help with partition problems

Hello, braddy!

A partition of a set $A$ divides $A$ into a number of disjoint subsets.

We must see if the following is true:
. . (a) There is no "overlap" among the subsets.
. . (b) All the elements of $A$ are used.

You can answer these questions with some Thinking.

I'll baby-talk through most of them . . .

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Which of these collections of subsets are partitions of the set of integers?

$I \;= \;\{\hdots\,\text{-}3,\,\text{-}2,\,\text{-}1,\,0,\,1,\,2,\,3,\,\hdots\}$

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1) $A$ = the set of even integers, $B$ = the set of odd integers

$A \:=\:\{\hdots\,\text{-}5,\,\text{-}3,\,\text{-}1,\,1,\,3,\,5,\,\hdots\}$
$B \:=\:\{\hdots,\,\text{-}6,\,\text{-}4,\,\text{-}2,\,0,\,2,\,4,\,6,\,\hdots\}$

. . (a) $A \cap B \:=\:\emptyset$ . . . There is no overlap
. . (b) $A \cup B \:=\:I$ . . . All of $I$ is used

It is a partition.

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2) $C$ = the set of positive integers, $D$ = the set of negative integers

$C\:= \:\{1,\,2,\,3,\,4,\,\hdots\}$
$D\:= \:\{\text{-}1,\,\text{-}2,\,\text{-}3,\,\text{-}4,\,\hdots\}$

. . (a) $C \cap D\:=\:\emptyset$ . . . They are disjoint
. . (b) $C \cup D \:\neq\:I$ . . . The $0$ is not included.

It is not a partition.

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3) $P$ = the set of integers divisible by 3,
. . $Q$ = the set of integers leaving a remainder of 1 when divided by 3,
. . $R$ = the set of integers leaving a remainder of 2 when divided by 3

You can think your way through this one.

When we divide an integer by 3, only three things can happen:
. . [1] the remainder is 0 . . . the integer is in $P.$
. . [2] the remainder is 1 . . . the integer is in $Q.$
. . [3] the remainder is 2 . . . the integer is in $R.$

. . (a) $P,\,Q,\,R$ are disjoint.
. . (b) $P \cup Q \cup R \:=\:I$

It is a partition.

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4) $A$ = the set of integers less than -100,
. . $B$ = the set of integers with absolute value not exceeding 100,
. . $C$ = the set of integers greater than 100.

This one sounds complicated, but let's take baby steps . . .

$A$ is easy: . $A \:=\:\{\hdots\,\text{-}104,\,\text{-}103,\,\text{-}102,\,\text{-}101\}$

$C$ is easy: . $C\:=\:\{101,\,102,\,103,\,104,\,\hdots\}$

$B$ has integers $n$ , where $|n| \leq 100$ . . . that is: . $-100 \leq n \leq 100$
. . Hence: . $B \:=\:\{\text{-}100,\,\text{-}99,\,\text{-}98,\,\hdots\,98,\,99,\,100\}$

. . (a) $A,\,B,\,C$ are disjoint.
. . (b) $A \cup B \cup C \:=\:I$ . . . all of $I$ is used

It is a partition.