# 2.3 Subtraction of Integers

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1 2.3 Subtraction of Integers The key to subtraction of real numbers is to forget what you might have learned in elementary school about taking away. With positive numbers you can reasonably say take 3 away from 7, to get 4, but the reasoning becomes difficult if we want to take 5 away from 8. For subtraction of integers, we will work with a more formal (or abstract) definition. If we denote y as the opposite of y, then subtraction can be defined as: x! y = x + (!y) That is, to subtract two real numbers, we add the opposite of the second number (number being subtracted) to the first number. Thus, to compute5! 9, we convert the problem to 5 + ( 9) (since 9 is the opposite of 9), and thus: 5! 9 = 5 + (!9) converting to addition =!4 using the number line Similarly, to compute 6! (!5), we convert the problem to (since 5 is the opposite of 5), and thus: 6! (!5) = converting to addition Thus subtraction of integers consists of two steps: = 11 using the number line 1. Convert subtraction to addition by adding the opposite of the second number. 2. Using the number line to perform the addition. The next example will provide additional practice. 78

3 Following this procedure:!5! 7! (!3) + 4! 6 + (!2) =!5 + (!7) (!6) + (!2) = (!5) + (!7) + (!6) + (!2) =! =!13 Note that only subtractions are rewritten as addition in the first step. Also note the second step, in which we reorganized the additions using the commutative property. Often in algebra you will see steps or reorganization, even though nothing was computed during the step. The third step involved adding all negative numbers and positive numbers separately, and the fourth step uses the number line to compute the final answer. Example 2 Compute the expression:!6! 8! (!5)! (!2) (!6)! 1 Solution First rewrite all subtractions as addition of opposites, then add positive and negative numbers separately, then use the number line for the final addition. The steps are:!6! 8! (!5)! (!2) (!6)! 1 =!6 + (!8) (!6) + (!1) = (!6) + (!8) + (!6) + (!1) =! =!10 Recall that solutions to equations are numbers which, when substituted for the variable in the equation, produce a true statement. We can now determine solutions to equations which involve subtraction, as the following example illustrates. 80

4 Example 3 Determine whether or not the given integer value is a solution to the equation. a. x! 4 = 2 ; x = 6 b. x! 3 =!5 ; x =!8 c. y! 3 =!5 ; y =!2 d. a! (!4) = 0 ; a = 4 Solution a. Substitute x = 6 into the equation and rewrite the subtraction as an addition: 6! 4 = (!4) = 2 2 = 2 Note the last addition was performed using the number line. Since this last statement (2 = 2) is true, x = 6 is a solution to the equation x! 4 = 2. b. Substitute x =!8 into the equation and rewrite the subtraction as an addition:!8! 3 =!5!8 + (!3) =!5!11 =!5 Note the last addition was performed using the number line. Since this last statement ( 11 = 5) is false, x =!8 is not a solution to the equation x! 3 =!5. c. Substitute y =!2 into the equation and rewrite the subtraction as an addition:!2! 3 =!5!2 + (!3) =!5!5 =!5 Since this last statement ( 5 = 5) is true, y =!2 is a solution to the equation y! 3 =!5. d. Substitute a = 4 into the equation and rewrite the subtraction as an addition: 4! (!4) = = 0 8 = 0 Since this last statement (8 = 0) is false, a = 4 is not a solution to the equation a! (!4) = 0. 81

5 Recall that an arithmetic sequence is a sequence of numbers (called terms) in which the same number (called the common difference) is added on to each term to obtain the next term. For example, with the sequence 5, 3, 1, 1, from the last section, the common difference was noted to be 2. Another way to find the common difference is by using subtraction. Note that:!3! (!5) =!3 + 5 = 2!1! (!3) =!1 + 3 = 2 1! (!1) = 1+ 1 = 2 Since each successive subtraction results in the number 2, the common difference is 2, and thus the next term in the sequence is = 3. This provides an alternate approach to finding terms of an arithmetic sequence. Example 4 Find the common difference and the next term for each arithmetic sequence. a. 2, 3, 8, b. 12, 10, 8, c. 17, 36, 55, d. 103, 76, 49, Solution a. Subtracting terms to find the common difference:!3! 2 =!3 + (!2) =!5!8! (!3) =!8 + 3 =!5 The common difference is 5, so the next term is!8 + (!5) =!13. b. Subtracting terms to find the common difference:!10! (!12) =! = 2!8! (!10) =! = 2 The common difference is 2, so the next term is!8 + 2 =!6. c. Subtracting terms to find the common difference:!36! (!17) =! =!19!55! (!36) =! =!19 The common difference is 19, so the next term is!55 + (!19) =!74. d. Subtracting terms to find the common difference:!76! (!103) =! = 27!49! (!76) =! = 27 The common difference is 27, so the next term is! =!22. 82

