Assignment 19

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course Mth 151

Your videos are labeled wrong for Chapters 4 and 5. I was watching the videos for Chapter 5 and trying to do the assignments for Chapter 4.

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.

019. Place-value System with Other Bases

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

The calculations of the preceding qa were done in our standard base-10 place value system. We can do similar calculations with bases other than 10.

For example, a base-4 calculation might involve the number 3 * 4^2 + 2 * 4^1 + 1 * 4^0. This number will be expressed as 321{base 4}.

What would this number be in base 10?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

3 * 10˄2 + 2 * 10˄1 + 1 * 10˄0.

3 * 100 + 2 * 10 + 1 * 0

300 + 20 + 1 = 321

confidence rating #$&*: 3

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Given Solution:

In base 10, 3 * 4^2 + 2 * 4^1 + 1 * 4^0 = 3 * 16 + 2 * 4 + 1 * 1 = 48 + 8 + 1 = 57.

STUDENT COMMENT:

I am not understanding this.

INSTRUCTOR RESPONSE

statement 1: 321{base 4} means 3 * 4^2 + 2 * 4^1 + 1 * 4^0.

statement 2: 3 * 4^2 + 2 * 4^1 + 1 * 4^0 = 57.

What is it you do and do not understand about the above two statements?

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Self-critique (if necessary): Your question asked us to do the problem in base 10, but you did your solution in base 4. I don’t understand that.

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Self-critique Rating: 2

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Everything up to the last two sentences above is background and definition.

The essential question can be isolated form the two sentences

"This number will be expressed as 321{base 4}.

What would this number be in base 10?"

'this number' refers to the number 321 {base 4 }, which is 3 ^ 4^2 + 2 * 4^1 + 1 * 4^0.

In base 10, this number is expressed as 57.

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Question: `q002. What would the number 213{base 4} be in base 10 notation?

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Your solution:

After the last question, I’m assuming you want me to do this in base 4:

2 * 4˄2 + 1 * 4˄1 + 3 * 4˄0

2 * 16 + 1 * 4 + 3 * 1

32 + 4 + 3 = 39

confidence rating #$&*: 3

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Given Solution:

213{base 4} means 2 * 4^2 + 1 * 4^1 + 3 * 4^0 = 2 * 16 + 1 * 4 + 3 * 1 = 32 + 4 + 3 = 39.

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Self-critique (if necessary): OK

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Self-critique Rating: 3

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Question: `q003. Suppose we had a number expressed in the form 6 * 4^2 + 7 * 4^1 + 3 * 4^0. In base 4 every term needs to be expressed in the highest possible power of 4. This is not the case for the given number, since for example the coefficient 7 can be expressed as 1 * 4^1 + 3 * 4^0.

How would the number 6 * 4^2 + 7 * 4^1 + 3 * 4^0 be expressed without using any coefficients greater than 3?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Can’t figure this one out.

confidence rating #$&*: 0

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Given Solution:

7 = 4 + 3 so 7 * 4^1 can be written as 4 * 4^1 + 3 * 4^1 = 4^2 + 3 * 4^1 Since 6 = 4 + 2, we have 6 * 4^2 = 4 * 4^2 + 2 * 4^2. Since 4 * 4^2 = 4^3, this is 4^3 + 2 * 4^2. Thus

6 * 4^2 + 7 * 4^1 + 3 * 4^0 =

(4 * 4^2 + 2 * 4^2) + (4 * 4^1 + 3 * 4^1) + 3 * 4^0

=4^3 + 2 * 4^2 + 4^2 + 3 * 4^1 + 3 * 4^0 =

1 * 4^3 + 3 * 4^2 + 3 * 4^1 + 3 * 4^0. This number would then be 1333 {base 4}.

STUDENT COMMENT

I understand the answer, but not the first paragraph of the explanation.

INSTRUCTOR RESPONSE

Here is an expanded version of the first line:

7 * 4^1 = (4 + 3) * 4^1 = 4 * 4^1 + 3 * 4^1.

Since 4 * 4^1 = 4^2, it follows that 7 * 4^1 = 4^2 + 3 * 4^1.

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Self-critique (if necessary): I can’t figure this out, because I’m not sure what a coefficient is. Even after looking up the definition, I’m not sure.

When you start off writing 7 as 4 * 4˄2, isn’t that first four a coefficient? If so, it is more than 3.

I wish you had done a video on this kind of problem.

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Self-critique Rating: 0

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To clarify the terminology of 'coefficient':

In the expression

6 * 4^2 + 7 * 4^1 + 3 * 4^0

6 is the coefficient of 4^2,

7 is the coefficient of 4^1 and

3 is the coefficient of 4^0.

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One source of confusion, for which I apologize, is that I wrote

6 * 4^2 + 7 * 4^1 + 3 * 4^1

instead of

6 * 4^2 + 7 * 4^1 + 3 * 4^0.

I've made that correction above and in the original document.

*@

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7 * 4^1 is not in base-4 notation, precisely because 7 is greater than 3.

So we write it as

4 * 4^1 + 3 * 4^1.

Now the 3 * 4^1 is OK, at least for the moment, but the 4 * 4^1 is still problematic.

So we just multiply 4 by 4^1, which gives us

4 * 4^1 = 4 * 4 = 4^2.

That is the plain unadorened 4^2 term in the step

4^3 + 2 * 4^2 + 4^2 + 3 * 4^1 + 3 * 4^0,

the one without a coefficient (which is understood to be 1):

We have two 4^2 terms, 2 * 4^2 and 4^2, which add up to 3 * 4^2. We see this term in the following step

1 * 4^3 + 3 * 4^2 + 3 * 4^1 + 3 * 4^0.

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Question: `q004. What would happen to the number 1333{base 4} if we added 1?

