# Systems of Counting

### Degrees of Accuracy

The system of counting that you use affects the accuracy of some calculations in what might seem to be an erratic way. One example can be seen in taking the square root of 3, with the decimal system. Follow the method from part 2 for taking the square root of 2.

## APPROXIMATIONS

Notice how each "place" in the decimal system, yields a closer approximation to the square root of 3. To test it, see how close squaring the root brings you to the square you started with: 3.

The first place is 1, which squared is only 1—an error of 2 from the true square of 3. Had 2 been used, the answer would have been closer: a square of 4 — an error of 1. But our rule is to stay below the true value. Another method could use the closest value before going to the next step.

The second place comes closer quickly. 1.7 squared is 2.89, reducing the error to 0.11. the third place, 1.73 yields a square of 2.9929 — an error of 0.0071. The fourth place, 1.732 comes a lot closer, making a square of 2.999824 — an error of 0.000176.

### Fractions in extended system counting

If you used a *septimal* (7s) *system*, a fraction of 1/7 would be 0.1 — completely accurate with only one place beyond the point (not a decimal point, if this system is a septimal). In the decimal system, the fraction that results from dividing by 7 isn't so easy.

Follow the same error-noting procedure. Though noting the progressive reduction of error is similar, of more interest is the kind of decimal fractions that repeat.

## FRACTIONS and DECIMALS

These problems should make you ask how accurate or reliable the figures are. What does an error of 1 part in 1 million mean? whether you happen (which is very unlikely) to be using a septimal system instead of a decimal one, just how precise is 1/7?

### Orders of magnitude

The orders of magnitude begin another whole new concept in mathematics. To show another angle of this concept, suppose you are approaching an area that consists of a perfect square. To get the area you need more accurately, you add or subtract a little bit to or from both dimensions. Starting with a square of dimension L each way, you either add or subtract small pieces S to or from each dimension. The change in area consists of two small, long slices (dimensions L by S) and one very much smaller piece that measures S both ways. The smaller S is, relative to L, the smaller S squared, relative to SL.

You could extend this concept to a similar adjustment on a cubic volume. Now, starting with a big cube, L each way, you add or subtract 3 slabs that are L square and S thick, three sticks that are L long by S square, and one very tiny cube that is S cubed. If S is 1/10 of L (and it might be much smaller), then S cubed is 1/1000 of L cubed.

## ORDERS OF MAGNITUDE

You can show the same progression algebraically. To do this, if a is a small fraction, then powers of a, a^{2}, a^{3}, a^{4}, etc., consist of a descending series of orders of magnitude. Notice that successive powers of a have a series of coefficients which, if you take the fourth power, are 1, 4, 6, 4, and 1.

Still lingering in our familiar decimal system, you substitute different values for a, and show how changing it changes the successive powers of (1 + a). If a is 0.1, successive powers begin to "spill over" into earlier "places." Up to the 4th power, the first two digits are 1.1, 1.2, 1.3, but at the 4th power, 1.5 would be nearer.

If a is 0.01, higher powers of a do not interfere with the first term, which is now in the second decimal place. Drop what follows the second place, the first two places are now 1.01, 1.02, 1.03, and 1.04. Further terms in that 4th power only make it 1.0406 in the 4th place.

However if a is 0.2, the later terms much sooner intrude into earlier ones.The blocked figures show this intrusion.

### Systems of counting

Before electronic digital devices were invented, we used counters with little wheels that carried numbers. The numbers that showed through the front window were like those that electronic digital devices display. If you took the cover off the rest of the wheel, you could see how it work, which helped you understand number systems.

The right-most wheel counted from 0 to 9 on a decimal system. When it came to 9, it would move from 9 to 0, and move the next wheel from 0 to 1. Every time the first wheel passed from 9 to 0, the next wheel would advance 1 more, until it returned to 9. Then, two wheels would read 99. As the first wheel moved from 9 to 0 this time, the next one would also move from 9 to 0, and the third wheel would move from 0 to 1, making it read 100.

### Duodecimal system

The decimal (base ten) system is not the only system that you could use. Years ago, some cultures used the *duodecimal system* — counting to twelve instead of ten. To use a wheel-counter system, you would need two more numbers on each wheel. In the wheels shown here, the extra symbols are t and e for ten and eleven. Modern digital systems more often use base 16, called the hexadecimal system.

The first six letters of the alphabet complete the single digit numbers up to what would be called 15.

Decimal

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Hexadecimal

0 1 2 3 4 5 6 7 8 9 A B C D E F

In the decimal system, " 10" (one zero) means ten. In the duodecimal system, " 10" means twelve. In hexadecimal, " 10" means sixteen. To get some exercise in different systems, use duodecimal for a bit. Vbu will see why calculators or computers use hexadecimal on the inside — they usually read out in decimal.

