assignment 21 qa

#$&*

course mth151

12/7/14 12:20

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.

026. Modular systems

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Question: `q001. There are 7 questions in this assignment.

We defined an operation as follows: x * y (mod 4) = remainder when x * y is divided by 4.

Find 3 * 9 (mod 4); 7 * 12 (mod 4) and 11 * 13 (mod 4).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

3*9 (mod4)

3*9 = 27

27/4 = 6 with a remainder of 3

therefore 3*9(mod4) = 3

7*12(mod4)

7*12=84

84/4 leaves a remainder of 0

7*12(mod4) = 0

11*13(mod4)

11*13 = 143

143/4 leaves a remainder of 3

11*13(mod4) = 3

confidence rating #$&*:3

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Given Solution:

3 * 9 (mod 4) is the remainder when 3 * 9 is divided by 4. Since 3 * 9 = 27 and 27 / 4 leaves remainder 3, we see

that 3 * 9 (mod 4) = 3.

7 * 12 (mod 4) is the remainder when 7 * 12 is divided by 4. Since 7 * 12 = 84 and 84 / 4 leaves remainder 0, we

see that 7 * 12 (mod 4) = 0.

11 * 13 (mod 4) is the remainder when 11 * 13 is divided by 4. Since 11 * 13 = 143 and 143 / 4 leaves remainder

3,we see that 11 * 13 (mod 4) = 3.

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Self-critique (if necessary):ok

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Self-critique Rating:ok

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Question: `q002. Make a table for the x * y mod 4 operation, which we will call '* mod 4', operating on the set

{0, 1, 2, 3}.

Determine which of the properties, including commutativity, associative, identity, inverse and closure properties,

are properties of this operation.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

* mod 4

0 1 2 3

0 0 0 0 0

1 0 1 2 3

2 0 2 0 2

3 0 3 2 1

if x is 0 then anything multiplied by x is 0 and therefore when divided by 4 has a remainder of 0

1*1=1 and when divided by 4 leaves a remainder of 1

1*2=2 when divided by 4 leaves remainder of 2

1*3=3 when divided by 4 leaves a remainder of 3

2*2=4 when divided by 4 leaves a remainder of 0

2*3=6 when divided by 4 leaves a remainder of 2

3*3=6 when divided by 4 leaves a remainder of 1

there is symmetry accross the diagonal line so the operation has the commutative property

we can also see that it has the closer property since in the set {0,1,2,3} the remainder will always be 0, 1, 2,

or 3

since 0 and 2 do no have an additive inverse in the table then the operation does not have the inverse property

this also comtains the identity property since the number 1, when provided, with any other element provide that

same element. 1*2=2, 1*0=0

multuiplication is associative and therefore the operation has the associative property

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

Whatever x is, 0 * x = x * 0 = 0, which when divided by 4 leaves remainder 0.

Whatever x is, 1 * x = x * 1 = x, and if x is in the set {0, 1, 2, 3} we have get remainder x when dividing by 4

(e.g., 4 divides into 0, 1, 2 or 3 zero times, leaving that number as the remainder) and x mod 4 = x. From this we

can see that 1 is the identity for this operation.

Multiplying 0, 1, 2, and 3 by 2 we get 0, 2, 4, and 6, which when divided by 4 leave remainders 0, 2, 0 and 2,

respectively.

Multiplying 0, 1, 2, and 3 by 2 we get 0, 3, 6, and 9, which when divided by 4 leave remainders 0, 3, 2 and 1,

respectively.

The table for this operation is therefore

* mod 4 0 1 2 3

0 0 0 0 0

1 0 1 2 3

2 0 2 0 2

3 0 3 2 1

We note that this operation does contain identity 1, but since neither 0 nor 2 can be combined with any of the

elements of the set to give us the identity, the operation on this set does not have the inverse property.

We do see from the symmetry of the table about the main diagonal that it has the commutative property, which we

could in any event have concluded from the fact that multiplication is commutative so that the product we get

before calculating the remainder is independent of the order of the two numbers.

In a similar matter we can reason that the operation is associative.

The operation is also closed, since the remainder upon dividing by 4 must always be 0, 1, 2 or 3 and hence in the

set {0, 1, 2, 3}.

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Self-critique (if necessary):ok

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Self-critique Rating:ok

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Question: `q003. Repeat the preceding exercise for the operation x * y mod 5, defined to give the remainder when x

* y is divided by 5, on the set {1, 2, 3, 4}. Determine which of the properties are exhibited by this operation.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

* mod 5

1 2 3 4

1 1 2 3 4

2 2 4 1 3

3 3 1 4 2

4 4 3 2 1

1*1 divided by 5 leaves 1

1*2 divided by 5 leaves 2

1*3 leaves 3

1*4 leaves 4

2*2 leaves 4

2*3 leaves 1

3*3 leaves 4

3*4 leaves 2

4*4 leaves 1

we can see that if we drew a diagonal line across the table there would be symmetry on either side so it has

commutative property

as before we can say that the operation has associative property because of the associative properties of

multiplication

the results stay within the set {1, 2, 3, 4} so it has the closure property

it has the identity 1 preoperty because 1 can be multiplied with any other of the numbers to reslut in that same

number, ex. 1*1=1, 1*4=4,...

we can see from the table that each number appears once in each column and each row and therefore has the inverse

property

confidence rating #$&*:3

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Given Solution:

First we might wish to do a couple of example calculations to get familiar with the operation. For example:

2 * 3 mod 5 = 6, which when divided by 5 gives us remainder 1.

3 * 4 mod 5 = 12 which when divided by 5 gives us remainder 2.

2 * 4 mod 5 = 8 which when divided by 5 gives us remainder 3.

The table is

* mod 5 1 2 3 4

1 1 2 3 4

2 2 4 1 3

3 3 1 4 2

4 4 3 2 1

We immediately see that all the results are in the set {1, 2, 3, 4}, so that the operation is closed.

This operation has identity 1, as we can see from the row and the column across from and beneath 1.

We easily see from the table that the identity appears exactly once in each row and in each column, which assures

us that the operation has the inverse property. Specifically we see that 1 * 1 mod 5 = 1 so that 1 is its own

inverse, that 2 * 3 mod 5 = 1 so that 2 and 3 are inverses, and that 4 * 4 mod 5 = 1, so that 4 is its own

inverse.

The associativity and commutativity of the operation follow from the associative and commutative properties of

multiplication on real numbers, as discussed in the preceding problem.

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Self-critique (if necessary):ok

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Self-critique Rating:ok

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Question: `q004. The equation 3x + 7 = 9 (mod 5) has an integer solution for x = 0, 1, 2, 3 or 4. Which value of x

is a solution to this equation?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

3x + 7 =9 (mod5) is 3x + 7 -9 = 0 (mod5)

3x-2=0 (mod5)

this means that we must solve 3x-2 and then divide that by 5 and get a remainder of 0

if x=0 then 3(0)-2=-2

x=1 then 3(1)-2=1

x=2 then 3(2)-2=4

x=3 then 3(3)-2=7

x=4 then 3(4)-2=10

the only one of these that is a multiple of 5 and would therefor give us a remainder of 0 would be x=4

3(4)+7=9 (mod5)

confidence rating #$&*:3

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Given Solution:

3x + 7 = 9 (mod 4) means that 3x + 7 - 9 = 0 (mod 5) so 3x - 2 = 0 (mod 5). If 3x - 2 = 0 (mod 5) then when we

divide 3x - 2 by 5 we should get remainder 0. So we substitute the different possible values for x into the

expression 3x - 2 until we get a number of which when divided by 5 gives us remainder 0.

If x = 0 then 3x - 2 = -2, and -2 (mod 5) = 3 (if you don't understand why -2 mod 5 = 3, think of the 5-hour clock

in the text; but for now it should be obvious that -2 is not a multiple of 5 so that you cannot get remainder 0

when dividing -2 by 5).

If x = 1 then 3x - 2= 1, and 1 (mod 5) = 1.

If x = 2 then 3x - 2= 4, and 4 (mod 5) = 4.

If x = 3 then 3x - 2= 7, and 7 (mod 5) = 2.

If x = 4 then 3x - 2= 10, and 10 (mod 5) = 0.

Thus x = 4 is a solution to the equation 3x + 7 = 9 (mod 5).

STUDENT QUESTION

What is that 5-hour clock?

INSTRUCTOR RESPONSE

A 5-hour clock is depicted below:

We label the usual 12:00 position 0. Moving 1 hour at a time in the clockwise direction we encounter hours 1, 2,

3 and 4 before finally returning after 5 hours to the 0 position.

