triclino Posted October 7, 2015 Share Posted October 7, 2015 In proving that :[latex]x\leq y\wedge 0\leq z\Longrightarrow xz\leq yz [/latex] we have the following proof: Let ,:[latex]x\leq y\wedge 0\leq z[/latex] and let ,[latex]\neg(xz\leq yz) [/latex].......................................................1 But from (1) and using the trichotomy law we have :yz<xz.And using the fact [latex]0\leq z[/latex] we have for 0= z, y0<x0 => 0<0 , a contradiction since ~(0<0) Hence [latex] xz\leq yz[/latex] Link to comment Share on other sites More sharing options...
Country Boy Posted August 18, 2016 Share Posted August 18, 2016 (edited) No this "proof" is not valid! In particular, the negation of "[math]xz\le yz[/math]" is NOT "[math]xz> yz[/math]" for all values of x, y, z satisfying the conditions. It is that [math]xz> yz[/math] for some such values of x, y, and z. Indeed, you cannot prove this unless you use some property of "inequality" such as "if [math]x\ge 0[/math] and [math]y\ge 0[/math] then [math]xy\ge 0[/math]". Edited August 18, 2016 by Country Boy Link to comment Share on other sites More sharing options...
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