2 Replies - 949 Views - Last Post: 17 September 2013 - 07:40 AM

#1 deprosun   User is offline

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Is there an easier way to count for this problem?

Posted 17 September 2013 - 07:05 AM

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?
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#2 sepp2k   User is offline

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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.
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#3 deprosun   User is offline

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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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