To calculate 3 + 4 on this clock, we start at 0 and move 3 hours, which puts us at position 0. Then we move 4

more hours, which takes us through positions 4, 0 and 1 before ending up at 2. Thus, on this clock, 3 + 4 = 2.

It's easy to see how this gives us a picture for the calculation (3 + 4) mod 5, and why the result is the

remainder when we divide 3 + 4 by 5.

To calculate 3 * 4 on this clock, we would start at 0 and move 4 hours to position 4, then 4 more hours to

position 3, then 4 more hours to position 2. Thus on this clock 3 * 4 = 2, and this gives us a picture of (3 *

4) mod 5. It's easy to see that we have moved 3 * 4 = 12 hours, and that where we end up is the remainder when 12

divided by 5.

To divide 3 by 4 on this clock, we observe that for ordinary base-10 division, (3 / 4) * 4 = 3. So we ask what

number multiplied by 4 on this clock will give us 3. If we start at 0 and move 4 hours around the clock we end up

at 4; if we move 4 more hours we end up at 3. Thus 4 * 2 = 3, and we can therefore say that 2 = 3/4. That is, 3

divided by 4 is 2.

To solve the equation 3x - 2 = 0:

Next we try different numbers for x:

If x = 1 we get 3 * 1 - 2 = 1.

If x = 2 we get 3 * 2 - 2 = 4.

If x = 3 we get 3 * 3 - 2 = 7. Since 7 hours on this clock takes us from our starting position at 0 to

position 2, we conclude that when x = 3 our result is 3 * 3 - 2 = 2.

If x = 4 we get 3 * 4 - 2 = 10. Since 10 hours on this clock takes us from our starting position at 0,

twice around the clock and back to position 0, we conclude that when x = 4 our result is 3 * 4 - 2 = 0. So x = 4

is a solution to our equation 3 * x - 2 = 0 (mod 5).

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Self-critique (if necessary):ok

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Self-critique Rating:ok

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Question: `q005. You see that x = 4 is a solution to the equation 3x + 7 = 9 (mod 5). One of the numbers x = 5, 6,

7, 8, 9 is also a solution. Which one is it?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

if x=5 then 3(5)-2=13

x=6 then 3(6)-2=16

x=7 then 3(7)-2=19

x=8 then 3(8)-2=22

x=9 then 3(9)-2=25

the only one that is a multiple of 5 and would therefore give us a remainder of 0 is x=9

confidence rating #$&*:3

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Given Solution:

We recall that 3x + 7 = 9 (mod 5) is equivalent to 3x - 2 = 0 (mod 5). We evaluate 3x - 2 (mod 5) for x = 5, 6, 7,

8 and 9 and we find that the results are 3, 1, 4, 2, and 0. So x = 9 is our next solution.

We might also note that the series of results 3, 1, 4, 2, 0 is the same as the series we got for x = 0, 1, 2, 3,

4. Our results therefore seem to indicate a repeating pattern in which the remainder 0 occurs every fifth number

starting with 4. This is in fact what happens, and you might wish to think about why this happens. However, you

should in a case remember that this is what happens.

In general when we have an equation of the form A x + B = C (mod n), integer solutions happen at intervals of n.

for some values of A, B and C integer solutions can also occur at shorter intervals, but they always do occur at

intervals of n.

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Self-critique (if necessary):ok

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Self-critique Rating:ok

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Question: `q006. What are the first five positive values of x which solve the equation 3x + 7 = 9 (mod 5) of the

preceding problem?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

in the previous problems we noted a pattern with the remainders being 3,1,4,2,0 in respects to x being 0,1,2,3,4

and again with x being 5,6,7,8,9

from this we can gather that the remainder will be 0 every fifth number starting from 0, so x= 4, 9, 14, 19, 24

confidence rating #$&*:3

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Given Solution:

We just saw that x = 4 and x = 9 are solutions, and we saw that because we are solving an equation mod 5, the

solutions have to occur at intervals of 5. Thus the first five solutions are x = 4, 9, 14, 19 and 24.

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Question: `q007. Can you find a solution to the equation 3x + 7 = 9 (mod 6)? Show how you reason out your

answer.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

from previous questions we know that 3x+7=9 (mod6) will be 3x-2=0 (mod6)

i cannot find a solution to this equation. everythings that going to be a multiple of 6 is also going to be a

multiple of 3 and then you subtract 2 and you will always have a remainder

for ex...

x=4 3(4)-2= 12-2=10 which leaves you with a remainder of 4

x=36 3(36)-2= 108-2=106 which again leaves a remainder of 4

and the list can go on and on

confidence rating #$&*:

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Self-critique Rating:ok

@&

Excellent reasoning.

