document.write( "Question 1042516: Prove that for n<0\n" ); document.write( "a. |m/n|=|m|/|n|
\n" ); document.write( "b. |mn|=|m||n|
\n" ); document.write( "c. -|m|<=m<=|m|
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Algebra.Com's Answer #657671 by robertb(5830)\"\" \"About 
You can put this solution on YOUR website!
I will prove only (a). You do the rest....\r
\n" ); document.write( "\n" ); document.write( "\"abs%28x%29+=+x\" if x > 0; \"abs%28x%29+=+-x\" if x < 0. (And of course, \"abs%280%29+=+0\" .)\r
\n" ); document.write( "\n" ); document.write( "Given: n < 0.\r
\n" ); document.write( "\n" ); document.write( "Case I. m > 0.
\n" ); document.write( "===> \"m%2Fn+%3C+0\", and so \"abs%28m%2Fn%29+=+-%28m%2Fn%29+=+m%2F%28-n%29+=+abs%28m%29%2Fabs%28n%29\"\r
\n" ); document.write( "\n" ); document.write( "Case II. m < 0.
\n" ); document.write( "===> \"m%2Fn+%3E+0\", and so \"abs%28m%2Fn%29+=+m%2Fn+=+%28-m%29%2F%28-n%29+=+abs%28m%29%2Fabs%28n%29\".\r
\n" ); document.write( "\n" ); document.write( "Case III. m = 0 is trivial after substitution of 0 for m to both sides of the equation.\r
\n" ); document.write( "\n" ); document.write( "Hope you get the drift...
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