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

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Stumbling Robot

A Fraction of a Dot
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Taking reciprocals is order-reversing

* Proof. * Since and are positive we have and , so and exist. Then, and implies and (by I.3.5, Exercise #4). Hence,
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### Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):

Prove that if and are positive reals with , then .

So, by Theorem I.19, we have

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