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Is the correct answer to this:

$A \times B = \{(2a),(3a),(4a),(2b),(3b),(4b)\}$



  • $\begingroup$ Given the sets $A$ and $B$, how is $A\ast B$ defined? $\endgroup$ Nov 22 '15 at 12:48
  • $\begingroup$ @JoãoVictorBateliRomão the title of this question is all I am provided with in the test paper I am working on $\endgroup$ Nov 22 '15 at 12:50
  • $\begingroup$ Isn't the notatin $A\times B$? $\endgroup$ Nov 22 '15 at 12:52
  • $\begingroup$ Yes, apologies. My head stays in programming language arithmetic operators. $\endgroup$ Nov 22 '15 at 12:53
  • $\begingroup$ For one thing, the elements of $A\times B$ are tuples, not products; e.g. $(2,a)$, not $(2a)$. $\endgroup$ Nov 22 '15 at 20:44

The terms of the ordered pairs must be separated by a comma. The right answer is $\{(2,a),(3,a),(4,a),(2,b),(3,b),(4,b)\}$.


From comments I see you meant $\times$, which is the standard notation for Cartesian product defined for two (or more) sets.

So for sets $A,B$ $$ A \times B = \{(a\color{red}{,}b) \mid \forall a \in A, \forall b \in B\}. $$

So, your answer is missing the commas.

I've added two notes about Cartesian product, since it seems to me, like this concept is new to you. Hopefully it will make it rather clear than confusing for you.

Note A

To be more precise, what does $(a,b)$ mean? It means that the pair $a,b$ is ordered, that is, if $a\neq b$, then $(a,b) \neq (b,a)$. But in set theory, what means ordered, e.g. $\{1, 2, 3\} = \{3, 1, 2\}$. There are probably many (infinitely many, I suppose) ways to define $(a, b)$ so that is satisfies the given condition, the standard way I met (called Kuratowski, thanks wiki) is the following $$ (a,b) := \{\{a\},\{a,b\}\}. $$ And it is true if $a\neq b$, that $$ \{\{a\},\{a,b\}\} = (a,b) \neq (b,a) = \{\{b\},\{a,b\}\}. $$

Note B

The second note is about associativity of $\times$. The question is, whether it is true that $A \times (B \times C) = (A \times B) \times C$, for $A,B,C$ sets. Suppose now $a \in A, b \in B, c \in C$, and look at both sides. To be precise, on the left we get $$ (a,(b,c)), $$ because first we do $B \times C$, and we get $(b,c)$ and then $(a,(b,c))$. However, on the right side we get $$ ((a,b),c). $$ We see, that $(a,(b,c)) \neq ((a,b),c)$, not even for $a=b=c$. But we see, that there is natural correspondence between those two, meaning, that there is natural bijection between $A\times(B\times C)$ and $(A\times B)\times C$, which is given by $$ (x,(y,z)) \mapsto ((x,y),z). $$ And therefore we can see $A\times(B\times C)$ and $(A\times B)\times C$ as the same thing, and instead writing $(a,(b,c)), ((a,b),c)$ we write $(a,b,c)$ . (We used three sets $A,B,C$, but this can be done for infinitely many.)


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