course mth 151 {虥|܀˟{assignment #004
......!!!!!!!!...................................
08:27:04 `qNote that there are 4 questions in this assignment. `q001. From the collection of letters a, b, c, d, e, how many smaller collections having at least one element may be formed?
......!!!!!!!!...................................
RESPONSE --> (a) (b) (c) (d) (e) ab, ac, ad, ae, bc, bd, be, cd, ce, de acd, ace, abe, abd, abc, ade, bcd, bce, bde, cde abcd, abde, acde, bcde, abce 5 - 1 element 10 - 2 element 10 - 3 element 5 - 4 element Total is 30 different smaller collections confidence assessment: 2
.................................................
......!!!!!!!!...................................
08:27:59 We will list the original collection by placing its elements between braces: { a, b, c, d, e }. The collection {a, b, c, d} is a smaller collection obtained by eliminating e from the original collection. Similarly we can eliminate d or c or b or a to get the 4-element collections {a, b, c, e}, {a, b, d, e}, { a, c, d, e} and {b, c, d, e}. Alternatively we could simply include either a or b or c or d or e in a 1-element collection, obtaining {a}, {b}, {c}, {d} and {e}. It should be clear that these are the only ways to form collections of 1 or 4 elements. To form a collection of 2 elements we could include a and one other element, obtaining { a, b}, { a, c }, { a, d } and { a, e }. Or we could include b and one other element (excluding a, since we already have the collection { a, b } which is identical to the collection { b, a } since it has exactly the same elements). We obtain { b, c }, { b, d } and { b, e }. {}Or we could include c and one other element (other than a or b, since these have already been listed) to obtain { c, d } and { c, e }. Finally we could include d and the only other element left, e, to get { d, e}. This gives us a complete listing of the 10 sets we can form with 2 of the original elements. This leaves us the 3-element sets, which can be formed by excluding the 2-element sets. Working in reverse order, we can exclude { d, e } to get { a, b, c }, or { c, e } to get { a, b, d }, etc.. The remaining sets we get in this fashion are { a, b, e}, { a, c, d }, { a, c, e}, { a, d, e}, { b, c, d}, {b, c, e}, {b, d, e}, {c, d, e}. We thus have 10 three-element sets. The total number of smaller sets containing at least one element is therefore 5 + 5 + 10 + 10 = 30.
......!!!!!!!!...................................
RESPONSE --> The collection (a,b,c,d,e) has 30 smaller element sets possible self critique assessment: 2
.................................................
......!!!!!!!!...................................
08:33:23 `q002. A one-to-one correspondence between two sets is a rule that associates each element of the each with exactly one element of the other. A natural one-to-one correspondence between the sets { a, b, c } and { 1, 2, 3 } would be to associate a with 1, b with 2, c with 3. This correspondence might be represented as [ a <--> 1, b <--> 2, c <--> 3 ]. This isn't the only possible one-to-one correspondence between these sets. Another might be [ a <--> 2, b <--> 1, c <--> 3 ]. In each case, every element of each set is associated with exactly one element of the other. Another correspondence between the sets might be [ a <--> 3, b<-->2, c<-->3 ]. This correspondence is not one-to-one. In what way does it fail to be a one-to-one correspondence (remember that a one-to-one correspondence is one in which every element of each set is associated with exactly one element of the other).
......!!!!!!!!...................................
RESPONSE --> because each set has to associated with exactly one element of the other and in this set a is associated to 3 and c is also associated to 3. confidence assessment: 2
.................................................
......!!!!!!!!...................................
08:34:19 [ a <--> 3, b<-->2, c<-->3 ] fails to be a one-to-one correspondence for two reasons. In the first place, 3 is associated with a and with c, and every element of each set is to be associated with exactly one element of the other. 3 is associated with two elements of the other set. It also fails because the element 1 of the second set is not associated with anything in the first set.
......!!!!!!!!...................................
RESPONSE --> also fails because the element 1 of the second set is not associated with anything in the first set. I did not think of this in this way. self critique assessment: 3
.................................................
......!!!!!!!!...................................
08:41:43 `q003. There are four possible one-to-one correspondences between the sets {a, b, c} and {1, 2, 3} which were not described in the preceding exercise. One of them would be [ a <--> 3, b <--> 2, c <--> 1 ]. What are the other three possible one-to-one correspondences?
......!!!!!!!!...................................
RESPONSE --> [a<->2, b<->1, c<->3] [a<->1, b<->3, c<->2] [a<->1, b<->2, c<->3] confidence assessment: 2
.................................................
......!!!!!!!!...................................
08:45:08 If we designate the correspondence [ a <--> 1, b <--> 2, c <--> 3 ] as the '123' correspondence, [a <--> 2, b <--> 1, c <--> 3 ] as the '213' correspondence and [a <--> 3, b <--> 2, c <--> 1 ] as the '321' correspondence, in each case listing the numbers associated with a, b, c in that order, we see that the remaining three correspondences could be designated 132, 231 and 312. These correspondences could of course be written out as [ a <--> 1, b <--> 3, c <--> 2 ], [ a <--> 2, b <--> 3, c <--> 1 ] and [ a <--> 3, b <--> 1, c <--> 2 ]. Note that 123, 132, 213, 231, 312, 321 represent the six ways of rearranging the digits 1, 2, 3 into a 3-digit number, listed in increasing order.
......!!!!!!!!...................................
RESPONSE --> Ok I'm looking at it in a different perspective, I can see what I done wrong self critique assessment: 2
.................................................
......!!!!!!!!...................................
08:46:26 `q004. Explain why it is not possible to put the sets { a, b, c} and {1, 2, 3, 4} into a one-to-one correspondence.
