_______________________________________________________________________________________

NEGATIVE NUMBERS. In the previous lesson, one of your answers involved a temperature below zero. The answer was given as a negative number. Sometimes it is easier to understand negative values if you use a device such as the number line represented below. Positive numbers are to the right of the zero; negative numbers are to the left of the zero. (Zero is usually classified as a "non-negative" number.) Both ends of the number line extends to infinity (without end). All whole numbers, fractions, decimals, and mixed numbers are represented on the number line. Negative number are denoted by a negative or minus sign (–) before the number. Positive numbers are denoted by a positive or plus sign (+) before the number or by having no sign before the number.

_______________________________________________________________________________________

You have reviewed how to add, subtract, multiply, and divide positive numbers in the preceding lessons. This lesson will give you rules for adding, subtracting, multiplying, and dividing when negative numbers are involved.

One concept that is helpful when working with negative numbers is "absolute value." Absolute value pertains to the numerical value of a figure without regard to whether it is a positive or negative number. Using the number line, it is the number's distance from zero without regard whether it is to the right or left of the zero. Another way of think of absolute value is that positive numbers stay positive and negative numbers become positive. The symbol for absolute value is two parallel lines with the number between the lines.

The absolute value of negative eight can be written as - -8 - .

If you add the absolute values of two numbers, you add the values of the numbers without regard to whether the numbers are positive or negative. For example

- ^–2 - + - ^–3 - = - ⁺2 - + - ⁺3 - = - ⁺2 - + - ^–3 - = - ^–2 - + - ⁺3 - = 5

The absolute value of negative five ( - ^–5 - ) is the same as the

absolute value of ____________________________ .

Solution to Frame 5-1.

negative

right

_______________________________________________________________________________________

(1) Change both numbers to absolute values.

(2) Add their absolute values

(3) Place a positive symbol (or no symbol) in front of

the sum.

Add: ⁺5 + ⁺3 Answer _____________

_______________________________________________________________________________________

(1) Change both numbers to absolute values.

(2) Add their absolute values.

(3) Place a negative symbol in front of the sum.

_______________________________________________________________________________________

(1) Change both numbers to absolute values.

(2) Subtract the smaller absolute value from the larger

absolute value.

(3) Place the original sign of the larger absolute value in front

of the difference (remainder).

a. Add: ⁺5 + ^–3 Answer _____________

Solution to Frame 5-4.

^–8

^–5 + ^–3 = - ^–5 - + - ^–3 - =

_______________________________________________________________________________________

In exercise "a" above, did you notice that adding a "smaller" negative number to a positive number is much like subtracting a smaller positive number from a larger positive number?

Also, the answers to "a" and "b" had the same absolute value. The difference is that the first problem resulted in a positive answer and the second resulted in a negative answer.

Remember: Add the absolute values if the signs of the two numbers are the same (both positive, both negative) and ____________ the absolute values if the signs of the two numbers are not the same (one positive, one negative).

a. ⁺2

⁺5 + ^–3 =

_______________________________________________________________________________________

Change the sign of the subtrahend (second or bottom

number), then add the two numbers according to the

rules for addition given in Frames 5-3, 5-4, and 5-5.

Subtract these numbers (both numbers are positive).

a. Subtract: ⁺5 – ⁺3 Answer _____________

b. Subtract: ⁺3 – ⁺5 Answer _____________

_______________________________________________________________________________________

Subtract these numbers (both numbers are negative).

a. Subtract: ^–5 – ^–3 Answer _____________

b. Subtract: ^–3 – ^–5 Answer _____________

a. ⁺2

⁺5 – ⁺3 =

⁺5 + ^–3.

(Add pos and neg)

- ⁺5 - – - ^–3 -

⁺ - 5 – 3 - = ⁺2

b. ^–2

⁺3 – ⁺5 =

⁺3 + ^–5.

(Add pos and neg)

- ^–5 - – - ⁺3 -

_______________________________________________________________________________________

Subtract these numbers (positive minus negative).

a. Subtract: ⁺5 – ^–3 Answer _____________

b. Subtract: ⁺3 – ^–5 Answer _____________

a. ^–2

^–5 – ^–3 =

^–5 + ⁺3.

(Add pos and neg)

- ^–5 - – - ⁺3 - =

_______________________________________________________________________________________

Subtract these numbers (negative minus positive).

a. Subtract: ^–5 – ⁺3 Answer _____________

b. Subtract: ^–3 – ⁺5 Answer _____________

a. ⁺8

⁺5 – ^–3 =

⁺5 + ⁺3.

(Add pos and pos)

- ⁺5 - + - ⁺3 -

⁺ - 5 + 3 - = ⁺8

b. ⁺8

⁺3 – ^–5 =

⁺3 + ⁺5.

(Add pos and pos)

- ⁺3 - + - ⁺5 -

_______________________________________________________________________________________

(1) Change both numbers to absolute values.

(2) Multiply the absolute values.

(3) Place the appropriate sign in front of the product.

(a) If the original numbers have the same sign (both positive or both negative), the sign of the product is positive.

(b) If the original numbers have different signs (one positive and one negative), the sign of the product is negative.

Solve these problems

a. Multiply: ⁺5 x ⁺3 Answer _____________

b. Multiply: ^–5 x ⁺3 Answer _____________

c. Multiply: ⁺5 x ^–3 Answer _____________

d. Multiply: ^–5 x ^–3 Answer _____________

a. ^–8

^–5 – ⁺3 =

^–5 + ^–3.

(Add neg and neg)

- ^–5 - + - ^–3 -

^– - 5 + 3 - = ^–8

b. ^–8

^–3 – ⁺5 =

^–3 + ^–5.

(Add neg and neg)

- ^–3 - + - ^–5 -

_______________________________________________________________________________________

(1) Change both numbers to absolute values.

(2) Divide the divisor into the dividend (just like both numbers were positive).

(3) Place the appropriate sign in front of the quotient.

(a) If the original divisor and dividend have the same sign (both positive or both negative), the sign of the quotient is positive.

(b) If the original divisor and dividend have different signs (one positive and one negative), the sign of the quotient is negative.

Solve the following problems.

a. ⁺8 ) ⁺4 Answer _____________

b. ^–8 ) ⁺4 Answer _____________

c. ⁺8 ) ^–4 Answer _____________

d. ^–8 ) ^–4 Answer _____________

a. ⁺15

(pos x pos = pos)

b. ^–15

(neg x pos = neg)

c. ^–15

(pos x neg = neg)

d. ⁺15

_______________________________________________________________________________________

SCIENTIFIC NOTATION. When you are looking at a very large or very small number, one with a lot of zeroes, do you ever wish that someone would count the zeroes for you so you would know what the number is?

Well, there is a system that does that for you. It is called "scientific notation." Scientific notation is a method of writing numbers in terms of the powers of 10. You have already studied the powers of 10 in Lessons 1 and 4 of this subcourse.

Basically 10^P (ten to the p-th power) is a "1" followed by "P" zeroes if the "P" is positive. If the "P" is negative, then "P" is the number of decimal places to the right of the decimal point [decimal point followed by (P-1) zeroes followed by the numeral one (1).]

NOTE: The superscripted (raised) P ("^P") is called the "exponent."

a. 10 ⁷ is written as __________________________________

b. 10! ⁷ is written as _________________________________

a. ⁺2

(pos ) pos = pos)

b. ^–2

(neg ) pos = neg)

c. ^–2

(pos ) neg = neg)

d. ⁺2

_______________________________________________________________________________________

Convert the following numbers to 10^P format.

a. 10,000 _____________

b. 0.00001 _____________

a. 10,000,000

_______________________________________________________________________________________

The above works for numbers that consist of one "1" and any number of zeroes, but how about other numbers like 32,700,000,000,000 or

0.000000000065?

In scientific notation, a number is reduced to a standard form. That form is a number between 1 and 10 (the number can be a decimal such as 3.475) followed by "x 10^P " with "P" being the power needed to restore the new number to the original number. The number before the "times" symbol (x) is sometimes called the "coefficient."

