course math 151 Not sure which is which I did so many at one time.
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13:52:25 `q001. There are no questions for this assignment.
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KĪg¸ϛӇ Student Name: assignment #007
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13:02:04
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13:35:03 `q001. Note that there are 7 questions in this assignment. Sketch three points A, B and C forming an equilateral triangle on a piece of paper, with point A at the lower left-hand corner, point B at the lower right-hand corner and point C at the top. Sketch the segments AB and AC. Now double the lengths of AB and AC, and place a point at each of the endpoints of these segments. Connect these new endpoints to form a new equilateral triangle. Two sides of this triangle will have three points marked while the new side will only have its two endpoints marked. Fix that by marking that middle point, so all three sides of your new triangle are marked the same. How many marked points were there in the original triangle, and how many are there in the new triangle?
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RESPONSE --> I was answering a question in assignment 7, and the answer was 13 more, so the next number was going to 13 added to the last one. A funny noise happened and the screen went to this question I don't think that I had finished assignment 7???? Please help!!!1 The answer to this q001 is 3 in the original triangle, and 2 for the new triangle, because it shares two points with the original triangle.
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13:37:23 The original triangle had the three points A, B and C. When you extended the two sides you marked the new endpoints, then you marked the point in the middle of the third side. So you've got 6 points marked. Click on 'Next Picture' to see the construction. The original points A, B and C are shown in red. The line segments from A to B and from A to C have been extended in green and points marked at the ends of these segments. The new endpoints have been connected to form the third side of a larger triangle, and an equally spaced point has been constructed at the midpoint of that side. Your figure should contain the three original points, plus the three points added when the new side was completed.
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RESPONSE --> I know what I did wrong- I went down with my extension of segement AC. I see the correct way now from the picture.
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13:40:09 You added the two new endpoints when you extended the sides. You then should have marked two new points on the new third side, so that each side contains 4 points including its endpoints. Your figure will now contain 10 marked points.
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. I think that I have added 4 more points to my drawing, giving a totla of 10 points, but I am confused. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.
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13:41:16 `q003. Continue the process for another step-extend each side by a distance equal to the original point-to-point distance. How many points do you have in the new triangle?
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RESPONSE --> I now have a mess on my paper, but I think that I just added 3 new points????
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13:43:06 You will add an endpoint to each newly extended side, so each of the new sides will contain 5 points. You will then have to add 3 equally spaced points to the new side, giving you a total of 13 points on the new triangle. In addition there are two marked points inside the triangle, for a total of 15 points. Click on 'Next Picture' to see the construction. The line segments along two sides of the triangle have again been extended and points marked at the ends of these segments. The new endpoints have been connected to form the third side of a larger triangle, and equally spaced points have been constructed along that side.
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RESPONSE --> I got the 3 new points correct, and now I see the 5 points all together on the picture on the screen.
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13:48:03 `q004. Continue the process for one more step. How many points do you have in the new triangle?
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RESPONSE --> I drew minw, and there are 6 points in the new triangle.
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13:48:31 You will add an endpoint to each newly extended side, so each of the new sides will contain 6 points. You will then have to add 4 equally spaced points to the new side, giving you a total of 15 points on the new triangle. There are also 5 marked points inside the triangle for a total of 21 marked points.
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RESPONSE --> Yeah!!! Praise the Lord!!! I get it.
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13:48:51 `q005. The sequence of marked points is 3, 6, 10, 15, 21. What do expect will be the next number in this sequence?
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RESPONSE --> should be 28.
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13:49:29 The differences between these numbers are 3, 4, 5, 6. The next difference, according to this pattern, should be 7, which would make the next number 28.
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RESPONSE --> I might make an A+ in this yet!!! Ha ha
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13:50:47 `q006. How can you tell, in terms of the process you used to construct these triangles, that the next number should be 7 greater?
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RESPONSE --> Because with each new triangle it increased by 1. You started with 3, then 4, then 5 points, then 6, and it should be 7 fot the next extension.
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13:50:58 When you extend the triangle again, you will add two new endpoints and each side will now have 7 points. The 7 points on the new triangle will be all of the new points.
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13:51:37 `q007. How do you know this sequence will continue in this manner?
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RESPONSE --> Ths shape of the triangle is staying the same and the other properties are staying the same.
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13:51:42 Each time you extend the triangle, each side increases by 1. All the new marked points are on the new side, so the total number of marked points will increase by 1 more than with the previous extension.
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nҤּyx Student Name: assignment #010
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ZɁy\} Student Name: assignment #010
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Ɓ~xqȱε_ Student Name: assignment #005 005. Infinite Sets
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12:33:15 `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.
