September 29, 2010
Today, as always, Mr. B. started off class with a practice problem on the smartboard. Today’s problem was:
The subscripts on the values were meant that those values are in a different base system. Our numeral system is based off of a base 10 system, or a decimal system. This means that each place value of each of the numbers that we write are in a multiple of ten; the ones place is 10^0, the tens place is 10^1, the hundreds place is 10^2, and so on. Each digit of a value expressed in base 10 notation shows how many of that power of 10 are in the value. For instance, in the number 2,345, there are two 10^3′s, three 10^2′s, four 10^1′s, and five 10^0′s. When these values are multiplied and then added together, we arrive at what we know as two thousand, three hundred and forty-five. The same principle holds true for different base systems. For instance, if we are using a base “x” system, the value furthest to the right is how many x^0′s are included in the value. The digit one to the left is how many x^1′s are involved, then x^2, x^3, and so on. Let’s use a base eight system, since it is involved in our problem. This means that in this system, 66 is actually equal to 54 in a base 10 system. This is because the 6 nearest to the right is in the “ones” place; that is, it is in the place counting how many 8^0′s are involved in this problem. Since 8^0=1, six ones are equal to 6. The six on the left is a bit more tricky. This spot is how many 8^1′s are involved in this values, so this 6 digit is in the “eights” spot. Six 8^1′s = 48. Since 48+6=54, 66 base 8 is equal to 54 in base 10 notation.
To continue with this problem, we will convert all of the rest of the additives to base 10 notation, since it is the one we are most familiar with, and then convert our sum to base 8 notation to come to our conclusion. 132 in base four notation is equal to 30 base 10;
2*4^0 + 3*4^1 + 1*4^2
= 2 + 12 + 16
=30 in base 10 notation.
1011 base 2 =
1*2^0 + 1*2^1 + 0*2^2 + 1*2^3
= 1 + 2 + 0 + 8
= 9 in base 10 notation.
When you add up all of these values (54+30+9) we find that these three values add to 95 base 10. However, we must convert this value into base 8 notation. There are no 8^3′s in this value; 8^3=512, and there are obviously no 512′s in 95. Therefore, let’s start with 8^2. 8^2=64, a number that clearly goes into 95 only once. This gives us a “1″ in the “sixty-fours” place:
1 _ _
Once we divide by 64, there is a remainder of 31 that still must be accounted for in our number. 8^1=8, and there are three 8^1′s in 31:
1 3 _
This leaves us with a remainder of 7 since 8*3=24, and 31-24=7. Our final place is represented by 8^0, or, in other words, 1. There are seven ones in seven, so the final “ones” place is a seven:
1 3 7 base 8.
This is our answer.
The rest of the class period was spent examining our test that we labored over yesterday and asking any questions about problems that we were confused on.
Here are a couple of notes regarding the different numeral systems:
- Each digit in a base system cannot be larger than the base value; that is, if we are using a base 4 system, every digit must be between 0 and 3. This is because once a value reaches 4, it carries over into the next place value.
- The binary system, or base two, and the hexadecimal, or base 16 systems are the ones most commonly used (besides base 10, of course). Base 16 uses the digits 0-9 and then the letters a-e to represent a value.
Here is a website in which all of the different notations of a lot of values are listed in different base systems for a quick reference.
http://home.comcast.net/~igpl/NXO.html
Hope this helped!!
The next scribe will be Allison T.