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course mth151 11/20/13 019. Place-value System with Other Bases.............................................
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): ------------------------------------------------ Self-critique Rating:
********************************************* Question: `q002. What would the number 213{base 4} be in base 10 notation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2*4^2+1*4^1+3*4^0= 2*16+1*4+3*1= 32+4+3= 39 confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
Given Solution: 213{base 4} means 2 * 4^2 + 1 * 4^1 + 3 * 4^0 = 2 * 16 + 1 * 4 + 3 * 1 = 32 + 4 + 3 = 39. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:
********************************************* Question: `q003. Suppose we had a number expressed in the form 6 * 4^2 + 7 * 4^1 + 3 * 4^0. In base 4 every term needs to be expressed in the highest possible power of 4. This is not the case for the given number, since for example the coefficient 7 can be expressed as 1 * 4^1 + 3 * 4^0. How would the number 6 * 4^2 + 7 * 4^1 + 3 * 4^0 be expressed without using any coefficients greater than 3? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1*4^3+3*4^2+3*4^1+3*4^0= 1,333 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
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): ------------------------------------------------ Self-critique Rating:
********************************************* Question: `q004. What would happen to the number 1333{base 4} if we added 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2*4^3+0*4^2+0*4^1+0*4^0= 2,000 base of 4 confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
Given Solution: Since 1 = 1 * 4^0, Adding one to 1 * 4^3 + 3 * 4^2 + 3 * 4^1 + 3 * 4^0 would give us 1 * 4^3 + 3 * 4^2 + 3 * 4^1 + 3 * 4^0 + 1 * 4^0 = 1 * 4^3 + 3 * 4^2 + 3 * 4^1 + 4 * 4^0. But 4 * 4^0 = 4^1, so we would have 1 * 4^3 + 3 * 4^2 + 3 * 4^1 + 1 * 4^1 + 0 * 4^0 = 1 * 4^3 + 3 * 4^2 + 4 * 4^1 + 0 * 4^0 . But 4 * 4^1 = 4^2, so we would have 1 * 4^3 + 3 * 4^2 + 1 * 4^2 + 0 * 4^1 + 0 * 4^0 = 1 * 4^3 + 4 * 4^2 + 0 * 4^1 + 0 * 4^0 . But 4 * 4^2 = 4^3, so we would have 1 * 4^3 + 1 * 4^3 + 0 * 4^2 + 0 * 4^1 + 0 * 4^0 = 2 * 4^3 + 0 * 4^2 + 0 * 4^1 + 0 * 4^0. We thus have the number 2000{base 4}. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:
********************************************* Question: `q005. How would the decimal number 659 be expressed in base 4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2*4^4+2*4^3+1*4^2+0*4^1+3*4^0 confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
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): ------------------------------------------------ Self-critique Rating:
********************************************* Question: `q006. Find the base-10 equivalent of the number 322{base 4}. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4*2^+4*1+4^2+2^4+0*4^1+3*4#$&*
course mth151 11/20/2013 *********************************************.............................................
Given Solution: 2 means 2 * 10^0 1 means 1 * 10^1 8 means 8 * 10^2 3 means 3 * 10^3 5 means 5 * 10^4 Thus the number 53812 means (5*10^4)+(3*10^3)+(8*10^2)+(1*10^1)+(2*10^0). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:
********************************************* Question: `q query 4.2.20 536 + 279 in expanded notation Write 536 + 279 in expanded notation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 15*10^0 = 10 *10^0+5*10^0= 10^1+5*10^09*10^2+1*10^1+5*10^0 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
Given Solution: `a** We write this sum as 5 * 10^2 + 3 * 10^1 + 6 * 10^0 + 2 * 10^2 + 7 * 10^1 + 9 * 10^0 ______________________________ 8 * 10^2 + 10* 10^1 + 15* 10^0. Since 10 * 10^1 = 10^2 we can write this as 9 * 10^2 + 0 * 10^1 + 15 * 10^0. Since 15 * 10^0 = 10 * 10^0 + 5 * 10^0 = 10^1 + 5 * 10^0 we rewrite this as 9 * 10^2 + 1 * 10^1 + 5 * 10^0. This result is expressed in our place-value system as 915. ** STUDENT QUESTION: When adding the 6*10^0 and the 9*10^0 I don’t carry the one like in regular math? INSTRUCTOR RESPONSE: We're not applying the rules for addition as we all learned them in elementary school, but reasoning our results out from the more basic perspective of a place-value system. 6 * 10^0 + 9 * 10^0 = 15 * 10^0. 15 * 10^0 means 10 * 10^0 + 5 * 10^0, and since 10 * 10^0 = 10^1 we conclude that our original expression 15 * 10^0 is equal to 1 * 10^1 + 5 * 10^0. This is the reason you 'carry the 1'. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:
" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: `gr31#$&* course mth151 11/20/13 019. Place-value System with Other Bases
