document.write( "Question 1207709: Prove that if 0 < a < b, then 0 < (1/b) < (1/a).\r
\n" ); document.write( "\n" ); document.write( "Can any of the proofs posted today be proven by using the Direct Proof Method? \n" ); document.write( "
Algebra.Com's Answer #845739 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "0 < a < b
\n" ); document.write( "a < b
\n" ); document.write( "a*(1/a) < b*(1/a)
\n" ); document.write( "1 < b/a
\n" ); document.write( "1*(1/b) < (b/a)*(1/b)
\n" ); document.write( "1/b < 1/a
\n" ); document.write( "0 < 1/b < 1/a\r
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\n" ); document.write( "\n" ); document.write( "Since a > 0, we know that 1/a > 0 also.
\n" ); document.write( "Multiplying both sides by a positive number will not flip the inequality sign.
\n" ); document.write( "The same applies for b as well.
\n" ); document.write( "b > 0 leads to 1/b > 0.\r
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\n" ); document.write( "\n" ); document.write( "Another approach\r
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\n" ); document.write( "\n" ); document.write( "0 < a < b
\n" ); document.write( "a < b
\n" ); document.write( "a*(1/(ab)) < b*(1/(ab))
\n" ); document.write( "1/b < 1/a
\n" ); document.write( "0 < 1/b < 1/a\r
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\n" ); document.write( "\n" ); document.write( "Note that a > 0 and b > 0 lead to ab > 0, and furthermore 1/(ab) > 0.
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