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I came across this question in my assignment...

Question 1. What would be the intersection of $B=\{x:x= 2n-1,n \in N\}$ and $D=\{x:x$ is a prime number$\}.$

My approach was pretty simple.

I first listed out the elements of both of these sets separately in roster form which looked like this.

$B=\{1,3,5,7,9,11,13,15...\}$ which would continue on and on forever. And

$D=\{2,3,5,7,11,13,17...\}$ which would continue on and on forever.

Next, I listed out the elements that were present both in sets $B$ and $D$ and concluded the answer by writing...$$B\cap D= \{3,5,7,11,13\}$$

When I looked at the solution, I found it to be...$$D-\{2\}$$

I do not understand why the solution is $D-\{2\}$. Does it mean that $B \cap D$ essentially contains all the elements of D except $2$? But how is it possible as we can see the apparent elements that are both present in $B$ and $D$?

Am I missing out something here? Or the question/answer is incorrect itself?

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    $\begingroup$ It means remove the element $2$ from $D$. Generally it is written as $D \setminus \{2\}$. $\endgroup$ – SJ. Nov 6 '17 at 13:13
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The solution is correct: $B$ is the set of positive odd integers. $D$ is the set of prime numbers. All prime numbers are odd except $2$, so $B\cap D=D-\{2\}$.

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  • $\begingroup$ But in set $B$ 9 isn't odd neither is 15. $\endgroup$ – Saksham Sharma Nov 6 '17 at 13:16
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    $\begingroup$ Yes, so they are not in the intersection. But all of the prime numbers (elements of $D$) apart from $2$ are odd, so they are in the intersection. $\endgroup$ – A. Goodier Nov 6 '17 at 13:17
  • $\begingroup$ $B$ is kind of acting as a universe here.Right? $\endgroup$ – Saksham Sharma Nov 6 '17 at 13:19
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    $\begingroup$ Yes, in the sense that the intersection is contained in $B$ $\endgroup$ – A. Goodier Nov 6 '17 at 13:21
  • $\begingroup$ Aren't positive odd integers always positive? Why have you written "positive odd integers"? $\endgroup$ – Saksham Sharma Nov 6 '17 at 13:29

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