1. If , show that the solution set of the inequality  consists of all numbers x for which .

Let us assume for contradiction that there exists x such that

, yet 

Since , x is negative for it is less than a negative number.
Then -x is a positive number and 

 [both sides are positive]. Therefore,
 which contradicts .

That's half the proof.

Now assume for contradiction that there exists x such that

 such that , yet 

Since ,  which contradicts  

Thus the proof is done, and 
 
,  implies .

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2. If , show that the solution set of the inequality 
consists of all numbers x for which  or .

Assume for contradiction that there exists x such that

 and , yet  

Then both  and x are negative numbers, and thus -x is a 
positive number.

 implies  thus 

which contradicts 

That's half the proof. 

Next, assume for contradiction that there exists x such that

 and , yet  

Then  implies  which contradicts .

Then ,  implies 


Edwin