#$&* course MTH 151 12:15 4/18/13 019. Place-value System with Other Bases
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Given Solution: In base 10, 3 * 4^2 + 2 * 4^1 + 1 * 4^0 = 3 * 16 + 2 * 4 + 1 * 1 = 48 + 8 + 1 = 57. STUDENT COMMENT: I am not understanding this. INSTRUCTOR RESPONSE statement 1: 321{base 4} means 3 * 4^2 + 2 * 4^1 + 1 * 4^0. statement 2: 3 * 4^2 + 2 * 4^1 + 1 * 4^0 = 57. What is it you do and do not understand about the above two statements? &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q002. What would the number 213{base 4} be in base 10 notation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Base 4: 2 * 4^2 + 1 * 4^1 + 3 * 4^0 Base 10: 2 * 4^2 + 1 * 4^1 + 3 * 4^0 =2 * 16 + 1 * 4 + 3 =32 + 4 + 3 =39 confidence rating #$&*: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Numbers 6 and 7 are greater than 4 so: 2+4=6 3+4= 7 6 * 4^2 + 7 * 4^1 + 3 * 4^1= (4 * 4^2 + 2 * 4^2) + (4 * 4 + 3 * 4) + 3 * 4^0 = 4^3 + 2 * 4^2 + 4^2 + 3 * 4 + 3 * 4^0 = 1 * 4^3 + 3 * 4^2 + 3 * 4^0 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 7 = 4 + 3 so 7 * 4^1 can be written as 4 * 4^1 + 3 * 4^1 = 4^2 + 3 * 4^1 Since 6 = 4 + 2, we have 6 * 4^2 = 4 * 4^2 + 2 * 4^2. Since 4 * 4^2 = 4^3, this is 4^3 + 2 * 4^2. Thus 6 * 4^2 + 7 * 4^1 + 3 * 4^1 = (4 * 4^2 + 2 * 4^2) + (4 * 4^1 + 3 * 4^1) + 3 * 4^0 =4^3 + 2 * 4^2 + 4^2 + 3 * 4^1 + 3 * 4^0 = 1 * 4^3 + 3 * 4^2 + 3 * 4^1 + 3 * 4^0. This number would then be 1333 {base 4}. STUDENT COMMENT I understand the answer, but not the first paragraph of the explanation. INSTRUCTOR RESPONSE Here is an expanded version of the first line: 7 * 4^1 = (4 + 3) * 4^1 = 4 * 4^1 + 3 * 4^1. Since 4 * 4^1 = 4^2, it follows that 7 * 4^1 = 4^2 + 3 * 4^1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): In the given solution you show exponents to the first power such as 4^1 is that necessary to show in this type of equation I understand raising it to the zero power because that would change the answer ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q004. What would happen to the number 1333{base 4} if we added 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If you added 1 to the number 1333 at base 4 you would have to go back to the previous question and add an exponent to each number line like: 1 = 1 * 4^0 4 * 4^1 = 4^2 4 * 4^2 = 4^3 If you notice the exponents go up by adding 1 in the previous question we worked it out and by adding 1 to this you would now come up with a number that is 2000 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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}. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q005. How would the decimal number 659 be expressed in base 4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I am really not sure how to work this problem I am looking in the section 4.2 but I do not see anything as to converting decimals into bases confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): As I read the given solution a light bulb went off in my head and everything seemed to make since. The solution was so detailed and gave me perfect insight on how to answer questions like this in the future!