The ternary relation TImes(i, j, k) to be the relation corresponding to the predicate "i times j equals k". This relation on the base set {0, 1, 2, 3, 4}. How many elements does the relation have?
Is there an easier way to count the number of elements rather than "doing it all out"? If so, how?
2 Replies - 949 Views - Last Post: 17 September 2013 - 07:40 AM
#1
Is there an easier way to count for this problem?
Posted 17 September 2013 - 07:05 AM
Replies To: Is there an easier way to count for this problem?
#2
Re: Is there an easier way to count for this problem?
Posted 17 September 2013 - 07:24 AM
If you have a fixed i and a fixed j, how many k can exist at most such that Times(i,j,k)? Will there always be at least one such k? If not, under which conditions will there be no k for a given i and j?
Answering those questions should lead you to some shortcuts you can use instead going through all 53 combinations.
Answering those questions should lead you to some shortcuts you can use instead going through all 53 combinations.
#3
Re: Is there an easier way to count for this problem?
Posted 17 September 2013 - 07:40 AM
Oh, I had this question in my Probability class. You're right! thanks:)
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