Metamath Proof Explorer


Theorem pell14qrdivcl

Description: Positive Pell solutions are closed under division. (Contributed by Stefan O'Rear, 18-Sep-2014)

Ref Expression
Assertion pell14qrdivcl ( ( 𝐷 ∈ ( ℕ ∖ ◻NN ) ∧ 𝐴 ∈ ( Pell14QR ‘ 𝐷 ) ∧ 𝐵 ∈ ( Pell14QR ‘ 𝐷 ) ) → ( 𝐴 / 𝐵 ) ∈ ( Pell14QR ‘ 𝐷 ) )

Proof

Step Hyp Ref Expression
1 pell14qrre ( ( 𝐷 ∈ ( ℕ ∖ ◻NN ) ∧ 𝐴 ∈ ( Pell14QR ‘ 𝐷 ) ) → 𝐴 ∈ ℝ )
2 1 recnd ( ( 𝐷 ∈ ( ℕ ∖ ◻NN ) ∧ 𝐴 ∈ ( Pell14QR ‘ 𝐷 ) ) → 𝐴 ∈ ℂ )
3 2 3adant3 ( ( 𝐷 ∈ ( ℕ ∖ ◻NN ) ∧ 𝐴 ∈ ( Pell14QR ‘ 𝐷 ) ∧ 𝐵 ∈ ( Pell14QR ‘ 𝐷 ) ) → 𝐴 ∈ ℂ )
4 pell14qrre ( ( 𝐷 ∈ ( ℕ ∖ ◻NN ) ∧ 𝐵 ∈ ( Pell14QR ‘ 𝐷 ) ) → 𝐵 ∈ ℝ )
5 4 recnd ( ( 𝐷 ∈ ( ℕ ∖ ◻NN ) ∧ 𝐵 ∈ ( Pell14QR ‘ 𝐷 ) ) → 𝐵 ∈ ℂ )
6 5 3adant2 ( ( 𝐷 ∈ ( ℕ ∖ ◻NN ) ∧ 𝐴 ∈ ( Pell14QR ‘ 𝐷 ) ∧ 𝐵 ∈ ( Pell14QR ‘ 𝐷 ) ) → 𝐵 ∈ ℂ )
7 pell14qrne0 ( ( 𝐷 ∈ ( ℕ ∖ ◻NN ) ∧ 𝐵 ∈ ( Pell14QR ‘ 𝐷 ) ) → 𝐵 ≠ 0 )
8 7 3adant2 ( ( 𝐷 ∈ ( ℕ ∖ ◻NN ) ∧ 𝐴 ∈ ( Pell14QR ‘ 𝐷 ) ∧ 𝐵 ∈ ( Pell14QR ‘ 𝐷 ) ) → 𝐵 ≠ 0 )
9 3 6 8 divrecd ( ( 𝐷 ∈ ( ℕ ∖ ◻NN ) ∧ 𝐴 ∈ ( Pell14QR ‘ 𝐷 ) ∧ 𝐵 ∈ ( Pell14QR ‘ 𝐷 ) ) → ( 𝐴 / 𝐵 ) = ( 𝐴 · ( 1 / 𝐵 ) ) )
10 pell14qrreccl ( ( 𝐷 ∈ ( ℕ ∖ ◻NN ) ∧ 𝐵 ∈ ( Pell14QR ‘ 𝐷 ) ) → ( 1 / 𝐵 ) ∈ ( Pell14QR ‘ 𝐷 ) )
11 10 3adant2 ( ( 𝐷 ∈ ( ℕ ∖ ◻NN ) ∧ 𝐴 ∈ ( Pell14QR ‘ 𝐷 ) ∧ 𝐵 ∈ ( Pell14QR ‘ 𝐷 ) ) → ( 1 / 𝐵 ) ∈ ( Pell14QR ‘ 𝐷 ) )
12 pell14qrmulcl ( ( 𝐷 ∈ ( ℕ ∖ ◻NN ) ∧ 𝐴 ∈ ( Pell14QR ‘ 𝐷 ) ∧ ( 1 / 𝐵 ) ∈ ( Pell14QR ‘ 𝐷 ) ) → ( 𝐴 · ( 1 / 𝐵 ) ) ∈ ( Pell14QR ‘ 𝐷 ) )
13 11 12 syld3an3 ( ( 𝐷 ∈ ( ℕ ∖ ◻NN ) ∧ 𝐴 ∈ ( Pell14QR ‘ 𝐷 ) ∧ 𝐵 ∈ ( Pell14QR ‘ 𝐷 ) ) → ( 𝐴 · ( 1 / 𝐵 ) ) ∈ ( Pell14QR ‘ 𝐷 ) )
14 9 13 eqeltrd ( ( 𝐷 ∈ ( ℕ ∖ ◻NN ) ∧ 𝐴 ∈ ( Pell14QR ‘ 𝐷 ) ∧ 𝐵 ∈ ( Pell14QR ‘ 𝐷 ) ) → ( 𝐴 / 𝐵 ) ∈ ( Pell14QR ‘ 𝐷 ) )