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April 30th, 2018, 10:49 AM  #1 
Newbie Joined: Apr 2018 From: Banovo Brdo Posts: 10 Thanks: 0  Union or intersection in inequalities?
I was solving this inequality $ \left\lvert\frac{x+2}{x+1}\right\rvert>1 $ and got $\begin{aligned}x\in(\frac{3}{2},1)\quad\lor\quad x\in(1,+\infty)\end{aligned}$ Apparently the solution should be the union of the two, not the intersection. How do I know which to do when? Both rules and intuitive approaches are much appreciated. Thanks! 
April 30th, 2018, 12:16 PM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,165 Thanks: 867 
Your real problem is that neither way, union or intersection, that answer is NOT correct! For example, if x= 2, $\displaystyle \left\frac{x+ 2}{x+ 1}\right= \frac{2+ 2}{2+ 1}= \frac{4}{3}> 1$. Here is how I would do it: x> a means either x> a or x< a. So $\displaystyle \left\frac{x+ 2}{x+ 1}\right> 1$ is the same as either I) $\displaystyle \frac{x+ 2}{x+ 1}> 1$ To solve that we need to multiply both sides by x+ 1. But multiplying both sides by a negative number changes the direction of the inequality so here we should consider two possibilities. a) x+ 1> 0. Then x+ 2> x+ 1. Subtracting x from both sides we have 2> 1. That is true for all x so this inequality is true for all x> 1. b) x+ 1< 0. Then x+ 2> (x+ 1)= x 1. Adding x to both sides and subtracting 2 from both sides, 2x> 3 so x> 3/2. Since this is for x< 1, it consists of x between 1 and 3/2. So far we have that the inequality is true for all x> 1 and for all x between 1 and 3/2. or II) $\displaystyle \frac{x+ 2}{x+ 1}< 1$. Again there are two possibilities. a) x+ 1> 0. Then x+ 2< (x+ 1)= x 1, 2x< 3, x< 3/2. b) x+ 1< 0. Then x+ 2> (x+ 1)= x 1. x> 3/2 That says that the inequality is satisfied all x> 1, all x between 1 and 3/2, and all x< 3/2. Of course, because the denominator is 0 at x= 1, the inequality is not satisfied for x= 1. Checking x= 3/2 in the $\displaystyle \left\frac{x+ 2}{x+ 1}\right= \left\frac{3/2+ 2}{3/2+ 1}\right= \left\frac{1/2}{1/2}\right= 1$, not less than 1. There are really three intervals here, x> 1, 1> x> 3/2, and x< 3/2. The inequality is satisfied by x in any of those so in the union of the three sets: $\displaystyle (\infty. 1)\cup (1, 3/2)\cup (3/2, \infty)$. You would use "intersection" here only if x would have to be in all three sets. But that's impossible because the three intervals are disjoint! There is NO value of x that is in all three. Last edited by Country Boy; April 30th, 2018 at 12:21 PM. 
April 30th, 2018, 12:18 PM  #3 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 188 Thanks: 57 Math Focus: Algebraic Number Theory, Arithmetic Geometry 
Saying $x$ is in the intersection of two intervals is saying that $x$ is in both of the intervals. (Note: in the example you've given, there are no points in both intervals. So saying the solutions to the inequality are the points in the intersection of the intervals is equivalent to saying there are no solutions to the inequality!) Saying $x$ is in the union of two intervals is saying that $x$ is in one interval or the other (or both). Perhaps you might like to look back over your argument with this distinction in mind. Last edited by cjem; April 30th, 2018 at 01:13 PM. 
April 30th, 2018, 12:39 PM  #4 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 188 Thanks: 57 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Be careful! What you've actually shown is that if $x$ satisfies the inequality, then $x$ is in $(\infty, \frac{3}{2})\cup (\frac{3}{2}, 1)\cup (1, +\infty)$. But you haven't shown the converse, that if $x$ is in $(\infty, \frac{3}{2})\cup (\frac{3}{2}, 1)\cup (1, +\infty)$, then $x$ satisfies the inequality. And in fact, this isn't even true! Indeed, if $x$ is in $(\infty, \frac{3}{2})$, then it does not satisfy the inequality (you might like to check this). However, if $x$ is in either of the other intervals, then it does satisfy the inequality (again, you might like to check this). This means the "union" version of the answer that the OP gave is correct. Finding the solutions to an equation/inequality in a variable $x$ means finding a set $A$ such that $x$ is in $A$ if and only if $x$ satisfies the equation/inequality. You've found a set that only satifies the "if" part! _______________________ Another example of this sort of issue: I shall (very inefficiently) solve $x  2 = 3$. Assume $x  2 = 3$. This implies $(x2)^2 = 3^2 = 9$, from which it follows that $x^2  4x  5 = 0$. Thus $(x+1)(x5) = 0$, and so $x = 1$ or $x = 5$. Here, what I've actually shown is that if $x  2 = 3$, then $x$ is in $\{1, 5\}$. But this doesn't answer the question  we still must check which elements of this set are actually solutions, and it's immediately clear that only $5$ is a solution. The moral of the story: if you're trying to solve an equality/inequality in $x$, it's not enough to assume the equality/inequality holds, and then deduce that $x$ lies in some set. You must then check which elements of this set actually satisfy the equation/inequality! Last edited by cjem; April 30th, 2018 at 01:15 PM. 

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inequalities, intersection, union 
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