Question

How do I put this in its simplest form?

Answer 1

`\sf\\\textsf{You may write this as R}-(-(13)/(7),-(18)/(12)]`

Step-by-step explanation:

`\sf\\\textsf{Here, }(-\infty,-(13)/(7)] \textsf{ contains all the numbers below }(-13)/(7)\textsf{ and including }(-13)/(7).\\\\(-(18)/(12),\infty)\textsf{ contains all the numbers greater than }-(18)/(12).`

`\sf\\\textsf{The union indicates that two intervals are joined together. So the union of these}\\\textsf{intervals do not contain the numbers in the interval: }\\(-(13)/(17),-(18)/(12))\\\textsf{This means that if we subtract this interval from the set of real numbers R, we}\\\textsf{would get the original interval.}`

`\sf\\\therefore\ R-(-(13)/(7),-(18)/(12)]=(-\infty,-(13)/(7)]\cup(-(18)/(12),\infty).`

Look the example below to understand this clearly:

`\sf\\\textsf{For example we have two intervals }(-\infty,3)\textsf{ and }(6,\infty).\\\textsf{The union of these intervals will be denoted as }(-\infty,3) \cup(6,\infty).`

`\sf\\\textsf{The first interval }(-\infty,3) \textsf{ contains all the numbers less than 3 and the second}\\\textsf{interval }(6,\infty)\textsf{ contains all the numbers greater than 6.}\\\\\textsf{So the union of these intervals would contain the numbers less than 3 and the }\\\textsf{numbers greater than 6. This means it won't contain 3, 4, 5 and 6 i.e. [3,6].}`

`\textsf{So if we subtract [3, 6] from the set of real numbers, we will again get the}\\\textsf{numbers greater than 6 and less than 3.}`

`\sf\\\therefore\ R-[3,6]=(-\infty,3)\cup(6,\infty).`