If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
021. There are
thirteen questions in this assignment.
Numeration and Operations
Question: `q001. If we define the
operation @ on two numbers x and y by
x @ y = remainder when the product x * y is multiplied by 2 then divided by 3, then find the following:
2 @ 5
3 @ 8
7 @ 13.
Your solution:
Confidence Assessment:
Given Solution:
By the definition, 2 @ 5 is the remainder when the product
2 * 5 is doubled then divided by 3. We
start with 2 * 5 = 10, then double that to get 20, then divide by three. 20 / 3 = 6 with remainder 2. So 2 @ 5 = 2.
We follow the same procedure to find 3 @ 8. We get 3 * 8 = 24, then double that to get 48, then divide by three. 48 / 3 = 12 with remainder 0. So 3 @ 8 = 0.
Following the same procedure to find 7 @ `2, we get 7 * 13 = 91, then double that to get 182, then divide by three. 182 / 3 = 60 with remainder 2. So 7 @ 13 = 2.
Self-critique (if necessary):
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Question: `q002. If we define the @ operation from the previous exercise just on the set {5, 6, 7} , we can use the same process as in the preceding solution to get 5 @ 5 = 2, 5 @ 6 = 0, 5 @ 7 = 1, 6 @ 5 = 0, 6 @ 6 = 0, 6 @ 7 = 0, 7 @ 5 = 1, 7 @ 6 = 0 and 7 @ 7 = 2. We can put these results in a table as follows:
@ 5 6 7
5 2 0 1
6 0 0 0
7 1 0 2.
Make a table for the @ operation, with the
values of x and y restricted to the
set {2, 3, 4}.
Your solution:
Confidence Assessment:
Given Solution:
Using the same process as in the solution to the preceding
problem we find that 2 @ 2 = 2, 2 @ 3 = 0, 2 @ 4 = 1, 3 @ 2 = 3 @ 3 = 3 @ 4 =
0, 4 @ 2 = 1, 4 @ 3 = 0 and 4 @ 4 = 2. The table is therefore
@ 2 3 4
2 2 0 1
3 0 0 0
4 1 0 2.
Self-critique (if necessary):
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Question: `q003. All the x and y
values for the table in the preceding problem came from the set {2, 3, 4}. From what set are the results x @ y taken?
Your solution:
Confidence Assessment:
Given Solution:
The results of the operation x @ y, which ultimately
consist of the remainder when some number is divided by 3, must all be
division-by-3 remainders. The only
possible remainders we can have when dividing by three are 0, 1 or 2. Thus all the results of the operation x @ y
are members of the set {0, 1, 2}.
Self-critique (if necessary):
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Question: `q004. Are the results
of the operation x @ y on the set {2, 3, 4} all members of the set {2, 3, 4}?
Your solution:
Confidence Assessment:
Given Solution:
The possible results of the operation, whose table is
@ 2 3 4
2 1 0 2
3 0 0 0
4 2 0 1
are seen from the table to be 0, 1 and 2. Of these possible results, only 2 is a member
of the set {2,3,4}. So it is not the
case that all the results come from the set {2, 3, 4}.
Self-critique (if necessary):
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Question: `q005. Since the operation x @ y on the set {2, 3, 4} can have results which are not members of the set, we say that the @operation is not closed on the set {2, 3, 4}.
Is the @ operation closed on the set S = {0, 1, 2}?
Is the @ operation closed on the set T = {0, 2}?
Is the @ operation closed on the set
R = {1, 2}?
Your solution:
Confidence Assessment:
Given Solution:
When operating on the set S = {0, 1, 2} the operation must as always give one of the division-by-3 remainders 0, 1 or 2. Thus the results are all in the set S.
When operating on the set T = {0, 2} the possible results are 0 @ 0 = 0 @ 2 = 2 @ 0 = 0, or 2 @ 2 = 2. Since the possible results are 0 and 2, both of which are in T, all the results come from the set T on which we are operating, and the operation is closed on the set T.
When operating on the set R = {1, 2} the possible results
are 1 @ 1 = 2, 1 @ 2 = 2 @ 1 = 1, or 2 @ 2 = 2. Since the possible results are 1 and 2 and both are in the set R, we can
say that all the results come from the set R on which we are operating, and the
operation is indeed closed on the set R.
Self-critique (if necessary):
Self-critique Rating:
Question: `q006. How can we tell
by looking at the table whether the operation is closed?
Your solution:
Confidence Assessment:
Given Solution:
If all the numbers in the table come from the far
left-hand column of the table--e.g., the column underneath the @ in the tables
given above, which lists all the members of the set being operated on --then
all the results of the operation are in that set and the operation is therefore
closed.
Self-critique (if necessary):
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Question: `q007. When calculating
x @ y for two numbers x and y, does it make a difference whether we calculate x
@ y or y @ x?