Remember: scientific notation begins with a positive single digit (1, 2, 3, 4, 5, 6, 7, 8, or 9) which may or may not be followed by a decimal point and additional digits.

For example: 32,700,000,000,000 can be converted into scientific notation.

Place a decimal point at the end of the number (following the zero in the unit's [one's] position.

Move the decimal point so that it falls after the "3" (the first non-zero digit of the original number starting from the left). Count the number of places you moved the decimal.

You moved the decimal point 13 places to the left to obtain

3.2700000000000. Therefore, the exponent "P" equals 13.

3.2700000000000 x 10 ¹³ = 32,700,000,000,000 .

The zeros are usually dropped as long as no non-zero digit follows.

32,700,000,000,000 = 3.27 x 10 ¹³. The scientific notation stands for the product of the following multiplication problem:

3.27 x 10 x 10 x 10 x10 x 10 x 10 x 10 x 10 x10 x 10 x 10 x 10 x 10.

Convert the following numbers to scientific notation.

a. 38,000,000 ________________

b. 40,100 ________________

a. 10 ⁴

b. 10 - ⁵

_______________________________________________________________________________________

In problem "b" of the previous frame, the zero between the four and the one is not dropped. Only zeroes following the last non-zero digit can be dropped without changing the value of the number.

However, you may want to round the number to make it easier to use; that is, you do not need a high degree of accuracy. For example,

4.01 x 10 ⁴ rounded to the nearest ten-thousand can be expressed as 4 x 10 ⁴.

Note: If you rounded 4.01 x 10 ⁴ to the nearest thousand, you would also get 4 x 10 ⁴.

a. Change 3,756 to scientific notation. ________________

b. Change 3,756 to scientific notation

rounded to the nearest thousand. ________________

(Hint: Round first, then change to scientific notation.)

a. 3.8 x 10 ⁷

_______________________________________________________________________________________

The preceding frames work fine for large numbers, but how about small numbers, like. 0.000000000065?

Begin at the decimal point.

Move the decimal point to the right until it passes the first non-zero number (the "6"). Count the number of places you are moving the decimal.

You moved the decimal point 11 places to the right to obtain 6.5. Therefore "P" equals "- 11." Remember, if you move the decimal point to the right, the exponent will be negative.

0.000000000065 = 6.5 x 10 ^–11 .

Convert the following numbers to scientific notation.

a. 0.0072 _____________

b. 0.101 _____________

c. 3.9 _____________

[Note: Remember that 10 (or any number) raised to the zero power is "1" and that any number times "1" remains the original number.]

a. 3.756 x 10 ³

_______________________________________________________________________________________

To convert a number in scientific notation back to normal, you move the decimal point based upon the power of 10 (the "P"). If the "P" is positive, move the decimal point "P" places to the right.

If the number in scientific notation form does not have a decimal point (that is, the coefficient is a whole single digit number), place the decimal point following the coefficient.

a. Convert 4.5 x 10 ⁴ to normal format. _________

b. Convert 7 x 10 ⁶ to normal format. _________

a. 7.2 x 10 ^–3

b. 1.01 x10 ^–1

c. 3.9 x 10 ⁰

[3.9 x 10 ⁰ =

_______________________________________________________________________________________

If the "P" is negative, convert a number in scientific notation to normal format by moving the decimal point "P" places to the left.

a. Convert 4.5 x 10 ^–4 to normal format. _________

b. Convert 7 x 10 ^–6 to normal format. _________

a. 45,000

_______________________________________________________________________________________

Scientific notation can make multiplication a little easier (or at least look neater).

Consider the problem: 32,000 x 10,200,000,000.

By using scientific notation, you can change the appearance of the problem to 3.2 x 10 ⁴ times 1.02 x 10 ¹⁰.

To solve the problem, you

(1) Multiply the coefficients (numbers in front).

(2) Add the exponents (the powers of 10). The sum will be

your new power of 10.

(3) Rewrite the answer so it is in scientific notation format,

if needed.