......!!!!!!!!...................................
RESPONSE --> Because in one-to-one correspondence you would have to take away one of the numbers or add a letter to the letter set. You have to have the same number of elements on each set. confidence assessment: 2
.................................................
......!!!!!!!!...................................
08:46:54 One set has 3 elements and the other has 4 elements. A 1-to-1 correspondence has to match each element of each set with exactly one element of the other. It would not be possible to find four different elements of the first set to match with the four elements of the second.
......!!!!!!!!...................................
RESPONSE --> ok I got that one self critique assessment: 2
.................................................
......!!!!!!!!...................................
08:53:29 `q001. Note that there are 8 questions in this assignment. The set { 1, 2, 3, ... } consists of the numbers 1, 2, 3, etc.. The etc. has no end. This set consists of the familiar counting numbers, which most of us have long known to be unending. This is one example of an infinite set. Another is the set of even positive numbers { 2, 4, 6, ... }. This set is also infinite. There is an obvious one-to-one correspondence between these sets. This correspondence could be written as [ 1 <--> 2, 2 <--> 4, 3 <--> 6, ... ], where the ... indicates as before that the pattern should be clear and that it continues forever. Give a one-to-one correspondence between the sets { 1, 2, 3, ... } and the set { 1, 3, 5, ... } of odd numbers.
......!!!!!!!!...................................
RESPONSE --> [1<->1,2<->3, 3<->5,...] confidence assessment: 2
.................................................
......!!!!!!!!...................................
08:53:47 This correspondence can be written [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ].
......!!!!!!!!...................................
RESPONSE --> That's what I got. self critique assessment: 2
.................................................
......!!!!!!!!...................................
08:56:56 `q002. Writing [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ] for the correspondence between { 1, 2, 3, ... } and { 1, 3, 5, ... } isn't bad, but the pattern here might be a bit less clear to the reader than the correspondence [ 1 <--> 2, 2 <--> 4, 3 <--> 6, ... ] given for { 1, 2, 3, ... } and { 2, 4, 6, ... }. That is because in the latter case it is clear that we are simply doubling the numbers in the first set to get the numbers in the second. It might not be quite as clear exactly what the rule is in the correspondence [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ], except that we know we are pairing the numbers in the two sets in order. Without explicitly stating the rule in a form as clear as the doubling rule, we can't be quite as sure that our rule really works. How might we state the rule for the correspondence [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ] as clearly as the 'double-the-first-number' rule for [ 1 <--> 2, 2 <--> 4, 3 <--> 6, ... ]?
......!!!!!!!!...................................
RESPONSE --> I was using the match the first element in the first set with the first element in the second set. confidence assessment: 2
.................................................
......!!!!!!!!...................................
08:57:32 We might say something like 'choose the next odd number'. That wouldn't be too bad. Even clearer would be to note that the numbers 1, 3, 5, ... are each 1 less than the 'double-the-counting-number' numbers 2, 4, 6. So our rule could be the 'double-the-first-number-and-subtract-1' rule. If we double each of the numbers 1, 2, 3, ... and subtract 1, we get 1, 3, 5, ... .
......!!!!!!!!...................................
RESPONSE --> ok there are different ways to approach this self critique assessment: 2
.................................................
......!!!!!!!!...................................
09:00:03 `q003. The 'double-the-number' rule for the correspondence [ 1 <--> 2, 2 <--> 4, 3 <--> 6, ... ] could be made even clearer. First we we let n stand for the nth number of the set {1, 2, 3, ... }, like 10 stands for the 10th number, 187 stands for the 187th number, so whatever it is and long as n is a counting number, n stands for the nth counting number. Then we note that the correspondence always associates n with 2n, so the correspondence could be written0 [ 1 <--> 2, 2 <--> 4, 3 <--> 6, ... , n <--> 2n, ... ]. This tells us that whatever counting number n we choose, we will associate it with its double 2n. Since we know that any even number is a double of the counting number, of the form 2n, this rule also tells us what each even number is associated with. So we can argue very specifically that this rule is indeed a 1-to-1 correspondence. In terms of n, how would we write the rule for the correspondence [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ]?
......!!!!!!!!...................................
RESPONSE --> [1<->1, 2<->3, 3<->5, ...,n<->3n,...} confidence assessment: 2
.................................................
......!!!!!!!!...................................
09:00:33 The rule for this correspondence is 'double and subtract 1', so n would be associated with 2n - 1. The correspondence would thus be [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... , n <--> 2n-1, ... ]. Note how this gives a definite formula for the rule, removing all ambiguity. No doubt is left as to how to figure which number goes with which.
......!!!!!!!!...................................
RESPONSE --> Ok I understand now. I would use 2n-1 self critique assessment: 2
.................................................
......!!!!!!!!...................................
09:02:17 `q004. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 5, 10, 15, ... }.
......!!!!!!!!...................................
RESPONSE --> [1<->5, 2<->10, 3<->15, ... n<->5n,...] confidence assessment: 2
.................................................
......!!!!!!!!...................................
09:03:14 It should be clear that each element of the second set is 5 times as great as the corresponding element the first set. The rule would therefore be n <--> 5n.
......!!!!!!!!...................................
RESPONSE --> n<-> 5n because the second set is 5 times greater as the corresponding element. self critique assessment: 2
.................................................
......!!!!!!!!...................................
09:04:53 `q005. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 7, 12, 17, ... }.
......!!!!!!!!...................................
RESPONSE --> [1<->7, 2<->12, 3<->17, ... n<-> 5n+2, ...] confidence assessment: 2
.................................................
......!!!!!!!!..................................."