Metamath Proof Explorer


Theorem 8th4div3

Description: An eighth of four thirds is a sixth. (Contributed by Paul Chapman, 24-Nov-2007)

Ref Expression
Assertion 8th4div3 ( ( 1 / 8 ) · ( 4 / 3 ) ) = ( 1 / 6 )

Proof

Step Hyp Ref Expression
1 ax-1cn 1 ∈ ℂ
2 8cn 8 ∈ ℂ
3 4cn 4 ∈ ℂ
4 3cn 3 ∈ ℂ
5 8re 8 ∈ ℝ
6 8pos 0 < 8
7 5 6 gt0ne0ii 8 ≠ 0
8 3ne0 3 ≠ 0
9 1 2 3 4 7 8 divmuldivi ( ( 1 / 8 ) · ( 4 / 3 ) ) = ( ( 1 · 4 ) / ( 8 · 3 ) )
10 1 3 mulcomi ( 1 · 4 ) = ( 4 · 1 )
11 2cn 2 ∈ ℂ
12 3 11 4 mulassi ( ( 4 · 2 ) · 3 ) = ( 4 · ( 2 · 3 ) )
13 4t2e8 ( 4 · 2 ) = 8
14 13 oveq1i ( ( 4 · 2 ) · 3 ) = ( 8 · 3 )
15 2t3e6 ( 2 · 3 ) = 6
16 15 oveq2i ( 4 · ( 2 · 3 ) ) = ( 4 · 6 )
17 12 14 16 3eqtr3i ( 8 · 3 ) = ( 4 · 6 )
18 10 17 oveq12i ( ( 1 · 4 ) / ( 8 · 3 ) ) = ( ( 4 · 1 ) / ( 4 · 6 ) )
19 9 18 eqtri ( ( 1 / 8 ) · ( 4 / 3 ) ) = ( ( 4 · 1 ) / ( 4 · 6 ) )
20 6cn 6 ∈ ℂ
21 6re 6 ∈ ℝ
22 6pos 0 < 6
23 21 22 gt0ne0ii 6 ≠ 0
24 4ne0 4 ≠ 0
25 divcan5 ( ( 1 ∈ ℂ ∧ ( 6 ∈ ℂ ∧ 6 ≠ 0 ) ∧ ( 4 ∈ ℂ ∧ 4 ≠ 0 ) ) → ( ( 4 · 1 ) / ( 4 · 6 ) ) = ( 1 / 6 ) )
26 1 25 mp3an1 ( ( ( 6 ∈ ℂ ∧ 6 ≠ 0 ) ∧ ( 4 ∈ ℂ ∧ 4 ≠ 0 ) ) → ( ( 4 · 1 ) / ( 4 · 6 ) ) = ( 1 / 6 ) )
27 20 23 3 24 26 mp4an ( ( 4 · 1 ) / ( 4 · 6 ) ) = ( 1 / 6 )
28 19 27 eqtri ( ( 1 / 8 ) · ( 4 / 3 ) ) = ( 1 / 6 )