course Mth 151 EŹȁУassignment #019
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19:52:13 `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*16=48 2*4=8 1*1=1 which all add up to 57. confidence assessment: 3
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19:52:26 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 --> I understand this. self critique assessment: 3
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19:53:17 `q002. What would the number 213{base 4} be in base 10 notation?
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RESPONSE --> 2*10^2+1*10^1+3*10^0. confidence assessment: 3
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19:56:21 213{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 --> I do understand this but I wrote it in base ten for some strange reason. 213 in base 4 would be written 2*4^2+1*4^1+3*4^0 which equals 32+4+3=39. I simply got in a hurry and wrote it in the wrong base. self critique assessment: 2
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20:04:13 `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 --> I am not really sure, but my guess would be that the place value would have to go up. For example:1*4^5+3*4^4+3*4^3+3*4^2+3*4^1+3*4^0. I don't know if this makes any sense or not. confidence assessment: 1
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20:07:33 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 don't understand this at all right now. I have read it over and over and just don't get what is going on at all. I really can't ask you a specific question to help me understand because I don't even understand what to ask.
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20:10:23 `q004. What would happen to the number 1333{base 4} if we added 1?
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RESPONSE --> You would have to regroup the numbers in order not to exceed the number 3. confidence assessment: 1
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20:14:07 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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RESPONSE --> I'm sorry I'm just very confused. I don't understand why base 10 and base 4 is being used throughout this explanation and I just don't get this. I read Section 4.2 in the textbook and understood it all and I also did well on the exercises in the book, but this isn't making sense to me. self critique assessment: 1
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20:22:09 `q005. How would the decimal number 659 be expressed in base 4?
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RESPONSE --> My initial thought was to do it like base ten because in base ten 6*10^2+5*10^1+9*10^0=659 but obviously this doesn't work correctly with base 4 because 6*4^2+5*4^1+9*4^0=125. I just can't seem to figure out how to convert the numbers correctly from base 10 to base 4. confidence assessment: 0
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20:27:31 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 --> For the first time during these excerises this solution actually makes sense to me. I know understand that I look for numbers multiplied by powers of 4 that do not exceed the number 659. I have to work backwards using guess and check type methods to come up with the next number multiplied by a base four number that again doesn't exceed the number and keep doing this all the way down to 4^0. self critique assessment: 2
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