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Your solution:

I could not figure out what you wanted me to do on this one until I looked at your given solution for a jump start. I didn’t know where you wanted to add the 1 (the beginning, end, etc,). After I looked down, I see that you want it added to the end (after) the problem has been written out. Otherwise, I would have wrote the number down as 13331. As you can see, one of my biggest problems is trying to interpret the question.

1 * 4˄3 + 3 * 4˄2 + 3 * 4˄1 + 3 * 4˄0 and then add 1 * 4˄0

1 * 64 + 3 * 16 + 3 * 4 + 3 + 1

64 + 48 + 4 + 3 + 1 = 120

@&

Unless you are trying to express the number in base-10 notation, wou don't want to multiply out 4^3, 4^2, etc.. Keep everything in powers of 4, written in exponential notation.

You have (amost) correctly expressed 1333 {base 4 } in base-10 notation, but that isn't the goal of this problem.

Note that the number is 1333 {base 4}, not 13331{base 4}.

*@

confidence rating #$&*: 0

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Given Solution:

Since 1 = 1 * 4^0, Adding one to 1 * 4^3 + 3 * 4^2 + 3 * 4^1 + 3 * 4^0 would give us

1 * 4^3 + 3 * 4^2 + 3 * 4^1 + 3 * 4^0 + 1 * 4^0 =

1 * 4^3 + 3 * 4^2 + 3 * 4^1 + 4 * 4^0.

But 4 * 4^0 = 4^1, so we would have

1 * 4^3 + 3 * 4^2 + 3 * 4^1 + 1 * 4^1 + 0 * 4^0 =

1 * 4^3 + 3 * 4^2 + 4 * 4^1 + 0 * 4^0 .

But 4 * 4^1 = 4^2, so we would have

1 * 4^3 + 3 * 4^2 + 1 * 4^2 + 0 * 4^1 + 0 * 4^0 =

1 * 4^3 + 4 * 4^2 + 0 * 4^1 + 0 * 4^0 .

But 4 * 4^2 = 4^3, so we would have

1 * 4^3 + 1 * 4^3 + 0 * 4^2 + 0 * 4^1 + 0 * 4^0 =

2 * 4^3 + 0 * 4^2 + 0 * 4^1 + 0 * 4^0.

We thus have the number 2000{base 4}.

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Self-critique (if necessary): Looking ahead and still cannot figure out what I’m supposed to be doing. I can follow you problem, I just cannot figure out what the problem is asking me to do.

So, I guess this means when we add 1 to the number, it turns it into 2000.

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You are adding 1 to 1333 {base 4}.

1333{base 4} = 4^3 * 3 * 4^2 + 3 * 4^1 + 3 * 4^0, so assing 1 (which we can express as 1 * 4^0) gives us

1 * 4^3 + 3 * 4^2 + 3 * 4^1 + 3 * 4^0 + 1 * 4^0.

Combining 3 * 4^0 and 1 * 4^0 we get

4 * 4^0,

which isn't in the right notation for a base-4 number, since the coefficient 4 is greater than 3.

However 4 * 4^0 = 4 * 1 = 4, which is 4^1, so we replace 3 * 4^0 + 1 * 4^0 with 4^1. Our expression is now

1 * 4^3 + 3 * 4^2 + 3 * 4^1 + 4^1.

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Do you see what happens next?

3 * 4^1 + 4^1 is 4 * 4^1, which is 4^2.

See if you can work out the rest.

*@

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Self-critique Rating: 0

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Question: `q005. How would the decimal number 659 be expressed in base 4?

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Your solution:

659 = 2 * 4˄4 + 2 * 4˄3 + 1 * 4˄2 + 0 * 4˄1 + 3 * 4˄0

confidence rating #$&*: 1

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Given Solution:

We need to express 659 in terms of multiples powers of 4, with the multiple not exceeding 3. The powers of 4 are 4^0 = 4, 4^1 = 4, 4^2 = 16, 4^3 = 64, 4^4 = 256, 4^5 = 1024. We could continue to higher powers of 4, but since 4^5 = 1024 already exceeds 659 we need not do any further.

The highest power of 4 that doesn't exceed 659 is 4^4 = 256. So we will use the highest multiple of 256 that doesn't exceed 659. 2 * 256 = 512, and 3 * 256 exceeds 659, so we will use 2 * 256 = 2 * 4^4.

This takes care of 512 of the 659, leaving us 147 to account for using lower powers of 4.

We then account for as much of the remaining 147 using the next-lower power 4^3 = 64. Since 2 * 64 = 128 is less than 147 while 3 * 64 is greater than 147, we use 2 * 64 = 2 * 4^3.

This accounts for 128 of the remaining 147, which now leaves us 19.

The next-lower power of 4 is 4^2 = 16. We can use one 16 but not more, so we use 1 * 16 = 1 * 4^2.

This will account for 16 of the remaining 19, leaving us 3. This 3 is accounted for by 3 * 4^0 = 3 * 1. Note that we didn't need 4^1 at all.

So we see that 659 = 2 * 4^4 + 2 * 4^3 + 1 * 4^2 + 0 * 4^1 + 3 * 4^0.

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Self-critique (if necessary): WOW! I understood this one and got it right.

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Self-critique Rating: 3

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Question: `q006. Find the base-10 equivalent of the number 322{base 4}.

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Your solution:

3 * 4˄2 + 2 * 4˄1 + 2 * 4˄0

48 + 8 + 1 = 57

confidence rating #$&*:

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Question: `q007. Find the base-4 equivalent of the number 487.

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Your solution:

1 * 4˄4 + 3 * 4˄3

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Very good.

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confidence rating #$&*:

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Self-critique Rating:

"

Self-critique (if necessary):

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Self-critique rating:

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