Thus, ad = bc
(ad)(bd)-1 = (bc)(bd)-1
(ad)(d-1b-1) = (cd)(b-1d-1) by BR5
((ad)d-1)b-1 = c(b(b-1d-1)) by BR6
(a(dd-1))b-1 = c((bd-1)d-1) by BR6
(a*1)b-1 = c(1*d-1) by BR5 and BR8
ab-1 = cd-1 by BR5 and BR7
a/b = c/d notation
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