Find the union and the intersection of the two sets and where
Rest are similar.
List all the possible arrangements of the four letters m,a,r and y. Let be the collection of the arrangements in which y is in the last position. Let be the collection of the arrangements in which m is in the first position. Find the union and intersection of and .
Solution: The arrangements are: mary, mayr, mray, mrya, myra, myar, amry, amyr, aymr, ayrm, arym, army, ryma, ryam, ramy, raym, rmya, rmay, yrma, yram, yamr, yarm, ymra, ymar.
. The rest is obvious.
If a sequence of sets is such that the sequence is said to be a nondecreasing sequence. Give an example of this kind of sequence of sets.
Solution: Let where n = 1, 2, 3...
If a sequence of sets is such that the sequence is said to be a nonincreasing sequence. Give an example of this kind of sequence of sets.
Solution: Let where n=1,2,3...
If are sets such that is defined as the union . Find if
If are sets such that is defined as the intersection . Find if
For every one dimensional set A let where . If and , find and .
Solution: and using the formulae of sum of a geometric series.
For every one dimensional set A for which the integral exists, let where , otherwise let be undefined. If and , find and .
Solution: can be found by integrating f(x) from to , as integral over a set of measure zero is 0 and can be found by integrating f(x) from 0 to 1.
Let for every two dimensional set A for which the integral exists; otherwise let be undefined. If and , find and .
Solution: , is zero since it represents volume of a sheet and if we introduce polar coordinates. This evaluates to .
To join a certain club, a person must be either a statistician or a mathematician or both. Of the 25 members in this club, 19 are statisticians and 16 are mathematicians. How many persons in the club are both a statistician and a mathematician?
Soluion: The number of persons which are statisticians or mathematicians = number of statisticians + number of mathematicians - number of persons which are both statisticians and mathematicians. This can be proved directly using properties of the counting measure. The general proof is as follows: Let A and B be two finite sets. Endow the natural numbers with the counting measure which we shall denote by n. We shall denote union by + and intersection by . and the context will make it clear whether we mean actual addition, multiplication or union, intersection. Now A-B (Here - represents the relative complement, AB and B-A are disjoint and as n is sigma additive so
- n(A)=n(A-B)+n(AB) which implies n(A-B)=n(A)-n(AB)
- n(B)=n(B-A)+n(AB) which implies n(B-A)=n(B)-n(AB)
Substituting n(A-B) and n(B-A) from (2) and (3) into (1) gives us n(A+B)=n(A)+n(B)-n(AB). So the answer is 10.
After a hard fought football game, it was reported that of the 11 starting players, 8 hurt a hip, 6 hurt an arm, 5 hurt a knee, 3 hurt both a hip and an arm, 2 hurt both a hip and a knee, 1 hurt both an arm and a knee, and no one hurt all three. Comment on the accuracy of the report.
Solution: We shall first establish that in the language of the previous problem:
Now if we let A, B and C to be the set of players with injured hips, knees and arms respectively, we would find the contradiction on substituting the appropriate terms in the above formula. Hence the report is not accurate.