Download 2.5 Division of Integers

2.5 Division of Integers
Recall that division is the inverse operation for multiplication. For example:
36 ÷ 12 = 3 since 3 • 12 = 36
15 ÷ 3 = 5 since 5 • 3 = 15
Now we consider division of integers. To compute !45 ÷ 5 , note that if !45 ÷ 5 = x , then
5 • x = !45 , thus x = !9 . Similarly, for 36 ÷ (!4) we have 36 ÷ (!4) = x , so !4 • x = 36 , and
thus x = !9 . Thus one negative number results in a quotient which is negative. If we are dividing
two negative numbers such as !28 ÷ (!7) , let !28 ÷ (!7) = x and so !7 • x = !28 , thus x = 4 .
So dividing two negative numbers results in a positive number. In summary, when dividing two
numbers, the sign is positive if the two numbers have the same sign, and negative if the two
numbers have different signs. This is an easy rule to remember, since this rule of signs is
identical to that of multiplication. An alternate notation used to represent division of integers is
a
a
, where = a ÷ b .
b
b
Example 1
Divide the two integers.
a.
b.
c.
d.
Solution
a.
!48 ÷ 6
!36 ÷ (!3)
108
!9
!68
!4
Since the quotient involves one negative number, the quotient will be
negative. Computing the value:
!48 ÷ 6 = ! ( 48 ÷ 6 ) rule for signs
= !8
b.
dividing
Since the quotient involves two negative numbers, the quotient will be
positive. Computing the value:
!36 ÷ (!3) = + ( 36 ÷ 3)
rule for signs
= 12
dividing
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c.
Since the quotient involves one negative number, the quotient will be
negative. Computing the value:
108
108
=!
rule for signs
!9
9
= !12
dividing
d.
Since the quotient involves two negative numbers, the quotient will be
positive. Computing the value:
!68
68
=+
rule for signs
!4
4
= 17
dividing
When the number 0 is involved with division, we must be careful. Beginning with the case
0÷6=
0
6
Recall we are looking for the value of x such that
0
=x
6
For this to be true, we must have
0=6•x
x=0
Thus 0 ÷ 6 =
0
= 0 . What happens if we reverse the situation? Consider
6
6÷0=
Again, we are looking for the value of x such that
6
=x
0
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6
0
For this to be true, we must have
6=0•x
6=0
Clearly this is impossible, so the desired value of x cannot exist, and we say that
6
is
0
undefined, or not defined. One more case to consider is
0÷0=
0
0
Using the same argument, we are looking for the value of x such that
0
=x
0
For this to be true, we must have
0=0•x
0=0
0
is
0
indeterminate, or not determined. We can summarize these division properties with 0 in the
following table, where x ≠ 0.
But this last statement is always true, so the desired value of x can be anything. We say that
General
0
0÷x=
x
x
x÷0=
0
0
0 ÷ 0=
0
Example
Conclusion
0÷8
0
5÷0
undefined
0÷0
indeterminate
We can now determine solutions to equations which involve division, as the following example
illustrates.
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Example 2
Determine whether or not the given integer value is a solution to the equation.
a.
b.
c.
d.
Solution
x
= 8 ; x = 32
4
x
= !9 ; x = 45
!5
36
= !4 ; y = !9
y
!32 ÷ a = !4 ; a = !8
a.
Substitute x = 32 into the equation and determine whether the statement is
true:
32
=8
4
8=8
Since this last statement (8 = 8) is true, x = 32 is a solution to the equation
x
= 8.
4
b.
Substitute x = 45 into the equation and determine whether the statement is
true:
45
= !9
!5
!9 = !9
Since this last statement (–9 = –9) is true, x = 45 is a solution to the
x
equation
= !9 .
!5
c.
Substitute y = !9 into the equation and determine whether the statement is
true:
36
= !4
!9
!4 = !4
Since this last statement (–4 = –4) is true, y = !9 is a solution to the
36
= !4 .
equation
y
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d.
Substitute a = !8 into the equation and determine whether the statement is
true:
!32 ÷ (!8) = !4
4 = !4
Since this last statement (4 = –4) is false, a = !8 is not a solution to the
equation !32 ÷ a = !4 .
Recall that a geometric sequence is a sequence of numbers (called terms) in which the same
number (called the common ratio) is multiplied by each term to obtain the next term. For
example, with the sequence 1, 3, 9, 27, … from the last section, the common ratio was noted to
be 3. Another way to find the common ratio is by using division. Note that:
3
9
27
= 3, = 3,
=3
1
3
9
Since each successive division results in the number 3, the common ratio is 3, and thus the next
term in the sequence is 27 • 3 = 81 . This provides an alternate approach to finding terms of a
geometric sequence.
Example 3
Find the common ratio and the next term for each geometric sequence.
a.
b.
c.
d.
Solution
–4, –8, –16, …
1, –3, 9, …
–5, 30, –180, …
–13, –156, –1872, …
a.
Dividing terms to find the common ratio:
!8
=2
!4
!16
=2
!8
The common ratio is 2, so the next term is !16 • 2 = !32 .
b.
Dividing terms to find the common ratio:
!3
= !3
1
9
= !3
!3
The common ratio is –3, so the next term is 9 • (!3) = !27 .
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c.
Dividing terms to find the common ratio:
30
= !6
!5
!180
= !6
30
The common ratio is –6, so the next term is !180 • (!6) = 1080 .
d.
Dividing terms to find the common ratio:
!156
= 12
!13
!1872
= 12
!156
The common ratio is 12, so the next term is !1872 • 12 = !22, 464 .
Because there is no commutative property for division, when interpreting language the order
must be carefully adhered to. If we say “find the quotient of –12 and 3”, we mean to compute
!12
= !4 . The next example will further illustrate the terminology used with division.
