𝐵 ↔ ( √ ‘ 𝐴 ) ≤ ( √ ‘ 𝐵 ) ) )