3.2 x 10 ⁴ X 1.02 x 10 ¹⁰ = 3.2 x 1.02 x 10 ⁴⁺¹⁰ = 3.264 x 10 ¹⁴

Solve this problem:

3 x 10 ⁵ X 7 x 10 ⁶ ___________________________

a. 0.00045

_______________________________________________________________________________________

NOTE: When you multiplied the coefficients to the problem in Frame 5-20, you got a new coefficient that was above 10 (21 x 10 ¹¹). However, Frame 5-15 states that the coefficient should be between 1 and 10. That is, the number to the left of the decimal point is to be a single digit.

Since 21 = 2.1 x 10 ¹

Then 21 x 10 ¹¹ = 2.1 x 10 ¹. x 10 ¹¹ = 2.1 x 10 ¹⁺¹¹ = 2.1 x 10 ¹².

Solve these two problems:

a. 3 x 10 ^–4 times 2 x 10 ^–3 = __________________

b. 5 x 10 ⁴ times 2 x 10 ^–6 = __________________

NOTE: Positive and negative exponents are added using the same rules for adding positive and negative numbers.

Solution to Frame 5-20.

21 x 10 ¹¹

_______________________________________________________________________________________

Scientific notation can also be used in division.

Consider the problem: 800,000 ) 200,000,000.

By using scientific notation, you can change the appearance of the problem to 8 x 10 ⁵ divided by 2 x 10 ⁸.

To solve the problem, you

(1) Divide the coefficients (numbers in front).

(2) Subtract the exponents. The difference will be your

new power of 10.

(3) Rewrite the answer so it is in scientific notation format,

if needed.

8 x 10 ⁵ ) 2 x 10 ⁸ = 8) 2 x 10 ^5–8 = 4 x 10 ^–3 (or 0.004)

Solve this problem:

5 x 10 ⁷ ) 4 x 10 ⁵ ___________________________

a. 6 x 10 ^–7

(3x2 x 10 ^–(4+3)

b. 10^–1 or 0.1

5x2 x 10 ^4–6

10 x 10^–2

1 x 10¹ X 10^–2

1 x 10^1–2

_______________________________________________________________________________________

Solve these division problems. Remember to use the rules for subtracting negative numbers.

a. 8 x 10 ⁷ ) 4 x 10 ^–5 ___________________________

b. 4 x 10 ^–7 ) 8 x 10 ^–5 ___________________________

c. 4 x 10 ^–7 ) 3 x 10 ³ ___________________________

(Hint: Round to the second place following the decimal.)

Solution to Frame 5-22.

a. 1.25 x10² or 125

5/4 x 10 ^7–5 =

_______________________________________________________________________________________

When a number is multiplied by itself, it is said to be "squared." For example, five times five equals twenty-five. This can be written as

5 x 5 = 25 or as 5² = 25. In the second method, the statement can be read as "five to the second power equals twenty-five" or as "five squared equals twenty-five."

The term "square" comes from the formula for determining the area of a square, which is s² (the length of one side of the square multiplied by itself). To "square" a number, multiply the number by itself.

Square the following numbers:

a. 15 ________

b. 0.03 ________

c. 1/2 ________

a. 2 x10¹²

8/4 x 10 ^{7– (– 5)} =

2 x 10 ⁷⁺⁵

b. 5 x 10 ^–3

4/8 x 10 ^{–7– (– 5)} =

0.5 x 10 ^–7+5 =

5 x 10 ^–1 x 10 ^–2

= 5 x 10 ^–(1+2)

c. 1.33 x 10 ^–10

4/3 x 10 ^{–7 – (+3)} =

1.3333 x 10 ^{–7 + (–3)} =

_______________________________________________________________________________________

Did you notice that the squares for "b" and "c" above are smaller than the original number?

For positive numbers, the "square" is larger than the original number if the original number is greater than 1 and is smaller than the original number if the original number is less than 1.

The square of a negative number will be a ____________ number.

a. 225

b. 0.0009

_______________________________________________________________________________________

In some problems, you have a number and need to know what number squared equals that number. This is called "square root." For example, five squared is twenty-five; therefore, the square root of twenty-five is five.

This is about the same as saying that the area of a square is 25 square meters. What is the length of one side of the square?

The "square root" is the reversal of the square function.

The symbol for square root is .