### Conversion from decimal to duodecimal

Why work in duodecimal when it's never used? Because something unfamiliar makes you think, it makes it easier to understand what is used. Hexadecimal is based on binary (base two), which is not as easy for systems that use a larger number base, because it's difficult to see something that is only two state (like yes or no) as counting. So, look at conversion from decimal to duodecimal.

To find how many times a number counts up to twelve, you divide the number by 12 in the familiar decimal system. The remainder at the bottom is the number of ones left over after a number of complete twelves in the quotient have been passed on to the twelves counter. Then, divide by 12 again. This time the remainder is eleven. In duodecimal, all numbers up to eleven must use a single digit, so e is used. "You can follow through the rest of this conversion. The duodecimal equivalent of decimal 143131 is 6t9e7.

**Convert decimal 143131 to duodecimal**

### Conversion from duodecimal to decimal

How do you convert from duodecimal to decimal? Simply reverse the process. Using duodecimal, divide the duodecimal number by ten however many times is necessary. You need at least the tens' column of the duodecimal multiplication table. You were probably familiar with the twelve times column—enough to do it fairly easily. However, this way you need to use the ten times column in the twelve system. This system is unfamiliar and it makes you think.

Go down the ten times column. Ten times two are 18. That means 1 twelve and 8, which you would more normally call twenty. Twelve and eight make twenty, don't they? Next, ten times 3 are 26, meaning 2 twelves and 6. Two twelves are 24, and six make what is normally called 30. Finish to the end of the column.

### DOUDECIMAL MULTIPLICATION TABLE

**Convert duodecimal 6+9e7 to decimal**

### Binary counting

The difficulty about working in binary is that each place has only two "states," which are 0 and 1 .You don't count "up to" something and then move to the next place. If you already have 1, the next 1 puts it back to 0 and passes a 1 to the next place. If you have a row of 1 s, then adding another 1 shifts them all back to 0, and passes a 1 to the next place (from right to left).

In the window panel here, the decimal number equivalent replaces the binary numbers. In the binary system, every place would be either a 1 or a 0.

### Converting decimal to binary

Here, at the top, values of places in binary that have a 1 instead of a 0, are listed as decimal. Start with the number in decimal form, 1546. First, the 11th column of binary, is 1024. that puts a 1 in the 11th column of binary. Subtract 1024 from 1546, leaving 522. Next, the 10th column in binary is 512, so subtract 512 from 522, leaving 10, and put a 1 in the 10th column of binary. With 10 left, the next binary digit that you can use is the 4th column, which is 8. So we pass over the columns from 9th to 5th, put a 1 in the 4th column, and subtract 8 from 10 (leaving 2). The 2 puts a 1 in the 2nd column of binary, which finishes the conversion.

To complete what the previous section began, the next table lists the binary equivalents for decimal numbers from 1 to 30.

### Binary multiplication

Although you enter data into your calculator or computer in the familiar decimal notation, they all use binary to perform all of the mathematical functions that they perform. Try running a sample multiplication, basically as your calculator does it. Suppose you multiply 37 by 27. First, it must convert each number to binary, which it does as you enter the numbers. I'll simplify it a bit by converting it to true binary, instead of one of the biquinary conversions that make it easier for the calculator, but more difficult for you to understand. That comes later.

Following are the conversions of 37 and 27 to pure binary.

Here is multiplication in binary, set out as you would set out ordinary long multiplication, but in a system where no numbers above 1 are "allowed." Every digit must be either 1 or 0. What it really amounts to is adding together the sequence of digits that represent 37 at every "place" where a 1-digit is in 27.

Four 1-digits are in 27, so the three 1-digits in 37 (with 0s interspersed) are entered 4 times in the proper places (to represent "27 times") and added. You can show them all added at once. However, the calculator does it. Every two Is return that place to 0 and pass a 1 to the next place to the left.

Working from the right, the first three places each have only one 1, which appears in the sum. The fourth place has two 1 s, which make a 0 in that place and pass 1 to the fifth place, which already has a 1 of its own, so it becomes 0 and passes a 1 to the sixth place. This place already has two 1 s, so that place goes to 1 again and passes a 1 to the seventh place, where again two 1 s are. This place now has a 1 and it passes a 1 to the eighth place. The eighth place has no Is, so the 1 passed is entered and that's the end of the "passing left." The remaining two places each have a single 1, which gets "brought down." The product, in binary, is 1111100111.