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Given Solution: In base 10, 3 * 4^2 + 2 * 4^1 + 1 * 4^0 = 3 * 16 + 2 * 4 + 1 * 1 = 48 + 8 + 1 = 57. STUDENT COMMENT: I am not understanding this. INSTRUCTOR RESPONSE statement 1: 321{base 4} means 3 * 4^2 + 2 * 4^1 + 1 * 4^0. statement 2: 3 * 4^2 + 2 * 4^1 + 1 * 4^0 = 57. What is it you do and do not understand about the above two statements? &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. What would the number 213{base 4} be in base 10 notation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2*4^2+1*4^1+3*4^0= 2*16+1*4+3*1= 32+4+3= 39 confidence rating #$&*: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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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. 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1*4^3+3*4^2+3*4^1+3*4^0= 1,333 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 7 = 4 + 3 so 7 * 4^1 can be written as 4 * 4^1 + 3 * 4^1 = 4^2 + 3 * 4^1 Since 6 = 4 + 2, we have 6 * 4^2 = 4 * 4^2 + 2 * 4^2. Since 4 * 4^2 = 4^3, this is 4^3 + 2 * 4^2. Thus 6 * 4^2 + 7 * 4^1 + 3 * 4^1 = (4 * 4^2 + 2 * 4^2) + (4 * 4^1 + 3 * 4^1) + 3 * 4^0 =4^3 + 2 * 4^2 + 4^2 + 3 * 4^1 + 3 * 4^0 = 1 * 4^3 + 3 * 4^2 + 3 * 4^1 + 3 * 4^0. This number would then be 1333 {base 4}. STUDENT COMMENT I understand the answer, but not the first paragraph of the explanation. INSTRUCTOR RESPONSE Here is an expanded version of the first line: 7 * 4^1 = (4 + 3) * 4^1 = 4 * 4^1 + 3 * 4^1. Since 4 * 4^1 = 4^2, it follows that 7 * 4^1 = 4^2 + 3 * 4^1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. What would happen to the number 1333{base 4} if we added 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2*4^3+0*4^2+0*4^1+0*4^0= 2,000 base of 4 confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. How would the decimal number 659 be expressed in base 4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2*4^4+2*4^3+1*4^2+0*4^1+3*4^0 confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We need to express 659 in terms of multiples powers of 4, with the multiple not exceeding 3. The powers of 4 are 4^0 = 4, 4^1 = 4, 4^2 = 16, 4^3 = 64, 4^4 = 256, 4^5 = 1024. We could continue to higher powers of 4, but since 4^5 = 1024 already exceeds 659 we need not do any further. The highest power of 4 that doesn't exceed 659 is 4^4 = 256. So we will use the highest multiple of 256 that doesn't exceed 659. 2 * 256 = 512, and 3 * 256 exceeds 659, so we will use 2 * 256 = 2 * 4^4. This takes care of 512 of the 659, leaving us 147 to account for using lower powers of 4. We then account for as much of the remaining 147 using the next-lower power 4^3 = 64. Since 2 * 64 = 128 is less than 147 while 3 * 64 is greater than 147, we use 2 * 64 = 2 * 4^3. This accounts for 128 of the remaining 147, which now leaves us 19. The next-lower power of 4 is 4^2 = 16. We can use one 16 but not more, so we use 1 * 16 = 1 * 4^2. This will account for 16 of the remaining 19, leaving us 3. This 3 is accounted for by 3 * 4^0 = 3 * 1. Note that we didn't need 4^1 at all. So we see that 659 = 2 * 4^4 + 2 * 4^3 + 1 * 4^2 + 0 * 4^1 + 3 * 4^0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. Find the base-10 equivalent of the number 322{base 4}. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4*2^+4*1+4^2+2^4+0*4^1+3*4
#$&* course mth151 11/20/2013 *********************************************
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Given Solution: 2 means 2 * 10^0 1 means 1 * 10^1 8 means 8 * 10^2 3 means 3 * 10^3 5 means 5 * 10^4 Thus the number 53812 means (5*10^4)+(3*10^3)+(8*10^2)+(1*10^1)+(2*10^0). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q query 4.2.20 536 + 279 in expanded notation Write 536 + 279 in expanded notation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 15*10^0 = 10 *10^0+5*10^0= 10^1+5*10^09*10^2+1*10^1+5*10^0 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** We write this sum as 5 * 10^2 + 3 * 10^1 + 6 * 10^0 + 2 * 10^2 + 7 * 10^1 + 9 * 10^0 ______________________________ 8 * 10^2 + 10* 10^1 + 15* 10^0. Since 10 * 10^1 = 10^2 we can write this as 9 * 10^2 + 0 * 10^1 + 15 * 10^0. Since 15 * 10^0 = 10 * 10^0 + 5 * 10^0 = 10^1 + 5 * 10^0 we rewrite this as 9 * 10^2 + 1 * 10^1 + 5 * 10^0. This result is expressed in our place-value system as 915. ** STUDENT QUESTION: When adding the 6*10^0 and the 9*10^0 I don’t carry the one like in regular math? INSTRUCTOR RESPONSE: We're not applying the rules for addition as we all learned them in elementary school, but reasoning our results out from the more basic perspective of a place-value system. 6 * 10^0 + 9 * 10^0 = 15 * 10^0. 15 * 10^0 means 10 * 10^0 + 5 * 10^0, and since 10 * 10^0 = 10^1 we conclude that our original expression 15 * 10^0 is equal to 1 * 10^1 + 5 * 10^0. This is the reason you 'carry the 1'. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: