Determine if each of the following sets is closed under addition. \( \begin{array}{ll}\text { a) } L=\{0\} & \text { c) } N=\{6,9,11\} \\ \text { b) } S=\{0,6,12,18,24 \ldots\} & \text { d) }\{x \mid x \in \text { integers, } x>10\}\end{array} \) \( \begin{array}{l}\text { a) } L=\{0\} \text { is closed under addition. } \\ \text { True } \\ \text { False }\end{array} \)
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L is closed under addition because the sum of 0 and 0 is still 0. So yes, \( \{0\} \) can only produce 0 when you add its elements. For S, which consists of multiples of 6, adding any two elements from this set (say, \( 6m \) and \( 6n \)) will also yield a multiple of 6, confirming that S is closed under addition. Thus, it's true for S as well. Now, for the other sets: N consists of the numbers 6, 9, and 11. However, adding 6 and 9 gives us 15, which is not in the set. So, N is not closed under addition. Lastly, for the set of integers greater than 10, if you take any two integers greater than 10, their sum will also be greater than 10. Hence, this set is closed under addition. In summary: - a) True - b) True - c) False - d) True
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