Base Number Equivalence TablesDate: 10/30/2001 at 13:51:39 From: Laurianne Brown Subject: Base Numbers Tables My son needs to show a table for base 3 and base 7. Help, please. I have looked at your examples on converting bases but I am still not clear. Date: 10/31/2001 at 11:30:09 From: Doctor Greenie Subject: Re: Base Numbers Tables Helo, Laurianne - I don't know whether you are looking for a table of addition or multiplication facts for base 3 and base 7, or just a table of the base-3 and base-7 equivalents of the first several counting numbers. I will guess that for a 12-year-old you are only looking for a table of the equivalents of our familiar base-10 numbers. One way to get a hands-on approach to understanding different number bases is to consider counting money. For the purposes of this exercise, for counting in base b, we will have bills in the denominations of 1 dollar, b dollars, b^2 (b times b) dollars, b^3 (b times b times b) dollars, and so on; and we will have only (b-1) bills of each denomination to work with. So in the familiar base 10, we have 9 1-dollar bills, 9 10-dollar bills (10 times $1), 9 100-dollar bills (10 times $10), and so on. To represent the counting numbers from 1 to 20 using these bills, we have number 10's / 1's total --------------------------------------- 1 0 1 0(10) + 1(1) = 1 2 0 2 0(10) + 2(1) = 2 3 0 3 0(10) + 3(1) = 3 ... 8 0 8 0(10) + 8(1) = 8 9 0 9 0(10) + 9(1) = 9 10 1 0 1(10) + 0(1) = 10 11 1 1 1(10) + 1(1) = 11 12 1 2 1(10) + 2(1) = 12 ... 18 1 8 1(10) + 8(1) = 18 19 1 9 1(10) + 9(1) = 19 20 2 0 2(10) + 0(1) = 20 When we reached the number 9, we had used all 9 of the 1-dollar bills we have; so when we went to the next counting number, we "traded in" the 1-dollar bills for one bill of the next denomination. Similarly, when we reached 19 (represented with 1 10-dollar bill and 9 1-dollar bills), we had again used all 9 of the 1-dollar bills we have, so when we went to the next counting number we again traded in the 1-dollar bills for a second bill of the next larger denomination. Now let's count from one to twenty in base 7. For base 7, we have 6 1-dollar bills, 6 7-dollar bills (7 times 1 dollar), 6 49-dollar bills (7 times 7 dollars), and so on. To make the numbers one to twenty using these bills, we have number number (base 10) 7's / 1's (base 7) total (base 10) --------------------------------------------------------- 1 0 1 1 0(7) + 1(1) = 0+1 = 1 2 0 2 2 0(7) + 2(1) = 0+2 = 2 3 0 3 3 0(7) + 3(1) = 0+3 = 3 4 0 4 4 0(7) + 4(1) = 0+4 = 4 5 0 5 5 0(7) + 5(1) = 0+5 = 5 6 0 6 6 0(7) + 6(1) = 0+6 = 6 7 0 0 10 1(7) + 0(1) = 7+0 = 7 8 1 1 11 1(7) + 1(1) = 7+1 = 8 9 1 2 12 1(7) + 2(1) = 7+2 = 9 10 1 3 13 1(7) + 3(1) = 7+3 = 10 11 1 4 14 1(7) + 4(1) = 7+4 = 11 12 1 5 15 1(7) + 5(1) = 7+5 = 12 13 1 6 16 1(7) + 6(1) = 7+6 = 13 14 2 0 20 2(7) + 0(1) = 14+0 = 14 15 2 1 21 2(7) + 1(1) = 14+1 = 15 16 2 2 22 2(7) + 2(1) = 14+2 = 16 17 2 3 23 2(7) + 3(1) = 14+3 = 17 18 2 4 24 2(7) + 4(1) = 14+4 = 18 19 2 5 25 2(7) + 5(1) = 14+5 = 19 20 2 6 26 2(7) + 6(1) = 14+6 = 20 In base 7, we used up all our 1-dollar bills when we reached the number 6; when we went to the next counting number, we traded in the 1-dollar bills for a single bill of the next higher denomination ($7). And when we reached the counting number 13 (base 10), represented in base 7 by 1 7-dollar bill and 6 1-dollar bills, we had again used all the 1-dollar bills, so when we went to the next counting number we again traded in the 1-dollar bills for a second bill of the next higher denomination. And now let's count from one to twenty in base 3. For base 3, we have 2 1-dollar bills, 2 3-dollar bills (3 times 1 dollar), 2 9-dollar bills (3 times 3 dollars), 2 27-dollar bills (3 times 9 dollars) and so on. To make the numbers one to twenty using these bills, we have number 9's / number (base 10) 3's / 1's (base 3) total (base 10) -------------------------------------------------------------------- 1 0 0 1 1 0(9) + 0(3) + 1(1) = 0+0+1 = 1 2 0 0 2 2 0(9) + 0(3) + 2(1) = 0+0+2 = 2 3 0 1 0 10 0(9) + 1(3) + 0(1) = 0+3+0 = 3 4 0 1 1 11 0(9) + 1(3) + 1(1) = 0+3+1 = 4 5 0 1 2 12 0(9) + 1(3) + 2(1) = 0+3+2 = 5 6 0 2 0 20 0(9) + 2(3) + 0(1) = 0+6+0 = 6 7 0 2 1 21 0(9) + 2(3) + 1(1) = 0+6+1 = 7 8 0 2 2 22 0(9) + 2(3) + 2(1) = 0+6+2 = 8 9 1 0 0 100 1(9) + 0(3) + 0(1) = 9+0+0 = 9 10 1 0 1 101 1(9) + 0(3) + 1(1) = 9+0+1 = 10 11 1 0 2 102 1(9) + 0(3) + 2(1) = 9+0+2 = 11 12 1 1 0 110 1(9) + 1(3) + 0(1) = 9+3+0 = 12 13 1 1 1 111 1(9) + 1(3) + 1(1) = 9+3+1 = 13 14 1 1 2 112 1(9) + 1(3) + 2(1) = 9+3+2 = 14 15 1 2 0 120 1(9) + 2(3) + 0(1) = 9+6+0 = 15 16 1 2 1 121 1(9) + 2(3) + 1(1) = 9+6+1 = 16 17 1 2 2 122 1(9) + 2(3) + 2(1) = 9+6+2 = 17 18 2 0 0 200 2(9) + 0(3) + 0(1) = 18+0+0 = 18 19 2 0 1 201 2(9) + 0(3) + 1(1) = 18+0+1 = 19 20 2 0 2 202 2(9) + 0(3) + 2(1) = 18+0+2 = 20 When we count in base 3, we only have 2 bills of each denomination; we use up both of the 1- and 3-dollar bills to make the counting number 8, so when we go to the next counting number, we need to trade in all the smaller bills for a single bill of the third denomination, $9. I hope this helps. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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