Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Say we have the $S$ which is the set of all compositions of n >= 0 with an odd num. parts.

Define $S_1$ to be the set of all compositions of n >= 0 with an odd num. of parts where at least one part is <= 9

Define $S_2$ to be the set of all compositions of n >= 0 with an odd num. of parts where each part is >= 10

Is $S_1$ union $S_2$ a partition of $S$? I can't seem to wrap my head around it...

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Yes they are. These sets are complements of each other.

If not all of the elements are greater than or equal to 10, then at least 1 element is less than or equal to 9.

If none of the elements are at most 9, then all of the elements are at least 10.

share|improve this answer
    
Hmm... $\{S_1,S_2\}$ would be a partition of $S$, but not $S_1 \cup S_2=S$. –  Douglas S. Stones Jan 31 '13 at 0:39
    
@DouglasS.Stones Can you explain why the union is not $S$? –  Calvin Lin Jan 31 '13 at 0:51
    
I'm saying the union is $S$, and that's the problem. To illustrate, $\{1\} \cup \{2,3\}=\{1,2,3\}$ is not a partition of $\{1,2,3\}$, but $\{\{1\},\{2,3\}\}$ is. –  Douglas S. Stones Jan 31 '13 at 0:55
    
@DouglasS.Stones Ah. I caught that too, and was thinking that it's just strange language on OP's part. Though, isn't $S$ a partition of $S$? That's why I felt it would be yes off yes. –  Calvin Lin Jan 31 '13 at 0:57
    
I'm guessing your answer is right, and it's a bug in the question. (I wouldn't say $S$ is a partition of $S$.) –  Douglas S. Stones Jan 31 '13 at 1:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.