2
$\begingroup$

I've come across a question in Discrete Mathematics, asking me to use set builder notation to describe the set of all odd numbers between 100 and 200.

The answer I had was: $$\{ p | p = 2n + 1, n \text{ (all numbers) } [50, 99], 100 < p < 200 \}$$

Although this should technically give the correct answer, the answers in the textbook have:

$$\{x\,|\,100<x<200\text{ and }2\not | x\}$$ I get the first part, however I have no clue what the end means (2 |/ x); what is that symbol called, and does that represent all odd numbers?

$\endgroup$
5
  • 1
    $\begingroup$ It seems to be $2$ does not divide. $x$ $\endgroup$
    – NECing
    Commented Mar 9, 2015 at 3:34
  • 1
    $\begingroup$ Usually the diagonal strike would go through the divides by vertical strike, so you would get $2 \not | x$ to show that $2$ does not divide $x$ The use of $[50,99]$ and $100 \lt p \lt 200$ is redundant, but still correct. $\endgroup$ Commented Mar 9, 2015 at 3:36
  • 2
    $\begingroup$ Don't use p. It usually means p is prime. $\endgroup$
    – Qudit
    Commented Mar 9, 2015 at 3:38
  • $\begingroup$ It'd likely be better to either not specify the range in which $n$ is in as $$\{p | p=2n+1,n\text{ is an integer},100<p<200\}$$ - since that bit is redundant. You could also not specify where $p$ lies - and even get rid of $p$ altogether - to get $$\{2n+1 | 50\leq n \leq 99\}.$$ (Your answer is correct, but it's not in its simplest form) $\endgroup$ Commented Mar 9, 2015 at 3:43
  • $\begingroup$ Thanks for your help guys. Some helpful tips and information. $\endgroup$
    – zuc0001
    Commented Mar 9, 2015 at 3:44

2 Answers 2

3
$\begingroup$

The symbol $\mid$ means 'divides'. Drawing a line through it to get $\not \mid$ means 'does not divide'. As for your answer, it is absolutely correct. There are many equally correct ways to write the set of odd numbers with setbuilder notation.

$\endgroup$
2
  • $\begingroup$ So 2 does not divide by x is another way of writing all odd numbers? $\endgroup$
    – zuc0001
    Commented Mar 9, 2015 at 3:37
  • $\begingroup$ That is correct. The odd numbers are precisely those not divisible by $2$. $\endgroup$ Commented Mar 9, 2015 at 3:39
0
$\begingroup$

As others have noted, the $|$ symbol means "divide". And when we put a slash through it, it means "does not divide".

What does it mean for $2$ to "divide" a number? It means $2$ is a factor of that number. For example, $2$ divides $6$ because $6 = 3 \cdot 2$ (so $2$ is a factor of $6$). Similarly, $2$ divides $100$, since $100 = 5 \cdot 5 \cdot 2 \cdot 2$. Meanwhile, $2$ does not divide $15$ since $15 = 3 \cdot 5$.

Similarly, $2$ divides every even number (isn't that how we define even numbers? as numbers having a factor of $2$?).

But if we define even numbers as numbers having a factor of $2$ in them, then odd numbers are numbers without a factor of $2$ in them. That means $2$ is not a factor of any odd number, which means $2$ does not "divide" any odd number.

So, when we say that $100 < x < 200$ and $2 \not | x$ (i.e., $2$ does not divide $x$), we are saying $x$ is between 100 and 200, and $2$ does not divide $x$, i.e., $x$ is not even, i.e., $x$ is odd.

By the way, here is an extra question for you. Could the answer also be $\{ x | 100 \leq x \leq 200 \text{ and } 2 \not | x \}$? Why or why not?

$\endgroup$
2
  • $\begingroup$ Very nice detailed answer. Thanks a lot! To your extra question: although it could be written as <=, its technically not correct because both 100 and 200 are even numbers, meaning that they will never be reached. (Is this correct? :P) $\endgroup$
    – zuc0001
    Commented Mar 9, 2015 at 3:52
  • $\begingroup$ @zuc0001 Well, I think you have the right idea. It is still correct to write the set with $\leq$, because the $100$ and $200$ might satisfy first condition, but as you suggested, they don't satisfy the second condition since both are even. So they wouldn't be in the set anyway. $\endgroup$
    – layman
    Commented Mar 9, 2015 at 3:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .