course Mth 151
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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Self-critique (if necessary):
Self-critique Rating:
Question: **** `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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Your solution:
3*10^2+2*10^1+1*10^0
Confidence Assessment:
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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.
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Self-critique (if necessary):
I am not understanding this.
321{base 4} means 3 * 4^2 + 2 * 4^1 + 1 * 4^0.
3 * 4^2 + 2 * 4^1 + 1 * 4^0 = 57.
What is it you do and do not understand about the above two statements?
Self-critique Rating:
Question: **** `q002. What would the number 213{base 4} be in base 10 notation?
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Your solution:
2*4^2+1*4^1+3*4^0=
2*16+1*4+3*1=32+7=39
Confidence Assessment:
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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):
Self-critique Rating:
Question: **** `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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Your solution:
6*4^2+7*4^1+3*4^0=
6=4+2
7=4+3
4*4^2+2*4^2+4*4^1+3*4^1+3*4^0=
good start
Confidence Assessment:
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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}.
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Self-critique (if necessary):
I only could get so far but now I understand why it is 1333 base 4 b/c of adding the powers together such as 4*4^2+2*4^2
you had most of it, and I think you see the rest now
Self-critique Rating:
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+4*4^0
1*4^3+3*4^2+3*4^1+4*4^1
Confidence Assessment:
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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):
I didn’t add on my 0*4^0 and finishing out each addition step.
Self-critique Rating:
Question: **** `q005. How would the decimal number 659 be expressed in base 4?
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Your solution:
Where’s the decimal?
'decimal' means 'base 10'
a 'decimal point' is the . we use to indicate negative powers of 10.
Confidence Assessment:
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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.
Calling it a decimal number confused me but I know see what you wanted.
&#Good responses. See my notes and let me know if you have questions. &#