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RESPONSE --> The obvoius ono-to-one correspondence between the familiar unending counting numbers and the unending odd numbers could be written as [ 1,1 <-->,3, 2, 5,5,<--> 7, 6....]
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12:33:47 This correspondence can be written [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ].
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12:34:29 This correspondence can be written [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ].
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RESPONSE --> So, it should be that way.
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12:39:35 `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, ... ]?
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RESPONSE --> We could state that we are taking into consideration the first set of counting numbers, while also doubling the odd numbers to represent the second set.
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12:41:41 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, ... .
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RESPONSE --> That is true, they are each the next odd number, or 1 less than the even ""double the counting numbers"". You could further say that they are 1 more than the previous ""double the counting number.""
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12:44:44 `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, ... ]?
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RESPONSE --> [1 <--> 1, 2 <--> 3, 3<--> 5, n <--> 2n,...
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12:45:35 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.
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RESPONSE --> That is right, I was wrong. I forgot to subtract 1 to get the odd numbers.
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12:47:34 `q004. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 5, 10, 15, ... }.
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RESPONSE --> [1 <--> 1, 2<--> 3, 3 <--> 5, n, n+5 <-->...
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12:48:10 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.
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RESPONSE --> I understand, I just put it as n +5.
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12:49:14 `q005. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 7, 12, 17, ... }.
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RESPONSE --> This would be the n +5, always increasing by 5. 1<--> 2, 2<--> 3, 3<-->7, n +5...
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12:50:18 First we note that the numbers in the second set are going up by 5 each time. This indicates that we will probably somehow have to use 5n in our formula. Just plain 5n gives us 5, 10, 15, ... . It's easy to see that these numbers are each 2 less than the numbers 7, 12, 17, ... . So if we add 2 to 5n we get the numbers we want. Thus the rule is n <--> 5n+2, or in a bit more detail [ 1 <--> 7, 2 <--> 12, 3 <--> 17, ..., n <--> 5n+2, ... ].
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RESPONSE --> Yes, that is correct also, but from 7 to 12 is 5, from 12 to 17 is 5.
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12:54:27 `q006. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 3, 10, 17, ... }.
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RESPONSE --> 5n +7
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12:55:14 The numbers in the second set are going up by 7 each time, so we will probably 7n in our formula. Just plain 7n gives us 7, 14, 21, ... . These numbers are each 4 greater than the numbers 3, 10, 17, ... . So if we subtract 4 from 7n we get the numbers we want. Thus the rule is n <--> 7n-4, or [ 1 <--> 3, 2 <--> 10, 3 <--> 17, ..., n <--> 7n-4, ... ].
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RESPONSE --> I see that now, but did not think of that before I answered the question
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12:56:06 `q007. Surprisingly, the numbers { 1, 2, 3, ... } can be put into correspondence with the set of all integer-divided-by-integer fractions (the set of all such fractions is called the set of Rational Numbers). This set would include the fractions 1/2, 1/3, 1/4, ..., as well as fractions like 38237 / 819872 and 232987 / 3. It is a bit surprising that this set could be in 1-1 correspondence with the counting numbers, because just the fractions 1/2, 1/3, 1/4, ... can be put into one-to-one correspondence with the set {1, 2, 3, ... }, and these fractions are less than a drop in the bucket among all possible fractions. These fractions all have numerator 1. The set will also contain the fractions of numerator 2: 2/1, 2/2, 2/3, 2/4, ... . And the fractions with numerator 3: 3/1, 3/2, 3/3, 3/4, ... . We could go on, but the idea should be clear. It certainly seems like there should be more fractions than counting numbers. But it isn't so, as you will see in the lectures and the text. Give a one-to-one correspondence between just the fractions 1/2, 1/3, 1/4, ... and the counting numbers {1, 2, 3, ... }.
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12:58:13 The correspondence would be [ 1 <--> 1/2, 2 <--> 1/3, 3 <--> 1/4, ... ]. The denominator of the fraction is always 1 greater than the counting number. So if the counting number is n, the denominator the corresponding fraction is n + 1. We would therefore write the correspondence as n <--> 1 / (n+1), or in a bit more detail [ 1 <--> 1/2, 2 <--> 1/3, 3 <--> 1/4, ... , n <--> 1/(n+1), ... ].
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RESPONSE --> I think it would be [ 1 <--> 1/2, 2 <--> 1/3, 3 <--> 1/4, 4....]
