If $a = \{1,2,3\}$ and $b = \{a,b,c\},\;$ FIND $\;n(a\times b)$

Or is it impossible to multiply these sets?

What will be the answer?

  • 1
    $\begingroup$ I'm assuming, given your last post, that you mean the cardinality of $a \times b$, to be $n(a\times b)$: that number is a scalar, the product of the cardinality of a with that of b. The set $a \times b$ is a set of ordered pairs. $\endgroup$ – Namaste May 22 '13 at 4:11
  • $\begingroup$ @amWhy I just want to make sure the answer can never be a whole number right? $\endgroup$ – MethodManX May 22 '13 at 4:13
  • $\begingroup$ @MethodManX Is that $n$ before $(a\times b)$ a typo? $\endgroup$ – Maazul May 22 '13 at 4:14
  • $\begingroup$ @Maazul no. That's in the question. $\endgroup$ – MethodManX May 22 '13 at 4:15
  • $\begingroup$ There is certainly something wrong with the expression $n(a\times b) = \{1,a\}, \{2,a\}, \{3,a\}...$ It should have been $n(a\times b) = n\{\{1,a\}, \{2,a\}, \{3,a\} \{1,b\}, \{2,b\}, \{3,b\} \{1,c\}, \{2,c\}, \{3,c\}\}$ $\endgroup$ – Maazul May 22 '13 at 4:16

Let's use capital letters for sets: so let $$A = \{1, 2, 3\},\;\;\text{ and} \;\; B = \{a, b, c\}$$ and $n(A) = |A| = 3,\;n(B) = |B| = 3$.

Then the Cartesian-product $\,A\times B\;$ is a set of all ordered pairs $$A \times B= \{(a_i, b_j)\mid a_i \in A, b_j \in B\}.$$

In this case, $$A \times B = \{(1, a), (1, b), (1, c), (2, a),(2, b), (2, c), (3, a), (3, b), (3, c)\}$$

In general, for two sets $P, Q$, $$\;|P\times Q| = |P| \times |Q|$$

So, if $n(A \times B) = |A\times B|,\;$ then $n(A \times B) = n(A) \times n(B) = 3 \times 3 = 9$

  • $\begingroup$ so was I correct when I said -> {1,a}, {2,a}, {3,a}, {2,b}... etc? $\endgroup$ – MethodManX May 22 '13 at 4:06
  • $\begingroup$ @MethodManX If I saw those pairs with out context, it would not be obvious that they indeed run through all ordered pairs $\endgroup$ – Ethan May 22 '13 at 4:07
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    $\begingroup$ I think that Cartesian product is a far more common nomenclature. Cross-product is something I've never seen used outside of multidimensional calculus. $\endgroup$ – Cameron Buie May 22 '13 at 4:07
  • $\begingroup$ And ordered pairs are represented with $(\;,\;)$. Given your sets, you'd have 9 ordered pairs. $\endgroup$ – Namaste May 22 '13 at 4:08
  • $\begingroup$ @CameronBuie It seems he used cross product instead of cartesian product, but his definition is the same. $\endgroup$ – Ethan May 22 '13 at 4:08

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