Question 1208932: I would like to see the prove for these two questions.
1. If a > 0, show that the solution set of the inequality x^2 < a consists of all numbers x for which -sqrt{a} < x < sqrt{a}.
2. If a > 0, show that the solution set of the inequality x^2 > a consists of all numbers x for which x < -sqrt{a} or x > sqrt{a}.
Answer by Edwin McCravy(20060)   (Show Source):
You can put this solution on YOUR website!
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
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