Convert the binary number back to decimal, by putting the decimal equivalent of each binary place where a 1 is. Adding the decimal equivalents comes to 999. To check, multiply 37 by 27, the old fashioned long way.

"Which is the long way?" you might ask. The binary way seems long to you. The only reason a calculator does it so quickly is that it performs millions of "operations" per second. It goes the long way around and calculates more quickly than you can via the short way that you are familiar with.

## Multiply 37 x 27 Binary

## BINARY MULTIPLICATION

### Alternative binary conversion

Here is another way to convert decimals to binary. It uses a table of binary equivalents for numbers from 1 to 9 in each decimal place. To illustrate its use, the two numbers for division which is following are converted to binary below the table.

Notice that the binary equivalents for a particular digit bear no relationship to one another — from one column to the next. You cannot shift a decimal point or multiply by ten by making a similar shift in binary. I'll return to what calculators or computers do about this problem in a minute.

### Binary division

Binary division shows what you learned in part 1 of this book in a rather dramatic way: division is really repetitive subtraction. Subtracting the binary for 37, which is 100101, in the top places of the dividend is exact with no remainder. What is left is the binary for 37 in the last place. So, the quotient, in binary, is 1000001.

To interpret the binary number back to decimal, use some more subtraction in binary and apply the table from the previous section. The first subtraction is the binary for 100, which leaves 11101. For the binary of 20, which leaves 1001, subtract the binary for 9. So working through binary, dividing 4773 by 37 leaves 129 as the quotient.

## ALTERNATIVE BINARY CONVERSION

## DIVIDE 4773 by 37

1001010100101 by 100101

### Special calculator binary

You noticed that binary digits for various digits in decimal changing with each decimal place makes conversion complicated. When you enter a digit on a calculator, the first digit appears on the right. When you enter the next digit, the first digit moves left and the new one appears to its right. If the calculator had to convert the digit to the new binary sequence for the next place, the system would be very complicated.

So, the calculator allocates 4 binary places for each decimal place, which requires very little more "room" in the calculator's memory than pure binary would. In effect, the calculator now "works" in decimal, but uses 4 "bits" of binary to convey each place of decimal.

### Indices

In any system of numbers, binary, octal, decimal, or hexadecimal (or even some others that are not in common use), the place of a number indicates a power of the number on which the system is based. In the binary system, according to where the 1 appears, it represents some power of 2. In the 4th place, it is the 3rd power of 2, which is 8. Here is a comparison between powers of 2 and powers of 10.

From this example, you can see some rules for using indices that help us take further short cuts in multiplication and division. First, remember that multiplication and division are short-cut methods for performing repeated addition and subtraction. Now, *indices* are short-cut methods for repeated multiplication and division.

Suppose you have to multiply x^{a} by x^{b}. The product is x^{(a+b)}. You can easily see it if you write x multiplied by itself a times, then multiply the product by x multiplied by itself b times. The total number of times that you multiply x by itself is a + b times. To illustrate, assume that a is 3 and b is 2; x^{3} multiplied by x^{2} makes x^{5}. Numerically, 2^{3} is 8, 2^{2} is 4, and 2^{5} is 32. 8 x 4 = 32. It checks.

Now try division. Dividing x^{a} by x^{b}, the quotient is x^{a-b}. You can check this answer by multiplying x by itself a times the numerator of a fraction and using x times itself b times for the denominator. You can cancel b times the number of x's in the numerator and leave a residue of x's in the numerator that is (a — b) times. To illustrate, make a = 5 and b = 2. x^{5} divided by x^{2} equals x^{3}. If you used 2 for x, x^{5} is 32, x^{2} is 4, and x^{3} is 8. 32 divided by 4 is 8.

### Roots: inverse of powers

Here, you must distinguish between the inverse of a number and the inverse of a power. A *minus index* is the inverse or reciprocal of the number raised to the power, indicated by the index. *Roots* are the opposite of powers. For example, because 2^{2} is 4,4^{1/2} is 2; 2^{3} is 8, so 8^{1/3} is 2; 2^{4} is 16, so 16^{1/4} is 2, and so on.

Fractional indices represent roots. The 3/2 power of 4 is 8, the square root of 4 is 2, and 2^{3} is 8. Reversing this process, 8^{2/3} is 4. You can find other numbers in roots by the process of square root. For example, 2^{1/2} (the square root of 2) is 1.414, etc.; 8^{1/2} is twice this. Why? Since 4^{1/2} is 2 and 2^{1/2} is 1.414, (2 times 4)^{1/2} is 8^{1/2} (twice 1.414), which is 2.828.