Your solution:
Confidence Assessment:
Given Solution:
Since the first step in calculating x @ y is to multiply x * y, it would make no difference whether we multiplied x * y or y * x. So in the first step it makes no difference whether we calculate x @ y or y @ x. Since all we do after that is double our result and calculate the remainder when dividing by 3, the order of x and y won't make a difference there either.
We conclude that for this operation x @ y must always equal y @ x.
This property of the operation is called the commutative property. Speaking roughly, an operation is commutative if order doesn't matter.
Self-critique (if necessary):
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Question: `q008. Does the
operation of subtraction of whole numbers have the commutative property?
Your solution:
Confidence Assessment:
Given Solution:
Subtraction of whole numbers does not have the commutative
property, because it is not true for most whole numbers x and y that x - y = y
- x. For example, 5 - 3 = 2 while 3 - 5
= -2. So for subtraction order usually
does matter, and the operation is not commutative.
Self-critique (if necessary):
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Question: `q009. Is the operation
of subtraction closed on the set of whole numbers?
Your solution:
Confidence Assessment:
Given Solution:
Whole numbers are the numbers in the set {0, 1, 2, 3, ...
}. If we subtract a smaller number from
a larger, we will again get a whole number. However if we subtract a smaller number from the larger, we will get a
negative result, which is not a whole number. Since it is possible to subtract two numbers in the set and to get a
result that is not in the set, the operation is not closed.
Self-critique (if necessary):
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Question: `q010. Is the operation
of addition closed and commutative on the set of whole numbers?
Your solution:
Confidence Assessment:
Given Solution:
When we add two numbers x and y it doesn't matter which one we add to which--it doesn't matter whether we do x + y or y + x--so order doesn't matter and we can say that the operation is commutative.
And if we add two whole numbers, which must both be at least
0, we get a whole number which is at least 0. So the operation is also closed.
Self-critique (if necessary):
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Question: `q011. When we multiply
a number by 1, what must be our result?
Your solution:
Confidence Assessment:
Given Solution:
Any number multiplied by 1 will give us the same
number. Any number is unchanged if we
multiply it by 1. That is 10 * 1 = 10,
or -37.27 * 1 = -37.27, or 72 * 1 = 72. Multiplication by 1 does not change any number.
Self-critique (if necessary):
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Question: `q012. A number which does not change any number with which it is combined using a certain operation is called the identity for the operation. As we saw in the preceding exercise, the number 1 is the identity for the operation of multiplication on real numbers.
Does the operation @ (which was defined in preceding exercises by x @ y = remainder when x * y is doubled and divided by 3) have an identity on the set {0, 1, 2}?
Does @
have an identity on the set {0, 2}?
Your solution:
Confidence Assessment:
Given Solution:
The table for @ on {0, 1, 2) is
@ 0 2 1
0 0 0 0
1 0 2 1
2 0 1 2.
We see from the row
across from 2 that 2 * 0 = 0, 2 * 1 = 1 and 2 * 2 = 2. We also see from the column beneath 2 that 0
* 2 = 0, 1 * 2 = 1 and 2 * 2 = 2. Thus,
no matter how we combine 2 with other numbers in the set {0, 1, 2}, we don't
change those other numbers. That is, for
any x in the set, 2 @ x = x @ 2 = x.
Therefore @ does indeed have identity 2 on the set {0, 1,
2}.
On the set {0, 1} the number 2 isn't included. Since 0 * 1 = 0, not 1, the number 0 can't be
the identity. Since 1 * 1 = 2, not 1,
the number 1 can't be the identity. Both
0 and 1 at least sometimes change the number they operate with, and the
identity can't do this.
Self-critique (if necessary):
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Question: `q013. Does the set of
whole numbers on the operation of addition have an identity?
Your solution:
Confidence Assessment:
Given Solution:
The identity must be a number that doesn't change the number with which it is combined. The number 0 has this property. Whenever we add to 0 is what we get. 0 doesn't change the number it is combined with under the operation of addition. For any x, 0 + x = x and x + 0 = x. Therefore we can say that 0 is the identity for addition on the set of whole numbers.
Question: `q014. Let the & operation be defined as follows: x & y is the remainder obtained when x is multiplied by y, then divided by 5. The table below is for the & operation on the set {1, 2, 3, 4}. The table is partially filled in.
& 1 2 3 4
1 1 * * *
2 * 4 * *
3 * * 4 *
4 * * * 1
We note that if 4 is combined with 4 the result is 1.
What number, if any, would be combined with 2 to obtain the result 1?
What number, if any, would be combined with 3 to obtain the result 1?
Give the complete table.
In what ways would the nature of the table change if the operation was applied to the set {1, 2, 3, 4, 5}?
Your solution:
Confidence Rating:
Question: `q015. The operation of multiplication on the whole numbers has identity 1.
Is there a whole number which when multiplied by 2 gives us the identity?
Can you two different whole numbers which when multiplied gives us the identity?
Your solution:
Confidence Rating:
Self-critique Rating: