2
$\begingroup$

A^C

I'm trying to figure out why the answer would be everything but 0,2,3 as the universal set is 0-9..

I don't know how to explain that the answer would be {1,4,5,6,7,8,9}

LETTER G

  • 0
    The answer to what part??? – 2017-01-13
  • 0
    @ThomasGrubb Sorry, letter g – 2017-01-13
  • 1
    The definition of $A^c$ is $U-A$ – 2017-01-13
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    so U excluding A meaning everything in A must be gone – 2017-01-13

2 Answers 2

1

$A^C$ = $U$ - $A$ = Elements which are in set $U$ but not in set $A$.

Hence, $A^C$ = $\{$$1, 4, 5, 6, 7, 8, 9$$\}$

1

The Universal Set is the set that contains everything. So just imagine we are given a a universal set $S$; therefore all we know to exists is in the set $S$. That being said, any set $A$ $\subseteq$ $S$ we know that $A^{C} \cup A = U$ (this should be clear, but if it is not try to prove it). Now that we have this rule it should be clear that if:

$U$ = $\{$$0, 1, 2, \dots, 9$$\}$, A = $\{$$0, 2, 3$$\}$

Then $A^C$ = $U$ \ $A$ = $\{$$1, 4, 5, 6, 7, 8, 9$$\}$

Of course a more formal proof exists, but I hope I have explained intuively why it is.