Discrete Math - question about sets

[XodoX]

1. The problem statement, all variables and given/known data

Use set builder notation to give a description of each of these sets.

a) { 0,3,6,9,12 }

b) { -3, -2, -1,0, 1, 2, 3 }

c) { m,n,o,p }







3. The attempt at a solution

X={x l x is an odd possitive multiplier of 3 less than 12 }



X is supposed to be the set. Can I just name it randomly? Also, can I say it like this? Is that ok? Not really sure about b and c.

[VeeEight]

You can name your set what you like.

For b, notice the elements are integers from -3 to 3; that is, x\inZ AND -3 \leq x \leq 3

For a, x does not have to be odd since 6 and 12 are in the set. You are right that they are positive multiples of 3 and less than 12 but can you describe that in mathematical notation?

[XodoX]

You can name your set what you like.

For b, notice the elements are integers from -3 to 3; that is, x\inZ AND -3 \leq x \leq 3

For a, x does not have to be odd since 6 and 12 are in the set. You are right that they are positive multiples of 3 and less than 12 but can you describe that in mathematical notation?


Oh yeah.. nevermind. They are not all odd.

And no, I don't know. I'd just simply say : X= { x\inZ l x is positive x*3 less than 13 }

[VeeEight]

If you denote 3Z as the set multiples of three, then the set in (a) consists of elements x \in3Z such that 0 \leq x \leq 12

[union68]

Alternatively, if you've seen quantifiers before you can write the set as

\left\{0,3,6,9,12\right\} = \left\{ x \in \mathbb{Z} \mid \exists y \in \mathbb{Z} \left(x=3y\right), 0 \leq x \leq 12 \right\}.

The first predicate essentially says that x is included in the set if and only if there exists an integer y such that x is three times y.

For example, the number 4 would not be included in the set because there is no integer that satisfies 4=3y. If you haven't seen quantifiers yet, then nevermind. :smile: