Question 1208932
<pre>

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)}}}.

Let us assume for contradiction that there exists x such that

{{{a > 0""" such that {{{x^2 < a}}}, yet {{{x < -sqrt(a)}}}

Since {{{x < -sqrt(a)}}}, x is negative for it is less than a negative number.
Then -x is a positive number and 

{{{-x > sqrt(a)}}} [both sides are positive]. Therefore,
{{{x^2 > a}}} which contradicts {{{x^2 < a}}}.

That's half the proof.

Now assume for contradiction that there exists x such that

{{{a > 0}}} such that {{{x^2 < a}}}, yet {{{x > sqrt(a)}}}

Since {{{x > sqrt(a)}}}, {{{x^2 > a}}} which contradicts {{{x^2 < a}}} 

Thus the proof is done, and 
 
{{{a > 0}}}, {{{x^2 < a}}} implies {{{-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)}}}.

Assume for contradiction that there exists x such that

{{{a > 0}}} and {{{x^2 > a}}}, yet {{{-sqrt(a)<x<0}}} 

Then both {{{-sqrt(a)}}} and x are negative numbers, and thus -x is a 
positive number.

{{{-sqrt(a)<x<0}}} implies {{{0<-x<sqrt(a)}}} thus {{{x^2<a}}}

which contradicts {{{x^2>a}}}

That's half the proof. 

Next, assume for contradiction that there exists x such that

{{{a > 0}}} and {{{x^2 > a}}}, yet {{{0<=x<sqrt(a)}}} 

Then {{{0<=x<sqrt(a)}}} implies {{{x^2<a}}} which contradicts {{{x^2>a}}}.

Then {{{a > 0}}}, {{{x^2 > a}}} implies {{{matrix(1,3,x<-sqrt(a), or, x>sqrt(a))}}}


Edwin</pre>