Why is (x+h)^2 < k the same as -k < x+h < k?

1 Answer
Feb 21, 2018

"Please see the proof below."

Explanation:

"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