inequality problem

[triclino]

Can we prove :

 

if (1/a)>0 ,then a>0 ??

[bob000555]

Take both sides to the negative first power.

[D H]

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.

[triclino]

. 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

[ajb]

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].

[D H]

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?

[shyvera]

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