*@

"

Self-critique (if necessary):

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Self-critique rating:

"

Self-critique (if necessary):

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Self-critique rating:

#*&!

&#This looks good. See my notes. Let me know if you have any questions. &#

assignment 21 qa

#$&*

course mth151

12/7 1:35

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.

026. `query 26

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Question: `q4.4.40 (was 5.4.12) What is [ (10+7) * (5+3) ] mod 10 and how did you get your result?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

[(10+7) * (5+3)] (mod10) the answer is the remainder when [(10+7) * (5+3)] is divided by 10

(17*8) (mod10)

136/10

13 with a remainder of 6

therefore [(10+7) * (5+3)] (mod10) is 6

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`a** x mod 10 is the remainder when x is divided by 10.

So [ (10+7) * (5 + 3) ] mod 10 = ( 17 * 8) mod 10 = 136 mod 10 = 6, since 136 / 10 leaves remainder 6. **

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Self-critique (if necessary):ok

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Self-critique Rating:ok

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Question: `qquery 4.4.20 2 / 3 on 5-hour clock

What is 2 divided by 3 on a 5-hour clock, and how did you obtain this result?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

2/3=x is equivilant to 3x=2

on a five hour clock you must turn the hand 4 times to get to 2

3(4)=2

therefore 2/3=4

confidence rating #$&*:3

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Given Solution:

`a** You have to turn this one into a multiplication problem to get the correct answer.

In decimal numbers, for example, 60 / 20 = 3 because 3 * 20 = 60.

Whatever you get when you divide 2 by 3, when you multiply it by 3 you get 2. That is, if 2 / 3 = x, then 3 x = 2.

So what would you multiply by 3 to get 2 on a 5-hour clock?

It turns out that 3 * 4 = 2. So it follows that 2 / 3 = 4. **

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Self-critique (if necessary):ok

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Question: `qquery 4.4.42 (3 - 27) mod 5

What is (3 - 27) mod 5, and how did you reason out your result?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(3-27) (mod5) = -24(mod5)

on the clock go backwards around the clock 5 times and then go forward one to put you at -24

it would put you at 1 on the clcok

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`a** (3-27) mod 5 = -24 mod 5.

You would go all the way around around backwards 5 complete times to get -25, then move forward 1 to get to -24. That would put you at 1 on the actual clock. **

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Self-critique (if necessary):ok

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Self-critique Rating:ok

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Question: `qquery 4.4.56 Find the positive integer solutions of the equation (5x-3) = 7 (mod 4)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

5x-3=7 (mod4)

we can go ahead and divide 7 by 4 which gives us a remainder of 3

therefore (5x-3) would have to be 3 after being divided by 4

if x=1 5(1)-3=2 rem. 2

x=2 5(2)-3=7 rem. 3

x=3 5(3)-3=12 rem. 0

x=4 5(4)-3=17 rem. 1

x=5 5(5)-3=22 rem. 2

x=6 5(6)-3=27 rem. 3

x=7 5(7)-3=32 rem. 0

x=8 5(8)-3=37 rem. 1

x=9 5(9)-3=42 rem. 2

x=9 5(10)-3=47 rem. 3

as we see here it equals 3 starting at x=2 and every fourth number after that

{2,6,10,14,18...}

confidence rating #$&*:3

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Given Solution:

`a** The solutions have to be integers, and the mod makes a difference in the algebra.

7 (mod 4) is 3.

Since (5x - 3) mod 4 = 7 mod 4, (5x - 3) mod 4 must be 3.

For x = 1, 2, 3, 4, ..., the expression 5x - 3 takes values 2, 7, 12, 17, 22, 27, 32, 37, ... .

These numbers, when divided by 4, give remainders 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, ... .

Thus every fourth number, mod 4, is equal to 3.

This starts with the second number, which occurs when x = 2.

Every fourth number, starting with 2, gives us the sequence 2, 6, 10, 14, ...

2 is the first solution, 4 is the difference between solutions.

Thus x can be any element in the set {2, 6, 10, 14, . . . }.