"The square root of 25" is written as .

Another way of indicating square root is using "1/2" as the exponent (power of 10). For example: 25 ^1/2 = 5.

See if you know the square roots of the following numbers.

a. 625 ____________

b. 0004 ____________

c. 1/81 ____________

[Hint: Take a guess, then square your guess. Adjust your guess higher or lower until you hit upon the answer.]

_______________________________________________________________________________________

You may have been able to guess the square roots of the preceding problems, but you can see that this would not be a good way to determine the square root of numbers like 37,594,227.0374.

The easiest way of finding the square root of a number is to use a hand-held calculator or a computer. There is a method for calculating the square root of a number using paper and pencil, but this method takes time and will not be presented here.

Sometimes you may just want a good guess as to the square root or you may just want to know where the decimal point goes.

In such cases, it is useful to have a method of determining the approximate square root of a number. First, review the squares of the integers from 1 to 9. Remember, the square root of the square is the original number.

1² = 1 therefore 1^1/2 = 1

2² = 4 therefore 4^1/2 = 2

3² = 9 therefore 9^1/2 = 3

4² = 16 therefore 16^1/2 = 4

5² = 25 therefore 25^1/2 = 5

6² = 36 therefore 36^1/2 = 6

7² = 49 therefore 49^1/2 = 7

8² = 64 therefore 64^1/2 = 8

9² = 81 therefore 81^1/2 = 9

NOTE: A number that is an exact square of an integer (whole number) is sometimes called a "perfect square." The square root of a perfect square is an integer.

If the square of 387 is 149,769, then the square root of 149,769

is _____________ .

a. 25

b. 0.02

_______________________________________________________________________________________

The methods for estimating square root differs slightly depending upon whether the number is greater than 1 or less than 1. Let's begin with numbers that are greater than 1.

(1) Pair off the digits of the number beginning at the decimal point (or where the decimal point would be if the number had one) and going to the left. Drop any digits to the right of the decimal point.

(2) Identify the last digit or pair of digits (the digit or pair of digits at the beginning of the number). If the number had an even number of digits to the left of the decimal, you will have a pair of digits. If the number had an odd number of digits to the left of the decimal, you will have a single digit.

(a) If the digit/pair identified in step 2 is a perfect square

(1, 4, 9, 16, 25, 36, 49, 64, or 81), replace the digit/pair with the square root of that digit/pair.

(b) If the digit/pair identified in step 2 is not a perfect square, identify the largest perfect square that is less than the digit/pair and replace the digit/pair with the square root of that perfect square.

(3) For each pair of digits following the digit or digits identified in step 2, substitute a zero.

(4) The resulting number is the estimated square root (low).

(5) Increase the left (first) digit of the estimated square root (low) by 1 to arrive at the estimated square root (high).

(6) The actual square root will be less than the estimated square root (high) and equal to or greater than the estimated square root (low).

NOTE: If the digit/pair identified in step 2 is a perfect square and the following pairs were all zeros originally and no non-zero digits followed the decimal point of the original number, then the estimated square root (low) is the actual square root.

Estimate the square root of the following numbers using the above rules.

a. 149,769

b. 640,000

c. 36,000

_______________________________________________________________________________________

In case you had difficulty with the square roots given in Frame 5-28, the problems are worked in greater detail below.

a. 149,769

(1) Pair off beginning at the decimal point

149,769. = (14)(97)(69)

(2) Largest perfect square not over 14 is 9. Square root

of 9 is 3.

(3) Replace

(4) low (14)(97)(69) à (3)(0)(0) à 300

(5) high (3+1)(0)(0) à 400

(6) The actual square root of 149,769 is more than 300 and less than 400. (See Frame 5-27)

b. 640,000

(1) Pair off beginning at the decimal point

640,000. = (64)(00)(00)

(2) Largest perfect square not over 64 is 64. Square root

of 64 is 8.

(3) Replace

(4) low (64)(00)(00) à (8)(0)(0) à 800

Stop calculations. Based upon the NOTE, 800 is the exact square root of 640,000.

c. 36,000

(1) Pair off beginning at the decimal point

36,000. = (3)(60)(00)

(2) Largest perfect square not over 3 is 1. Square root of 1 is 1.

