 |
||
homepage
previous -- view my teaching portfolio here
|
|
Math Notes
Introduction The numeral system we use in the Western World is a system called the Decimal System or a Base-10 system. This numeral system comes to us pretty naturally because we're used to it, so it may seem a little weird to explain it, but in order to understand how other systems work, it helps to see how those systems are similar to the one we're used to. So let's break it down a bit. Let's consider the number 123. Starting from the right side, this number represents 3 x 1 + 2 x 10 + 1 x 100, or, using scientific/exponential notation, 3 x 100 + 2 x 101 + 1 x 102. For larger numbers, we keep going multiplying each digit by ten (since ten is the base) as we go to the left. So, another example: 98,765,432 = 2 x 100 + 3 x 101 + 4 x 102 + 5 x 103 + 6 x 104 + 7 x 105 + 8 x 106 + 9 x 107. Notice that if you want to start writing from the left, count the number of digits in the number, then the first exponent is one less since we have to end at zero and not one. We can also go the other way. Suppose we are given the following and asked to write the number it represents: 3 x 106 + 4 x 105 + 1 x 102 + 6 x 101 + 9 x 100. Be careful with this one. Whenever you skip a power of ten, it's the same thing as adding 0 x 104 (or whatever power of ten you're missing). So when you are building the number, replace the missing power of ten with a zero. We end up 3,400,169. The powerful feature of such a system is that we can write a number of any size with only ten symbols: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. These are the Arabic numerals.
Worked Examples Let's consider a couple simple problems and see how the base-10 system works in arithmetic. (You will want to refer back to these examples when you work in other bases. Simply replace the 10 everywhere with the base you are using.) A. 16 + 19. When we add, we start on the right side and add the last two digits. 6 + 9 = 15. This number is larger than nine, and thus cannot be written with a single digit. We should subtract 10 (or some multiple of ten) until we have a number remaining which is nine or smaller. That is the number in the ones digit. Record it. The tens that we subtracted must now be added to the next column, since it's a ten and that column records the number of tens we have. Now we have 1 + 1 from the original problem and + 1 that we had leftover from the addition of the ones digits. 1 + 1 + 1 = 3. So we end up with 16 + 19 = 35 by recording the ones digit result in the ones (100) place, and the tens (101) digit result in the tens place. B. Let's try a subtraction problem like 41-28. We wish to begin in the rightmost digit as we did with addition. 1-8. Well, we're going to end up with a negative number and we can't write numbers with negative digits. So we need to adjust the first number so that it's large enough to do the subtraction with. So we borrow a ten from the nearby tens digit. 10 + 1 = 11. So now 11 - 8 = 3. That's our result for the ones digit. For the next digit, we already borrowed away a ten, so the 41 is now not 40 + 1, but 30 + 11. So now subtract 30 - 20 or 3 - 2. We get a 1 for the tens digit. Now put them together. 41 - 28 = 13. C. Multiplication works very similarly to addition, at least with respect to the base. So let's consider division. 3285 ÷ 9. In the case of division, we start on the left. 9 divides into 3 not at all, so we take another digit. 9 divides into 32 three times. 32 - (9*3) = 32 - 27 = 5. Since the 32 actually represents 3200 (32 x 100), our 3 represents 300 (3 x 100), and our 5 represents 500 (5 x 100). We still have this 500 and the 85 we haven't accounted for yet. We start over dividing 9 into 585 and repeat. 9 doesn't divide into 5, so we take another digit. 9 divides into 58 a total of 6 times and then some. The 58 represents 580 (58 x 10), so our 6 represents 60 (6 x 10). And our leftover is 580 - (9*60) = 580 - 540 = 40 (4 x 10); i.e. our remainder on this step is 4. Again, we need to repeat. We have 40 leftover from this step, and a 5 still unaccounted for, or 45. 9 divides into 45 a total 5 times, and evenly. This was 45 ones, so we can stop. Now add them up. 5 ones + 6 tens + 3 hundreds or 365. 9 * 365 = 3285.
Problem Solving Tips For writing the expanded form, the important thing to remember is that the farthest right digit is the one's digit or the base raised to the zero power (100). When you are converting the expanded form into digits, it helps to write out all the possible powers from zero to the largest one given, and then fill in the digits in the table. Wherever there is a blank space, replace that with a zero. When working with bases other than base-10, refer back to these base-10 procedures and replace 10 everywhere with the base that you are using. Then follow the same steps.
Additional Problems with Answers 1. Answers: 1. Links to Outside Sources Arabic
numerals - Wikipedia, the free encyclopedia Links to Supporting Topics
Numbers II -- Binary, Hexadecimal and Other
Bases History Numeral system
- Wikipedia, the free encyclopedia |
| (c) 2006, 2005, 2004 by Betsy McCall, all
rights reserved To contact the webmistress, email betsy@pewtergallery.com |
||