7 d. With the phrase the difference of, we interpret the subtraction in exact order as the numbers are given. The expression and computation are:!5! 8 =!5 + (!8) rewrite as an addition =!13 computed on the number line e. Again, interpreting the difference of in exact order, we start with 7 then subtract 15. The expression and computation are:!7! (!15) =! rewrite as an addition = 8 computed on the number line f. The phrase diminish 7 by 4 indicates we should start with 7, then subtract 4 from it. The expression and computation are:!7! (!4) =!7 + 4 rewrite as an addition =!3 computed on the number line Terminology subtraction of integers Exercise Set 2.3 Subtract the two integers. Remember to rewrite all subtractions as addition for your first step ( 3) 6. 8 ( 4) ( 9) ( 7) ( 40) ( 32) ( 19) ( 48) ( 17) ( 18) 84

8 Compute each expression ! 8! (!3) ! 9! (!6) !8! 5! (!3) + (!2) 24.!12! 7! (!6) + (!4) 25.!7 + (!3)! 4! (!8) 26.!9 + (!6)! 7! (!12) 27. 4! 12 + (!5)! 5! (!6) 28. 6! 9 + (!4)! 9! (!2) 29.!5! 3 + (!6)! 11! 6! (!8) 30.!9! 13 + (!9)! 7 + 5! (!3) 31.! (!5)! (!3)! 7! 13! (!6) 32.! (!7)! (!5)! 6! 8! (!9) 33.!5 + (!2)! 5!! 4 34.!14 + (!9)! 12!! 7 35.! 15!! 7! ! 13!! 19! 37 Determine whether or not the given integer value is a solution to the equation. 37. x! 5 = 4; x = x! 5 = 4; x =!1 39. x! 6 = 3; x =!3 40. x! 6 = 3; x = y! 12 =!7; y =! y! 12 =!7; y = a! 7 =!15; a =!8 44. a! 7 =!15; a =! t! (!6) = 6; t = t! (!6) = 6; t = s! (!9) =!3; s = s! (!9) =!3; s =!12 Find the common difference and the next term in each sequence , 6, 8, 50. 5, 7, 9, 51. 8, 5, 2, 52. 9, 5, 1, , 6, 2, 54. 8, 4, 0, 55. 1, 5, 9, 56. 8, 13, 18, , 39, 53, , 62, 87, , 116, 85, , 133, 105, Convert each written phrase to a mathematical computation, then compute the required quantity. 61. Subtract 18 from Subtract 25 from Subtract 12 from Subtract 15 from Subtract 8 from Subtract 14 from Reduce 12 by Reduce 5 by Reduce 14 by Reduce 25 by Reduce 15 by Reduce 18 by The difference of 4 and The difference of 6 and The difference of 8 and The difference of 9 and 3. 85

9 77. The difference of 15 and The difference of 23 and Diminish 24 by Diminish 26 by Diminish 14 by Diminish 13 by Diminish 14 by Diminish 56 by less than less than less than less than less than less than 22. Answer the following questions. Give complete answers and explain your answer. 91. Is subtraction a commutative property? That is, does a! b = b! a for all integers a and b? Give examples in your explanation. 92. Is subtraction an associative property? That is, does ( a! b)! c = a! ( b! c) for all integers a, b, and c? Give examples in your explanation. 93. When will it be the case that x! y > 0 for two integers x and y? Give examples in your explanation. 94. When will it be the case that x! y < 0 for two integers x and y? Give examples in your explanation. 95. What number can be added to 8 to yield 12? 96. What number can be added to 14 to yield 9? 97. What number can be subtracted from 14 to yield 18? 98. What number can be subtracted from 23 to yield 15? 86

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