Assignment19

course Mth 151

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assignment #019

019. Place-value System with Other Bases

Liberal Arts Mathematics I

07-10-2007

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13:35:27

`q001. There are 5 questions in this set.

The preceding calculations have been 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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RESPONSE -->

3 * 4^2 = 3 * 16 = 48

2 * 4^1 = 2 * 4 = 8

1 * 4^0 = 1 * 1 = 1

48+8+1=57

confidence assessment: 3

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13:35:32

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

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RESPONSE -->

self critique assessment: 3

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13:37:18

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

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RESPONSE -->

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

2* 4^2= 2 * 16 = 32

1 * 4^1 = 1 * 4 = 4

3 * 4^0 = 3 * 1 = 3

32 + 4 + 3 = 39

confidence assessment: 3

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13:37:23

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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RESPONSE -->

self critique assessment: 3

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13:41:31

`q003. Suppose we had a number expressed in the form 6 * 4^2 + 7 * 4^1 + 3 * 4^0. This number isn't quite in the form needs to be if it is to be expressed in base 4. This is because we have the numbers 6 and 5, which exceed 4. How would this number be expressed without using any numbers 4 or greater?

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RESPONSE -->

6 * 4^2 can be divided into (4 * 4^2) + (2 * 4^2)

7 * 4^1 can be divided into (4 * 4^1) + (3*4^1)

The whole problem is

(4 * 4^2) + (2 * 4^2) + (4 * 4^1) + (3 * 4^1) + (3 * 4^0) = 64 + 32 + 16 + 12 + 3 = 127

confidence assessment: 3

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13:43:22

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}.

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RESPONSE -->

I added all of the sums together, instead of taking the first number of each segment. I see why the answer is 1333

self critique assessment: 2

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13:46:33

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

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RESPONSE -->

The base is written

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

If we add one it is

(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)

1334 in base 4

confidence assessment: 3

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13:48:30

Since 1 = 1 * 4^0, Adding one to 1 * 10^3 + 3 * 4^2 + 3 * 4^1 + 3 * 4^0 would give us

1 * 10^3 + 3 * 4^2 + 3 * 4^1 + 3 * 4^0 + 1 * 4^0 =

1 * 10^3 + 3 * 4^2 + 3 * 4^1 + 4 * 4^0.

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

1 * 10^3 + 3 * 4^2 + 3 * 4^1 + 1 * 4^1 + 0 * 4^0 =

1 * 10^3 + 3 * 4^2 + 4 * 4^1 + 0 * 4^0 .

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

1 * 10^3 + 3 * 4^2 + 1 * 4^2 + 0 * 4^1 + 0 * 4^0 =

1 * 10^3 + 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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RESPONSE -->

I think I understand. I had trouble with 4 * 4^0 + 4^1 , 4 * 4^1 = 4^2 and so on.

self critique assessment: 2

4 * 4^0 = 4 * 1 = 4.

4 * 4 = 16, which can be written as 4 * 4^1 = 4^2.

4 * 4^2 = 4 * 16 = 64, which can be written as 4 * 4^2 = 4^3.

Every time you multiply by 4 you get a higher power.

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13:54:38

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

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RESPONSE -->

(6 * 4^2) + (5 * 4^1) + (9 * 4^0)

But 6, 5, and 9 are all greater than 4

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

confidence assessment: 3

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13:57:48

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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RESPONSE -->

I wasn't sure what the best way to simplify this problem was. I was a little confused about what you wanted - but I understand how you arrived at the conclusion.

self critique assessment: 2

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Assignment19

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