(3) Replace

(4) low (3)(60)(00) à (1)(0)(0) à 100

(5) high (1+1)(0)(0) à 200

(6) The actual square root of 36,000 is more than 100 and less than 200. (Were you tricked because the square root of 36 is 6? Remember, you must begin paring off starting at the decimal point.)

a. between 300 and 400

b. 800 (exact)

(actual approximate square root is

_______________________________________________________________________________________

The method below is for decimal numbers less than one. If you have a fraction, change the fraction to a decimal and use the procedure given below. There are methods for calculating the square root of a fraction, but they are not covered in this subcourse.

(1) Pair off the digits of the number beginning at the decimal point and going to the right.

(2) Identify the first digit pair with a non-zero digit.

(a) The identified pair must contain two digits, not just one.

(b) If the first non-zero is at the end of the number and there is an odd number of digits to the right of the decimal, then you must add a zero to the end of the number to make the last digit part of a pair of digits.

(c) Drop all digits (if any) following this digit pair.

(3) Replace the remaining pairs of digits.

(a) If the pair identified in step 2 is a perfect square

(1, 4, 9, 16, 25, 36, 49, 64, or 81), replace the pair with the square root of that number.

(b) If the pair identified in step 2 is not a perfect square, identify the largest perfect square that is less than the pair and replace the pair with the square root of that perfect square.

(c) For each pair of double zero digits between the decimal point and the digit pair identified in step 2, substitute a zero.

(4) The resulting number is the estimated square root (low).

(5) Increase the last digit of the estimated square root (low) by 1 to arrive at the estimated square root (high).

(6) The actual square root will be less than the estimated square root (high) and equal to or greater than the estimated square root (low).

NOTE: If the pair of digits identified in step 2 is a perfect square and there were no non-zero digits following the pair in the original number, then the estimated square root (low) is the actual square root.

Estimate the square root of the following numbers.

Solution to Frame 5-29

_______________________________________________________________________________________

In case you had difficulty with the square roots given in Frame 5-28, the problems are worked in greater detail below.

a. 0.00004

(1) Pair off beginning at the decimal point

(2) 0.00004. = (00)(00)(40)

[You must add a zero to the end of the number so that

the "4" is part of a pair.]

(3) Largest perfect square not over 40 is 36. The square root

of 36 is 6.

(4) (low) (00)(00)(40) à (0)(0)(6) à 0.006

(5) (high) (0)(0)(6+1) à 0.007

b. 0.0004645775

(1) Pair off beginning at the decimal point

(2) 0004645775. = (00)(04)(64)(57)(75) à (00)(04)

[The digits after the first non-zero pair of digits are

dropped.]

(3) Largest perfect square not over 04 is 4. The square root

of 4 is 2.

(4) (low) (00)(04) à (0)(2) à 0.02

(5) (high) (0)(2+1) à 0.03

Solution to Frame 5-30

a. between 0.006 and 0.007

b. between 0.02 and 0.03

_______________________________________________________________________________________

Estimate the square root of the following numbers.

a. 0.0036

b. 0.1

c. 0.01

Solution to Frame 5-31

_______________________________________________________________________________________

If you feel that you need more review, look over the appropriate frames again. Then work the following self-test exercises shown below. The solutions are found on the following page.

1. Work the following problems.

a. ^– 3 + ^– 6 = ________

b. ^– 3 – ^– 6 = ________

c. ^– 3 + ⁺ 6 = ________

d. ^– 3 – ⁺ 6 = ________

e. ^– 3 x ⁺ 6 = ________

f. ^– 3 x ^– 6 = ________

g. ^– 3 ) ^– 6 = ________

h. ⁺ 3 ) ^– 6 = ________

Solution to Frame 5-32

a. 0.06 (exact)

.(00) (36)

(0)(6)

b. between 0.3 and 0.4

0.1 = 0.10

.(10)

.(3) low

.(3+1) high

[0.3² = 0.09

0.4² = 0.16]

c. 0.1 (exact)

.(01)