Assignment 19

course Mth 151

?????€g?o?????assignment #019019. Place-value System with Other Bases

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

Liberal Arts Mathematics I

07-25-2007

......!!!!!!!!...................................

13:59:50

`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?

......!!!!!!!!...................................

RESPONSE -->

3*100 + 2*10 + 1*1

Or

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

confidence assessment: 3

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

......!!!!!!!!...................................

14:06:01

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

......!!!!!!!!...................................

RESPONSE -->

I got this wrong. I just did the number 321 in base 10 not by base 4. I see my error.

self critique assessment: 3

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

......!!!!!!!!...................................

14:10:38

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

......!!!!!!!!...................................

RESPONSE -->

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

= 2*16 + 1*4 + 3*1

= 32+4+3

=39

confidence assessment: 3

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

......!!!!!!!!...................................

14:10:54

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

......!!!!!!!!...................................

RESPONSE -->

I answered this correctly

self critique assessment: 3

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

......!!!!!!!!...................................

14:33:06

`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?

......!!!!!!!!...................................

RESPONSE -->

Ok I am lost. I have looked through my book and checked my cd but I can't find anything that I missed to help me on how to do this so I am moving on to the answer to help me understand.

confidence assessment: 0

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

......!!!!!!!!...................................

14:46:13

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}.

......!!!!!!!!...................................

RESPONSE -->

Ok I understand now. I thought the process would be something like this but I did not know exactly how to break it down from this format (exp. 4 * 4^1 + 3 * 4^). Now I see how that is fairly easy. I will do the problem myself now.

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

= 1333

self critique assessment: 3

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

......!!!!!!!!...................................

14:50:23

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

......!!!!!!!!...................................

RESPONSE -->

It would completely change the number.

confidence assessment: 0

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

......!!!!!!!!...................................

15:04:06

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}.

......!!!!!!!!...................................

RESPONSE -->

I didn't do out the math but I knew it would change the number. I was unclear that you wanted us to work out the problem. My mistake, sorry.

self critique assessment: 3

As long as you understand, it's fine. No need to apologize.

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

......!!!!!!!!...................................

15:30:01

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

......!!!!!!!!...................................

RESPONSE -->

First lets break down our numbers since they are all greater than 4

6= 4+2

6= 4*4^2 + 2*4^2

6= 4*4^2 + 2*4^2 isn't a true statement; you could say something like

6 in the 4^2 place means 4*4^2 + 2*4^2, which I believe is your intent.

However the given number is a decimal number, so 6 represents 6 * 10^2, not 6 * 4^2.

Your reasoning is good, but it's the reasoning you would use to express 659 {base 4} in base 10. Of course you wouldn't write 659 in base 4 (wouldn't use digits bigger than 3, for reasons you appear in preceding problems to understand), but this would be the sequence of reasoning.

6= 4^3 + 2*4^2

5=4+1

5=4*4^1 + 1*4^1

5=4^2 + 1*4^1

9=4+4+1

9=4*4^0 + 4*4^0 + 1*4^0

9=4^1 + 4^1 + 1*4^0

9=4^2 + 1*4^0

So we have:

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

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

= 4^3 + 16^2 + 4^1 + 4^0

= 64 + 256 + 16 + 4

= 340

confidence assessment: 2

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

......!!!!!!!!...................................

15:33:56

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.

......!!!!!!!!...................................

RESPONSE -->

I got this wrong and I am still not sure why this process was different. Hopefully you can help me to see my error.

self critique assessment: 1

See my note and see if it helps. If you want to ask further questions, just submit a copy of the problem, your solution, the given solution and any questions. I'll be glad to answer.

................................................."

&#

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