The general term of this sequence is 2 + 4 n. So we can also say that x = 2 + 4 n, where n = 0, 1, 2, 3, . . .

Checking these results:

If n = 0 then x = 2 + 4 * 0 = 2.

If n = 1 then x = 2 + 4 * 1 = 6.

If n = 2 then x = 2 + 4 * 2 = 10.

If n = 3 then x = 2 + 4 * 3 = 14.

etc.

These are the solutions obtained above to the equation. **

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Self-critique (if necessary):ok

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Self-critique Rating:ok

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Question: `qquery 5.4.47 Give the table for addition mod 7 and list the properties of operation; show how the properties are proven

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

+ mod 7

0 1 2 3 4 5 6

0 0 1 2 3 4 5 6

1 1 2 3 4 5 6 0

2 2 3 4 5 6 0 1

3 3 4 5 6 0 1 2

4 4 5 6 0 1 2 3

5 5 6 0 1 2 3 4

6 6 0 1 2 3 4 5

it is closed since all numbers are between 0 and 6

it has the communative property because its symmetric if you drew a diagonal line

it has identity of 0 because you can add 0 to any number and still be the same

it has the inverse property because every number is in each column and row

its associative because of the associative property of addition

confidence rating #$&*:3

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Given Solution:

`a** Your table should read

0 1 2 3 4 5 6

0 0 1 2 3 4 5 6

1 1 2 3 4 5 6 0

2 2 3 4 5 6 0 1

3 3 4 5 6 0 1 2

4 4 5 6 0 1 2 3

5 5 6 0 1 2 3 4

6 6 0 1 2 3 4 5

The operation is closed, since all numbers mod 7 are between 0 and 6. The only numbers on the table are 0, 1, 2, 3, 4, 5, 6.

The operation has an identity, which is 0, because when added to any number 0 doesn't change that number. We can see this in the table because the row corresponding to 0 just repeats the numbers 0123456, as does the column beneath 0.

The operation is commutative--order doesn't matter because the table is symmetric about the main diagonal..

The operation has the inverse property because every number can be added to another number to get the identity 0: 0+7 = 0, 1+6=0, 2+5=0, 3+4=0. These numbers form pairs of inverses. This property can be seen from the table because the identity 0 appears exactly once in every row.

The operation is associative, since for any a, b, c we know that (a + b ) + c = a + ( b + c), and it follows that [ (a + b ) + c ] mod 7 must equal [ a + ( b + c) ] mod 7. **

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Self-critique (if necessary):ok

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Question: `q4.4.50 table for mult mod 4 and properties of operation

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Your solution:

*mod4

0 1 2 3

0 0 0 0 0

1 0 1 2 3

2 0 2 0 2

3 0 3 2 1

closed bcause all of the numbers stay within 0-3

associative because of the associative property of multiplication

communitative because it has symmetry if we drew a diagonal line

it has identity 1 because 1 can be multiplied by any other number to get the same number

does not have inverse property because 0 an 2 do not have inverses

confidence rating #$&*:3

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Given Solution:

`a** The correct table is

0 1 2 3

0 0 0 0 0

1 0 1 2 3

2 0 2 0 2

3 0 3 2 1

For example the row across from 2 is obtained as follows: 2 * 0 = 0 and 2 * 1 = 2, as always. Then 2 * 2 mod 4 = 4 mod 4, which is 0 and 2 * 3 mod 4 = 6 mod 4, which is 2.

the operation is closed because the results all come from the set {0, 1, 2, 3} being operated on

1 is the identity because the row and column for 1 both have 0,1,2,3 in that order, so 1 doesn't change a number when multiplied by that number.

0 and 2 lack inverses--they can't be combined with anything else to get 1--so the operation lacks the inverse property.

symmetry about the main diagonal implies commutativity

associativity follows from associativity of multiplication of real numbers**

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Self-critique (if necessary):ok

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Question: `qquery 5.4.66 y + [[ (y-1)/4 ]] - [[ (y-1) / 100 ]] + [[ (y-1) / 400 ]]; day of jan 1, 2002; smallest b with a = b (mod 7); b=0 Sunday etc.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

2002 + [[(2002-1)/4]] - [[(2002-1)/100]] + [[(2002-1)/400]]

2002 +[[500.25]] - [[20.01]] + [[5.0025]

2002 + 500 - 20 + 5

2487

2487(mod7) = 2

sunday is 0

???How do you get that Monday is 1, Tuesday is 2? What do you mean ""smallest b""? Also i understand what you are doing with the [[Q]] thing but where are you getting that? how are we supposed to know what you are talking about without looking at the answer for guidance???

confidence rating #$&*:2

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Given Solution:

`a** The calculation is

2002 + [[ 2002-1/4 ]] - [[ 2002-1/100 ]]+ [[ 2002-1/400 ]].