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13:00:32 We could alternate between the two sets, using odd numbers with fractions whose numerator is 1 and even numbers with fractions whose numerator is 2. The correspondence would be [ 1 <--> 1/2, 2 <--> 2/2, 3 <--> 1/3, 4 <--> 2/3, 5 <--> 1/4, 6 <--> 2/4, ... ]. It would be a little bit tricky, but not all that difficult, to write this rule in terms of n. However, we won't go into that here.
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. I thought that I was answering q008???? Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. ގPFҞݫz Student Name: assignment #006 006. Sequences and Patterns
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13:01:25 `q001. Note that there are 6 questions in this assignment. Find the likely next element of the sequence 1, 2, 4, 7, 11, ... .
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RESPONSE --> the next likely is 16
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13:07:21 `q002. Find the likely next two elements of the sequence 1, 2, 4, 8, 15, 26, ... .
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. My answer to q002 was correct. The answer to the next question is to add 5, then add 6, so the next two numbers would be 16, and 22.I hope that I am getting these anawers n the right place. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.
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13:09:58 The difference between 1 and 2 is 1; the difference between 2 and 4 is 2, the difference between 4 and 8 is 4; the difference between 8 and 15 is 7; the difference between 15 and 26 is 11. The differences form the sequence 1, 2, 4, 7, 11, ... . As seen in the preceding problem the differences of this sequence are 1, 2, 3, 4, ... . We would expect the next two differences of this last sequence to be 5 and 6, which would extend the sequence 1, 2, 4, 7, 11, ... to 1, 2, 4, 7, 11, 16, 22, ... . If this is the continuation of the sequence of differences for the original sequence 1, 2, 4, 8, 15, 26, ... then the next two differences of this sequence would be 16 , giving us 26 + 16 = 42 as the next element, and 22, giving us 42 + 26 = 68 as the next element. So the original sequence would continue as 1, 2, 4, 8, 15, 26, 42, 68, ... .
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RESPONSE --> Yeah!! I got those correct
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13:11:36 `q003. What would be the likely next element in the sequence 1, 2, 4, 8, ... . It is understood that while this sequence starts off the same as that in the preceding exercise, it is not the same. The next element is not 15, and the pattern of the sequence is different than the pattern of the preceding.
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RESPONSE --> I think the next one is 16. Because 1 + 1=2, 2+2=4, 4+4=8, then 8+8 would be 16.
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13:12:05 One obvious pattern for this sequence is that each number is doubled to get the next. If this pattern continues then the sequence would continue by doubling 8 to get 16. The sequence would therefore be 1, 2, 4, 8, 16, ... .
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RESPONSE --> Yeah!! I'm on a roll, just don't want to fall off!!! Ha
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13:24:11 `q004. There are two important types of patterns for sequences, one being the pattern defined by the differences between the numbers of the sequence, the other being the pattern defined by the ratios of the numbers of the sequence. In the preceding sequence 1, 2, 4, 8, 16, ..., the ratios were 2/1 = 2; 4/2 = 2; 8/4 = 2; 16/8 = 2. The sequence of ratios for 1, 2, 4, 8, 16, ..., is thus 2, 2, 2, 2, a constant sequence. Find the sequence of ratios for the sequence 32, 48, 72, 108, ... , and use your result to estimate the next number and sequence.
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RESPONSE --> The ratio is 3/2, so the next number would be 162.
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13:26:06 The ratios are 48/32 = 1.5; 72 / 48 = 1.5; 108/72 = 1.5, so the sequence of ratios is 1.5, 1.5, 1.5, 1.5, ... . The next number the sequence should probably therefore be 108 * 1.5 = 162.
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RESPONSE --> I got the right answer, although I compared it 3: 2. Do you agree that it is also 3: 2 as well as 1 1/2?
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ꝣ[y]Ƨ Student Name: assignment #015
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18:57:33 `q001. There are 6 questions in this set. The proposition p -> q is true unless p is true and q is false. Construct the truth table for this proposition.
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18:58:23
The proposition will be true in every case except the one where p is true and q is false, which is the TF case. The truth table therefore reads as follows:
p q p -> q
T T T
T F F
F T T
F F T
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RESPONSE -->
I do not understand this ?????
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19:01:03
`q002. Reason out, then construct a truth table for the proposition ~p -> q.
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~p->q.
