# "Just a minor thing -- what you asked, as stated in not correct."#
# "But there is a natural correction, which is what I think you" #
# "meant. Let me take this as what was meant:" #
# "Why is" \ \ ( x + h )^2 < k \ \ "the same as" \ - sqrt{k} < x + h < sqrt{k} \ \ "?"#
# "We'll show that. Let's start with the forward direction. We"#
# "see:" #
# \qquad \qquad \qquad \qquad \qquad ( x + h )^2 < k \quad => \quad ( x + h )^2 < ( sqrt{k} )^2. #
# "So here we have now:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad ( x + h )^2 - ( sqrt{k} )^2 < 0 #
# "So using the difference of two squares, we can factor the" #
# "left-hand side of the previous inequality, and we get:" #
# \qquad \qquad \qquad \quad [ ( x + h ) + ( sqrt{k} ) ] cdot [ ( x + h ) - ( sqrt{k} ) ] < 0. \qquad \qquad \qquad \ (1) #
# "Now if the product of 2 (real) numbers is negative, what can" #
# "we say about them ? They must have opposite signs --" #
# "one negative, the other positive." #
# "This is the situation in the inequality in (1). So we conclude:" #
# \qquad [ ( x + h ) + ( sqrt{k} ) ] < 0 \qquad "and" \qquad [ ( x + h ) - ( sqrt{k} ) ] > 0 \qquad \ \ \ (a) #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad "or" #
# \qquad [ ( x + h ) + ( sqrt{k} ) ] > 0 \qquad "and" \qquad [ ( x + h ) - ( sqrt{k} ) ] < 0. \qquad \ \ (b) #
# "Now look at the first pair inequalities -- (a), and analyze them:" #
# \qquad \quad [ ( x + h ) + ( sqrt{k} ) ] < 0 \qquad "and" \qquad [ ( x + h ) - ( sqrt{k} ) ] > 0 #
# \qquad \qquad \quad \ ( x + h ) < - ( sqrt{k} ) \qquad "and" \qquad ( x + h ) > + ( sqrt{k} ) #
# \qquad \qquad \qquad \qquad \quad x + h < - sqrt{k} \qquad "and" \qquad x + h > sqrt{k} #
# \qquad :. \qquad \qquad \qquad \qquad \qquad \quad \ sqrt{k} \ \ < x + h < - sqrt{k}. #
# "Note that the previous triple inequality is impossible, for it" #
# "would mean that:" \ \ sqrt{k} \ \ < - sqrt{k}; \ "implying a positive number" #
# "could be smaller than a negative number. Thus, the inequality" #
# "in (a) is impossible. So we conclude that only the inequality" #
# "in (b) can be true. Hence:" #
# \qquad \quad \ [ ( x + h ) + ( sqrt{k} ) ] > 0 \qquad "and" \qquad [ ( x + h ) - ( sqrt{k} ) ] < 0. #
# "Analyzing:" #
# \qquad \qquad \quad \ ( x + h ) > - ( sqrt{k} ) \qquad "and" \qquad ( x + h ) > + ( sqrt{k} ) #
# \qquad \qquad \qquad \qquad \quad x + h > - sqrt{k} \qquad "and" \qquad x + h < sqrt{k} #
# \qquad :. \qquad \qquad \qquad \qquad \quad -sqrt{k} \ \ < x + h < + sqrt{k}. #
# "Thus we conclude, finally, that:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \quad -sqrt{k} \ \ < x + h < + sqrt{k}. #
# "So, stating things from beginning to end here, we have shown:" #
# \qquad \qquad \qquad \quad ( x + h )^2 < k \quad => \quad -sqrt{k} \ \ < x + h < + sqrt{k}. \qquad \quad \quad \ (2) #
# "This shows the forward direction." #
# "Now we show the reverse. We see:" #
# \qquad \qquad \qquad \quad \ - sqrt{k} < x + h < sqrt{k} \quad => \quad | x + h | < sqrt{k}. #
# "So here we have now:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad | x + h | < sqrt{k}. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad \ (4)#
# "As the absolute value of any quantity is positive or zero, we" #
# "may multiply both sides of the inequality in (4) by" \ | x + h |, "and just change it to" \ <= \ "[since" \ |x + h | \ "can be zero]:" #
# \qquad \qquad \qquad \qquad \qquad \qquad | x + h | cdot color{red}{ | x + h | } <= sqrt{k} cdot color{red}{ |x + h | } #
# \qquad :. \qquad \qquad \qquad \qquad \qquad \ | x + h |^2 <= |x + h | cdot sqrt{k}. \qquad \qquad \qquad \qquad \qquad \qquad \ (c) #
# "As" \ sqrt{k} \ \ "is positive, we may multiply both sides of the" #
# "inequality in (4) by" \ sqrt{k} ":" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad | x + h | cdot color{red}{ sqrt{k} } < sqrt{k} cdot color{red}{ sqrt{k} } #
# \qquad :. \qquad \qquad \qquad \qquad \qquad \quad| x + h | cdot sqrt{k} < ( sqrt{k} )^2 #
# \qquad :. \qquad \qquad \qquad \qquad \qquad \quad | x + h | cdot sqrt{k} < k. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (d) #
# "Combining (c) and (d), we see:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \ | x + h |^2 \quad <= \quad |x + h | cdot sqrt{k} \quad < \quad k. #
# "So we conclude from the previous triple inequality that:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad | x + h |^2 < k. #
# "As" \ \ | x |^2 \ "is always the same as" \ x^2, "we have:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ( x + h )^2 < k. #
# "So, from beginning to end here, we have shown:" #
# \qquad \qquad \qquad \quad - sqrt{k} < x + h < sqrt{k} \quad => \quad ( x + h )^2 < k. \qquad \qquad \qquad \quad \ \ (5) #
# "This shows the reverse direction." #
# "Combining the results in (2) and (5), we see:" #
# ( x + h )^2 < k \qquad "is precisely the same as" \quad - sqrt{k} < x + h < sqrt{k}. #
# "This is what we wanted to establish." \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad square #
