Home » Blog » Taking reciprocals is order-reversing Taking reciprocals is order-reversing by RoRi June 30, 2015 Prove that if and are positive reals with , then . Proof. Since and are positive we have and , so and exist. Then, and implies and (by I.3.5, Exercise #4). Hence, So, by Theorem I.19, we have Related
November 19, 2016 at 3:14 pm Jay says: Alternative solution: We know (b-a)>0 (by def), a*b>0 (axiom 7) and [1/(a*b)]>0 (I.3.5, Exercise #4) So (b-a)*[1/(a*b)]>0 (axiom 7) (b-a)*[1/(a*b)]=((1*b)-(a*1))/(a*b) (Thm 1.14 and axiom 4) ((1*b)-(a*1))/(a*b)= (1/a)-(1/b) (1.3.3, Exercise #10) (1/a)-(1/b)>0 iff (1/b)<(1/a) (by def) By I.3.5, Exercise #4, 0<(1/b)<(1/a). QED
Alternative solution:
We know (b-a)>0 (by def), a*b>0 (axiom 7) and [1/(a*b)]>0 (I.3.5, Exercise #4)
So (b-a)*[1/(a*b)]>0 (axiom 7)
(b-a)*[1/(a*b)]=((1*b)-(a*1))/(a*b) (Thm 1.14 and axiom 4)
((1*b)-(a*1))/(a*b)= (1/a)-(1/b) (1.3.3, Exercise #10)
(1/a)-(1/b)>0 iff (1/b)<(1/a) (by def)
By I.3.5, Exercise #4, 0<(1/b)<(1/a).
QED