T T= T
T F = T
F T = T
F F = F
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19:01:46
This proposition will be false in the T -> F case where ~p is true and q is false. Since ~p is true, p must be false so this must be the FT case. The truth table will contain lines for p, q, ~p and ~p -> q. We therefore get
p q ~p ~p -> q
T T F T since (F -> T) is T
T F F T since (F -> F) is T
F T F T since (T -> T) is T
F F T T since (T -> F) is F
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RESPONSE -->
I do understand this, now
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19:03:33
To evaluate the expression we must first evaluate p ^ ~q and ~p -> ~q.
p ^ ~q is evaluated by first determining the values of p and ~q. If p is false and q true, then ~q is false. Thus both p and ~q are false, and p ^ ~q is false.
~p -> ~q will be false if ~p is true and ~q is false; otherwise it will be true. In the FT case p is false to ~p is true, and q is true so ~q is false. Thus it is indeed the case the ~p -> ~q is false.
(p ^ ~q) U (~p -> ~q ) will be false if (p ^ ~q) and (~p -> ~q ) are both false, and will otherwise be true. In the case of the FT truth values we have seen that both (p ^ ~q) and (~p -> ~q ) are false, so that (p ^ ~q) U (~p -> ~q ) is false.
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RESPONSE -->
Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.
I tried to respond to 'q003- but this screen came up instead,
Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.
mhf}v̓S˝
Student Name:
assignment #010
010. No questions at present.
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13:53:39
`q001. There are no questions for this assignment.
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RESPONSE -->
I like this one with NO Questions!!!!!!!!
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ΗˑP勻ΛGՑ
Student Name:
assignment #012
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18:45:22
`q001. Note that there are 4 questions in this assignment.
Suppose I tell you 'If it rains today, I'll give you $100.' Under which of the following circumstances can you claim that I was not telling the truth?
1. It rains and I give you $100.
2. It rains and I don't give you $100.
3. It doesn't rain and I give you $100.
4. I doesn't rain and I don't give you $100.
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RESPONSE -->
I think that #2 is the answer. You can't control the rain!!!!
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18:46:24
`q002. Suppose that tell you 'It will rain today and I will give you $100'. Under which of the following circumstances can you claim that I was not telling the truth?
1. It rains and I give you $100.
2. It rains and I don't give you $100.
3. It doesn't rain and I give you $100.
4. I doesn't rain and I don't give you $100.
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RESPONSE -->
Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.
I choose # 2 which was correct. I said that you could not control the rain!!!
Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.
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18:47:29
It should be clear that situation #1 completely fulfills the conditions of my statement. Both of the things that I say will happen do happen.
In situation #2, it rains but you don't get the $100. I said two things were going to happen and one of them didn't. In that case you would have to say that I wasn't telling truth.
In situation #3, again one of the things I say is going to happen does but the other doesn't, so again you would have to say that I wasn't telling truth.
In situation #4, neither of the things I say will happen does and certainly it would have to be said that I wasn't telling truth.
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RESPONSE -->
Yes, that is true for # 3, and # 4, but you cannot control the rain!!!
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18:52:21
`q003. Suppose that tell you 'It will rain today or I will give you $100, but not both'. Under which of the following circumstances can you claim that I was not telling the truth?
1. It rains and I give you $100.
2. It rains and I don't give you $100.
3. It doesn't rain and I give you $100.
4. I doesn't rain and I don't give you $100.
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RESPONSE -->
This question ( q003) is not worded incorrectly???? It reads ""Suppose that tell you 'It will rain today....""
My answer is # 4.
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18:53:27
In situations 2 and 3, one of the things happens and the other doesn't, so you would not be able to say that I wasn't telling the truth. However in situation 1, both things happen, which I said wouldn't be the case; and in situation 4 neither thing happens. In both of these situations you would have to say that I was not telling truth.
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RESPONSE -->
I didn't choose # 1 along with # 4. I just choose # 4, but I do see where you would not be telling the truth in # 1.
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18:54:15
`q004. Suppose that tell you 'It will rain today or I will give you $100'. Under which of the following circumstances can you claim that I was not telling the truth?
1. It rains and I give you $100.
2. It rains and I don't give you $100.
3. It doesn't rain and I give you $100.
4. I doesn't rain and I don't give you $100.
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RESPONSE -->
Maybe, the same as the last question- the answers are # 1, and # 4.
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18:55:40
At first this might seem to be the same as the preceding problem. But in the preceding problem we specifically said '... but not both.' In this case that qualification was not made. Therefore we have regard the statement as true as long as at least one of the conditions is fulfilled. This is certainly the case for situation 1: both conditions are true we can certainly say that at least one is true. So in situation #1 we have to regard the present statement as true. So situation #1 would not be included among those in which I could be accused of not telling the truth.
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RESPONSE -->
I was wrong- I did not understand that one of the conditions were fulfilled.
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