triclino Posted October 21, 2009 Share Posted October 21, 2009 Can we prove : if (1/a)>0 ,then a>0 ?? Link to comment Share on other sites More sharing options...
bob000555 Posted October 22, 2009 Share Posted October 22, 2009 Take both sides to the negative first power. Link to comment Share on other sites More sharing options...
D H Posted October 22, 2009 Share Posted October 22, 2009 Take both sides to the negative first power. How do you know that doing this preserves the inequality? Can we prove : if (1/a)>0 ,then a>0 ?? Yes. Multiplying both sides of an inequality of the form x>0 by any positive number b does preserve the inequality. In other words, if x>0 and b>0 then xb>0. Similarly, if x>0 and b≥0, then xb≥0. There is a particular non-negative number that transforms 1/a > 0 to a ≥ 0. Now all you have to deal with is the nasty case a=0. a must be non-zero because 1/0 is undefined. Link to comment Share on other sites More sharing options...
triclino Posted October 22, 2009 Author Share Posted October 22, 2009 . a must be non-zero because 1/0 is undefined. In that case the theorem to prove is: if [math]a\neq 0[/math] and (1/a)>0,then a>0 ,and not: if (1/a)>0,then a>0 Link to comment Share on other sites More sharing options...
ajb Posted October 22, 2009 Share Posted October 22, 2009 In that case the theorem to prove is: if [math]a\neq 0[/math] and (1/a)>0,then a>0 ,and not: if (1/a)>0,then a>0 Same thing as by definition [math]\frac{1}{a}[/math] is only defined for [math]a \neq 0[/math]. Link to comment Share on other sites More sharing options...
D H Posted October 22, 2009 Share Posted October 22, 2009 triclino, this looks like homework. Are you asking how to prove this, or are you posing this as a puzzle to which you already know the answer? Link to comment Share on other sites More sharing options...
shyvera Posted May 16, 2010 Share Posted May 16, 2010 (edited) Can we prove : if (1/a)>0 ,then a>0 ?? You know that [math]a\cdot\left(\frac1a\right)=1[/math] and [math]1>0.[/math] If [math]a<0[/math] then, since [math]\frac1a>0,[/math] the LHS would be negative. As 1 is positive, and [math]a\ne0,[/math] you must have [math]a>0.[/math] Edited May 16, 2010 by shyvera 1 Link to comment Share on other sites More sharing options...
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