You are not restricted to square roots, or indeed to any specific roots. Now, a whole new field of numbers is opened.

### Surds and indices

Introducing surds is actually going back to a virtually obsolete way of writing roots. Before the fraction index notation introduced in the previous section came into vogue, it was customary to use a surd in front of a number to indicate its square root. Thus, a surd in front of x represented the square root of x, the same as xl/ . Putting a 3 in front of the surd indicated the cube root of x, instead of the square root. Putting a small n or any other letter or number in front of the surd likewise represented a specific root. If the number under the surd has a power b and an a in front of the surd, the expression can be written as: x^{b/a}. A surd followed by a *vinculum* over (line over the top of) a^{2} + b^{2} is the root of the whole expression. This expression can be written: (a^{2} + b^{2})^{1/2}.

### Questions and problems

Note: The questions and problems here are not in graded order. They assume a knowledge of earlier parts of this book. If you have difficulty with a problem, try others first and then return to the one that is difficult. These questions are designed so that you have to exercise some initiative in applying the principles that have been introduced up to this point.

1. Find the decimal equivalent of the fraction 1/37. Determine the error that occurs when this decimal equivalent is found to three significant digits.

2. Using the binary system, multiply 15 by 63 and convert back to the decimal system. Check your result by directly multiplying the decimal numbers.

3. Using the binary system, divide 1922 by 31 and convert back to the decimal system. Check your result by directly dividing the decimal numbers.

4. Find the values of the following expressions:

(a) 16^{3/4} (b) 243^{0,8} (c) 25^{1,5}

(d) 64^{2/3} (e) 343^{4/3}

5. Convert the following numbers from decimal to binary. As a check, convert them back.

(a) 62 (b) 81 (c) 111

(d) 49 (e) 98 (f) 222

(g) 650 (h) 999 (i) 2000

6. Convert the following numbers from binary to decimal. As a check, convert them back.

(a) 101 (b) 1111 (c) 10101

(d) 111100 (e) 110111000110

7. Multiply 129 by 31 in the decimal system. Multiply the binary equivalents of these numbers. Suppose an error is made in the second digit from the right in the second number in the decimal product, so 129 is multiplied by 41 instead of by 31. Suppose a similar error occurs in the binary system, so the second digit from the right in the second number is reversed. Compare the relative error in the decimal system with the error in the binary system.

8. Evaluate the expression (a^{2} + b^{2})^{1/2} for the following values:

(a) a = 4 and b = 3 (b) a = 12 and b = 5

(c) a = 24 and b = 7 (d) a = 40 and b = 9

(e) a = 60 and b = 11 (f) a = 84 and b = 13

(g) a= 112 and 6= 15

What does each pair have in common?

9. Evaluate the expression (a^{2} + b^{2})^{1/2} for the following values:

(a) a = 8 and b = 6 (b) a = 15 and b = 8

(c) a = 24 and b = 10 (d) a = 35 and b = 12

(e) a = 48 and b = 14 (f) a = 63 and b = 16

What does each pair have in common?

10. Write as simple decimal numbers, without fractions, the following expressions:

(a) 100^{2} (b) 100^{1/2}

(c) 100^{-2} (d) 100^{-1/2}

From these four values, find the values of the following expressions by the method of adding and subtracting indices:

(e) 100^{3/2} (f) 100^{5/2}

(g) 100^{-3/2} (h) 100^{-5/2}

11. Using a calculator's square root function button only, evaluate the following to at least three decimal places:

(a) 100^{1/4} (b) 100^{1/8}

(c) 100^{1/16} (d) 100^{1/32}

12. As the value of the exponent in the previous problem is repeatedly cut in half, that is, 1/64, 1/128, 1/256, 1/512, and so on, what number will the expression approach? Why?

13. Find values, correct to three decimal places, for the following:

(a) 32^{0,1} (b) 32^{0,2} (c) 32^{0,3}

(d) 32^{0,4} (e) 32^{0,5} (f)32^{0,6}

(g) 32^{0,7} (h) 32^{0,8} (i) 32^{0,9}

14. Evaluate the following expressions, using a calculator if you wish. Where applicable, render expressions to at least three decimal places:

(a) (10^{2} - 2^{6})^{1/2} (b) (36^{2} - 8^{3})^{1/2} (c) (28^{2} - 21^{2})^{1/3}

(d) (5^{2} - 3^{2})^{1/4} (e) (17^{2} - 15^{2})^{1/6} (f) 6561^{1/2}

(g) 6561^{-1/2} (h) 6561^{1/4} (i) 6561^{-1/4}

(j) 6561^{1/8} (k) 6561^{-1/8}