[[ Q ]] means the greatest integer contained in Q.

So we get

2002+ [[500.25]] - [[20.01]] + [[5.0025]]

= 2002 + 500 - 20 + 5

= 2487.

Now 2487 mod 7 is 2.

Sunday is 0, Monday is 1 so Tuesday is 2.**

"

Self-critique (if necessary):

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Self-critique rating:

@&

Unfortunately I don't have the current edition of the text handy. It is possible that the text 'dumbed down' by eliminating this problem and the idea of the greatest integer function (this text is better than most at maintaining standards, but at some point it becomes impossible to buck the national trend, which I consider to be a terrible disservice to capable students such as yourself).

I'll try to check this out on campus tomorrow.

*@

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Question: `qquery 5.4.66 y + [[ (y-1)/4 ]] - [[ (y-1) / 100 ]] + [[ (y-1) / 400 ]]; day of jan 1, 2002; smallest b with a = b (mod 7); b=0 Sunday etc.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

2002 + [[(2002-1)/4]] - [[(2002-1)/100]] + [[(2002-1)/400]]

2002 +[[500.25]] - [[20.01]] + [[5.0025]

2002 + 500 - 20 + 5

2487

2487(mod7) = 2

sunday is 0

???How do you get that Monday is 1, Tuesday is 2? What do you mean ""smallest b""? Also i understand what you are doing with the [[Q]] thing but where are you getting that? how are we supposed to know what you are talking about without looking at the answer for guidance???

confidence rating #$&*:2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`a** The calculation is

2002 + [[ 2002-1/4 ]] - [[ 2002-1/100 ]]+ [[ 2002-1/400 ]].

[[ Q ]] means the greatest integer contained in Q.

So we get

2002+ [[500.25]] - [[20.01]] + [[5.0025]]

= 2002 + 500 - 20 + 5

= 2487.

Now 2487 mod 7 is 2.

Sunday is 0, Monday is 1 so Tuesday is 2.**

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

@&

Unfortunately I don't have the current edition of the text handy. It is possible that the text 'dumbed down' by eliminating this problem and the idea of the greatest integer function (this text is better than most at maintaining standards, but at some point it becomes impossible to buck the national trend, which I consider to be a terrible disservice to capable students such as yourself).

I'll try to check this out on campus tomorrow.

*@

#*&!

*********************************************

Question: `qquery 5.4.66 y + [[ (y-1)/4 ]] - [[ (y-1) / 100 ]] + [[ (y-1) / 400 ]]; day of jan 1, 2002; smallest b with a = b (mod 7); b=0 Sunday etc.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

2002 + [[(2002-1)/4]] - [[(2002-1)/100]] + [[(2002-1)/400]]

2002 +[[500.25]] - [[20.01]] + [[5.0025]

2002 + 500 - 20 + 5

2487

2487(mod7) = 2

sunday is 0

???How do you get that Monday is 1, Tuesday is 2? What do you mean ""smallest b""? Also i understand what you are doing with the [[Q]] thing but where are you getting that? how are we supposed to know what you are talking about without looking at the answer for guidance???

confidence rating #$&*:2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`a** The calculation is

2002 + [[ 2002-1/4 ]] - [[ 2002-1/100 ]]+ [[ 2002-1/400 ]].

[[ Q ]] means the greatest integer contained in Q.

So we get

2002+ [[500.25]] - [[20.01]] + [[5.0025]]

= 2002 + 500 - 20 + 5

= 2487.

Now 2487 mod 7 is 2.

Sunday is 0, Monday is 1 so Tuesday is 2.**

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

@&

Unfortunately I don't have the current edition of the text handy. It is possible that the text 'dumbed down' by eliminating this problem and the idea of the greatest integer function (this text is better than most at maintaining standards, but at some point it becomes impossible to buck the national trend, which I consider to be a terrible disservice to capable students such as yourself).

I'll try to check this out on campus tomorrow.

*@

#*&!#*&!

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