Assignment 22

#$&*

course MTH 151

10:34, 4/8/14

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

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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.

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Your solution:

2 * 5 = 10 - 10 * 2 = 20 / 3 = 6 remainder 2, because 6 * 3 = 18, with 2 left over. 2 @ 5 = 2.

3 * 8 = 24 - 24 * 2 = 48 / 3 = 16 remainder 0. 3 @ 8 = 0.

7 * 13 = 91 - 91 * 2 = 182 / 3 = 60 remainder 2, because 60 * 3 = 180, with 2 left over. 7 @ 13 = 2.

confidence rating #$&*: 3

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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.

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Self-critique (if necessary):

I’m finally feeling like I’m beginning to understand how the remainders work.

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Self-critique Rating: 3

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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}.

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Your solution:

2 @ 2 = 2 * 2 = 4 * 2 = 8 / 3 = 2 remainder 2, because 3 * 2 = 6. 2 @ 2 = 2.

2 @ 3 = 2 * 3 = 6 * 2 = 12 / 3 = 4 remainder 0. 2 @ 3 = 0.

2 @ 4 = 2 * 4 = 8 * 2 = 16 / 3 = 5 remainder 1, because 3 * 5 = 15. 2 @ 4 = 1.

3 @ 3 = 3 * 3 = 9 * 2 = 18 / 3 = 6 remainder 0. 3 @ 3 = 0.

3 @ 4 = 3 * 4 = 12 * 2 = 24 / 3 = 8 remainder 0. 3 @ 4 = 0.

4 @ 4 = 4 * 4 = 16 * 2 = 32 / 3 = 10 remainder 2, because 10 * 3= 30. 4 @ 4 = 2.

Values: 2 3 4

Results (below)

2 2 0 1

3 0 0 0

4 1 0 2

confidence rating #$&*: 2

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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.

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Self-critique (if necessary):

Still not quite sure that my table is correct.

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Self-critique Rating: 3

@&

I believe your table is completely correct, though for clarity you should have labeled the top line as well:

@ 2 3 4

*@

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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?

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Your solution:

The results for the last problem were 0, 1, and 2, because those were the remainders.

confidence rating #$&*: 2

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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}.

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Self-critique (if necessary):

ok

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Self-critique Rating: 3

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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}?

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Your solution:

No, because there are different numbers for both.

confidence rating #$&*: 2

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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}.

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Self-critique (if necessary):

ok

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Self-critique Rating: 3

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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}?

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Your solution:

All three sets would work, because the results from {2 3 4} all include the numbers 0, 1, and 2.

confidence rating #$&*: 3

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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.

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Self-critique (if necessary):

ok

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Self-critique Rating: 3

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Question: `q006. How can we tell by looking at the table whether the operation is closed?

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Your solution:

If the numbers in the table all match the results of the set.

confidence rating #$&*: 2

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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.

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Self-critique (if necessary):

ok

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Self-critique Rating: 3

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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?

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Your solution:

It would not, because either way, the sum would still be the same. For example, 2 * 3 and 3 * 2 both have the same answer, even if they are arranged differently.

confidence rating #$&*: 3

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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.

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Self-critique (if necessary):

ok

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Self-critique Rating: 3

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Question: `q008. Does the operation of subtraction of whole numbers have the commutative property?

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Your solution:

No, because the order of the numbers can make a difference in the sum. For example, 5 - 1 = 4, but 1 - 5 = -4.

confidence rating #$&*: 3

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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.

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Self-critique (if necessary):

ok

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Self-critique Rating: 3

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Question: `q009. Is the operation of subtraction closed on the set of whole numbers?

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Your solution:

It would depend on what the whole numbers were, and as long as they could be subtracted without getting a negative number.

confidence rating #$&*: 2

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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.

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Self-critique (if necessary):

ok

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Self-critique Rating: 3

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Question: `q010. Is the operation of addition closed and commutative on the set of whole numbers?

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Your solution:

I would assume that it would be closed, as long as all whole numbers are used.

confidence rating #$&*: 2

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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.

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Self-critique (if necessary):

ok

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Self-critique Rating: 3

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Question: `q011. When we multiply a number by 1, what must be our result?

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Your solution:

When any number is multiplied by 1, the answer is the same as the number being multiplied. For example, 3 * 1 = 3, 4 * 1 = 4, and so on.

confidence rating #$&*: 3

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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.

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Self-critique (if necessary):

ok

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Self-critique Rating: 3

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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}?

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Your solution:

I’m still not understanding identities completely, but to me it seems like both sets would have identities of 0 and 2.

confidence rating #$&*: 1

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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.

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Self-critique (if necessary):

ok

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Self-critique Rating: 3

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Question: `q013. Does the set of whole numbers on the operation of addition have an identity?

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Your solution:

It would depend on which whole numbers are being added.

confidence rating #$&*: 1

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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.

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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}?

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Your solution:

3 could be combined with 2 to result in 1, 2 can be combined with 3 to result in 1.

& 1 2 3 4

1 1 * *

2 4 1 *

3 1 4

4 * * 1

@&

The question asked for the complete table. The rest of the *'s should also be filled in.

It seems clear that you could do this correctly, but let me know if you aren't confident.

*@

confidence rating #$&*:

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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?

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Your solution:

Not really understanding the identities, but because you said 0 is the identity for whole numbers, because the sum is always the same when multiplied, I would assume it would be 0.

@&

0 is the identify for addition, since when you add zero to any number n you get n.

0 is not the identity for multiplication, because 0 * n = 0, not n.

The identify for multiplication is 1, since 1 * n = n for any whole number.

*@

@&

As to the current question, what number would you have to multiply by 2 to get the multiplicative identity 1? That number would be the multiplicative inverse of 2.

Is this number a whole number? If it is, then the multiplicative inverse exists in the set of whole numbers. If not, then this is not the case.

*@

3 and 0.

confidence rating #$&*:

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Self-critique Rating: ok

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Question: `q013. Does the set of whole numbers on the operation of addition have an identity?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

It would depend on which whole numbers are being added.

confidence rating #$&*: 1

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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.

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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}?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

3 could be combined with 2 to result in 1, 2 can be combined with 3 to result in 1.

& 1 2 3 4

1 1 * *

2 4 1 *

3 1 4

4 * * 1

@&

The question asked for the complete table. The rest of the *'s should also be filled in.

It seems clear that you could do this correctly, but let me know if you aren't confident.

*@

confidence rating #$&*:

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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?

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Your solution:

Not really understanding the identities, but because you said 0 is the identity for whole numbers, because the sum is always the same when multiplied, I would assume it would be 0.

@&

0 is the identify for addition, since when you add zero to any number n you get n.

0 is not the identity for multiplication, because 0 * n = 0, not n.

The identify for multiplication is 1, since 1 * n = n for any whole number.

*@

@&

As to the current question, what number would you have to multiply by 2 to get the multiplicative identity 1? That number would be the multiplicative inverse of 2.

Is this number a whole number? If it is, then the multiplicative inverse exists in the set of whole numbers. If not, then this is not the case.

*@

3 and 0.

confidence rating #$&*:

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Self-critique Rating: ok

#*&!

&#Good responses. See my notes and let me know if you have questions. &#