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It might be a stupid question, but can you give me some example of multiplication not being closed in some set?

I could find a case in "addition"(e.g., a set of odd numbers is not closed under addition) but am struggling to find an example for multiplication.

Rowing0914
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  • For instance, a set of imaginary numbers? – Rowing0914 Dec 28 '21 at 23:51
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    The set of all real numbers between $-2$ and $2$ – FShrike Dec 28 '21 at 23:53
  • I also found that the irrational numbers are not closed too.. – Rowing0914 Dec 28 '21 at 23:54
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    The singleton set containing 2 – Andrea B. Dec 28 '21 at 23:54
  • I am going to post another question but this question came to my mind when I was thinking about the axioms of rings! I feel that people often assume that binary operations are closed under multiplication in ring definition such as https://math.stackexchange.com/questions/1361466/should-a-ring-be-closed-under-multiplication. But i feel like this is vacuous... so I was thinking when multiplication is not closed – Rowing0914 Dec 28 '21 at 23:58
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    @AmejiB. That was exactly what I was going to put, but you beat me to it (2 it?). – Mark Saving Dec 29 '21 at 00:02
  • Take ANY example of set (with at least two members) that IS closed, choose two members $a$ and $b$, then REMOVE $ab$. – Randall Dec 29 '21 at 00:20

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Consider the set of negative integers, this set has the property that if you multiply any two negative integers you will never get another negative integer.

Steven Creech
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@Rowing0914 gave a nice example where multiplication acts to produce a different type of object.

Consider the set of all prime numbers $p_i$. By definition, none of these share any common factors.

Annika
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