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Taking reciprocals is order-reversing

Prove that if a and b are positive reals with a < b, then 0 < b^{-1} < a^{-1}.

Proof. Since a and b are positive we have a \neq 0 and b \neq 0, so a^{-1} and b^{-1} exist. Then, a > 0 and b > 0 implies a^{-1} > 0 and b^{-1} > 0 (by I.3.5, Exercise #4). Hence,

    \[ a^{-1} b^{-1} > 0 \]

So, by Theorem I.19, we have

    \begin{align*}  &&a(a^{-1} b^{-1}) &< b(a^{-1}b^{-1}) \\ \implies&& b^{-1} &< a^{-1}. \qquad \blacksquare \end{align*}

One comment

  1. 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).

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