Assignment 19

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course MTH 151

11:27pm, 4/1/14

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.

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?

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Your solution:

3 * 10^2 + 2 * 10^1 + 1 * 10^0 = 321. I’m not really feeling like I’m understanding this completely. To me, it seems easier to see 3 as X, 2 as Y and 1 as Z, and to plug those numbers into an equation using the 10 as the number in the equation, so X * 10^Y + Y * 10^Z + Z * 10^0. Could this method be used to figure out the bases? For example, if the problem read 543 {base 5}, could you read this problem as 5 * 5 ^4 + 4 * 5^3 + 3 * 5^0 = 3628?

321 {base 4} = 3 * 4^2 + 2 * 4^1 + 1 * 4^0

321 {base 10} = 3 * 10^2 + 2 * 10^1 + 1 * 10^0

confidence rating #$&*: 2

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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):

How do you know what numbers to multiply, and put to the power of 2, etc.? Does 321 {base 4} mean that you’re plugging in the 3, 2 and 1 into an equation like x * 4^y + 2 * 4^z + 1 * 4^0, like if 3 represented x, 2 represented y and 1 represented z? Where do the + 2 and + 1 come from, and for each base equation, are they always 2 and 1, or is that because they are the same as the numbers being plugged in before them (for example, 2 + 2, and 1 + 1)? Would the 10 replace the 4 as far as the base (for example, instead of 321{base 4} means 3 * 4^2 + 2 * 4^1 + 1 * 4^0, it would read 321 {base 10}, so 3 * 10^2 + 2 * 10^1 + 1 * 10^0? But when I do this, I don’t get the same answer for 321 {base 10}, I got 213.

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Self-critique Rating: 3

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321(base 4) expressed as a base-10 number would be 57.

321(base 10) would be 3 ^ 10^2 + 2 * 10^1 + 1 * 10^0, which is equal to 321, not 223.

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

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Your solution:

213 {base 4} = 2 * 4^2 + 1 * 4^1 + 3 * 4^0 = 39, so 213 {base 10} would equal 2 * 10^2 + 1 * 10^1 + 3 * 10^0 = 213, correct?

confidence rating #$&*: 2

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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):

Could you use a basic “equation” to find the base of numbers? For example, for 213 {base 4}, could you write out this problem as First number * base number^2 + second number * base number^1 + third number * base number^0?

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Self-critique Rating: 3

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All of your statements are correct.

However the question did not ask for 213{base 10), it asked for the base-10 representation of 213{base 4}.

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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?

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Your solution:

6 * 4^2 + 7 * 4^1 + 3 * 4^0 = 127

4 * 4^2 + 2 * 4^2 + 4 * 4^1 + 3 * 4^1 + 3 * 4^0 = 127

confidence rating #$&*: 2

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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 * 4^1 =

(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):

Had a hard time understanding this completely, and I got a different answer than you. I’m going to need to rewatch the DVD and reread the chapter to gain a better understanding of the bases, etc.

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Self-critique Rating: 3

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You did well, but stopped a couple of steps short:

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

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

and both are equal to the base-10 number 127.

4 * 4^2 is 4^3, and 4 * 4^1 is 4^2, so your expression

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

is equal to

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

and 2 * 4^2 + 4^2 = 2 * 4^2 + 1 * 4^2 = +3 * 4^2, so the above becomes

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:

1 * 4^3 + 3 * 4^2 + 3 * 4^1 + 3 * 4^0 = 127

1 * 4^3 + 3 * 4^2 + 3 * 4^1 + 3 * 4^0 + 1 = 128

confidence rating #$&*: 2

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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):

Got a different answer, I got 128. Could you possibly explain how you got 2000{base 4} in a different way? The explanation helps me a little, but is still somewhat confusing.

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Self-critique Rating: 3

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you switched over to base 10.

You can verify that 2 * 4^3 + 0 * 4^2 + 0 * 4^1 + 0 * 4^0 = 128, so the result in base 4 must be 2000.

However that doesn't help you work it out in base 4.

Start with the fact that 3 * 4^0 + 1 = 3 * 4^0 + 1 * 4^0 = 4 * 4^0, which is equal to 4^1 (plus, of course 0 * 4^0).

Now 3 * 4^1 + 1 * 4^1 = 4^2, which must be added to 3* 4^2 to give you 4 * 4^2, which is equal to 4^3.

This must be added to the 1 * 4^3 we already have to give you 2 * 4^3.

Think about this, then take another look at the given solution.

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

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Your solution:

I was really stumped by this problem. I think that it would still be written out as 6 * 4^2 + 5 * 4^1 + 9 * 4^0, but when you get the answer (125) just be written in the decimal form, like .125?

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):

Really needing to review this

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Self-critique Rating: 3

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The bottom line, literally in the case, is that 659 = 2 * 4^4 + 2 * 4^3 + 1 * 4^2 + 0 * 4^1 + 3 * 4^0, which you can easily verify.

Thus the decimal number 659 is equal to 22103 {base 4}.

That leaves the question of how to get this result, given just the 659.

To do so you have to find the largest power of 4 that 'fits into' 659 (i.e., the largest power of 4 that is less than 659).

Then you have to find out how much of the 659 is left over when you take this power of 4 out of the number.

You then find how much of the difference can be expressed as a multiple of the next-smaller power of 4.

You keep repeating this until you are out of powers of 4.

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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 = 58

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Good.

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confidence rating #$&*:

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

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Your solution:

4 * 4^2 + 8 * 4^1 + 7 * 4^0 = 103

confidence rating #$&*:

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This isn't correct.

487 means 4 * 10^2 + 8 * 10^1 + 7 * 10^0, not 4 * 4^2 + 8 * 4^1 + 7 * 4^0.

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Self-critique Rating:

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Good try. You'll get this. Check my notes.

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