3
Example 4
Convert each written phrase to a mathematical computation, then compute the
required quantity.
a.
b.
c.
d.
Solution
a.
b.
Find the quotient of –48 and 4.
Divide –32 by –8.
Find the quotient of –125 and –5.
Divide –6 into 120.
Performing the division:
!48
48
=!
4
4
= !12
rule for signs
Performing the division:
!32
32
=+
!8
8
=4
rule for signs
dividing
dividing
102
c.
d.
Performing the division:
!125
125
=+
!5
5
= 25
rule for signs
dividing
Performing the division:
120
120
=!
rule for signs
!6
6
= !20
dividing
The final topic in this section is finding the average of a group of numbers. The average of a
group of numbers is defined to be the sum of that group divided by the amount of numbers in the
group. For example, given the numbers 2, 5, 7, 6, their sum is 2 + 5 + 7 + 6 = 20, so the average
is given by
average =
Example 5
Find the average of each set of numbers.
a.
b.
c.
d.
Solution
2 + 5 + 7 + 6 20
=
=5
4
4
5, 6, 8, 9
–4, 8, 2, 6
–3, 5, –6, 2, –8
6, –3, 8, –4, 8
a.
There are 4 numbers, so the average is:
5 + 6 + 8 + 9 28
average =
=
=7
4
4
b.
There are 4 numbers, so the average is:
!4 + 8 + 2 + 6 12
average =
=
=3
4
4
c.
There are 5 numbers, so the average is:
!3 + 5 + (!6) + 2 + (!8) !10
average =
=
= !2
5
5
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d.
There are 5 numbers, so the average is:
6 + (!3) + 8 + (!4) + 8 15
average =
=
=3
5
5
Terminology
division of integers
indeterminate
undefined
average
Exercise Set 2.5
Divide the two integers. If a quotient is undefined or indeterminate, be sure to state so.
1.
3.
5.
7.
9.
11.
13.
15.
17.
!56 ÷ 8
72 ÷ (!3)
0 ÷ (!5)
!14 ÷ 0
!45 ÷ (!9)
!24
!4
!450
30
600
!25
0
0
2.
4.
6.
8.
10.
12.
14.
16.
18.
!48 ÷ 12
50 ÷ (!5)
!7 ÷ 0
0 ÷ (!23)
!78 ÷ (!2)
!120
!15
!150
15
400
!25
!68
0
Determine whether or not the given integer value is a solution to the equation.
19.
21.
23.
25.
x
= 23 ; x = 115
5
x
= !8 ; x = !32
!4
x
= 12 ; x = !108
!9
!48
= !4 ; x = !12
x
20.
22.
24.
26.
104
x
= 23 ; x = !115
5
x
= !8 ; x = 32
!4
x
= 12 ; x = 108
!9
!48
= !4 ; x = 12
x
27.
29.
!102
= 6 ; y = !17
y
a
= !3 ; a = !12
!4
28.
30.
!102
= 6 ; y = 17
y
a
= !3 ; a = 12
!4
Find the common ratio and the next term for each geometric sequence.
31.
33.
35.
37.
39.
41.
43.
3, 6, 12, …
2, –4, 8, …
–1, 3, –9, …
5, 60, 720, …
4, –100, 2500, …
–3, 48, –768, …
–4, –52, –676, …
32.
34.
36.
38.
40.
42.
44.
1, 5, 25, …
–2, –4, –8, …
–1, 4, –16, …
8, 120, 1800, …
5, –65, 845, …
–6, 72, –432, …
–2, 2, –2, …
Convert each written phrase to a mathematical computation, then compute the required quantity.
45.
47.
49.
51.
53.
55.
Find the quotient of –65 and –5.
Divide 240 by –15.
Divide –13 into 273.
Find the quotient of 900 and –25.
Divide –1620 by 45.
Divide 12 into –456.
46.
48.
50.
52.
54.
56.
Find the quotient of –48 and –12.
Divide 800 by –40.
Divide 14 into –378.
Find the quotient of –345 and 15.
Divide –2925 by –45.
Divide –15 into 390.
58.
60.
62.
64.
66.
68.
32, 14, 25, 21
–13, 5, –4, –8
5, 13, –23, –14, –26
4, –8, 5, –3, 2
8, –14, –13, 5, 2, –6
46, –59, –42, –38, 16, –61
Find the average of each set of numbers.
57.
59.
61.
63.
65.
67.
13, 12, 14, 25
8, –7, –5, –12
12, 8, –7, –16, –32
1, –5, –6, 2, –12
12, –13, –14, 5, –6, –8
24, –13, 38, –47, 52, –36
105
Answer each question as true or false, where x and y represent integers. If it is false, give a
specific example to show that it is false. If it is true, explain why.
x
< 0 , then x < 0 or y < 0.
70.
y
x
71. If > 0 , then x > 0 or y > 0.
72.
y
x
73. If = 0 , then x = 0.
74.
y
x
75. If y = 0 and x ≠ 0, then = 0 .
76.
y
x
77. If x = 0 and y = 0, then is undefined.
y
x
78. If x = 0 and y = 0, then is indeterminate.
y
69. If
x
< 0 , then x < 0 and y < 0.
y
x
If > 0 , then x > 0 and y > 0.
y
x
If = 0 , then y = 0.
y
x
If x = 0 and y ≠ 0, then = 0 .
y
If
Answer each question.
79. A start-up internet company loses $525,000 during their first five years. How much
did they lose (on average) each year?
80. A new business loses $5,520 during their first year of operation. How much did they
lose (on average) each month?
81. Dennis is a frequent trader of stocks. During one month he had gains of $3240 and
$1570, and losses of $675, $1450, and $2185. What was his average gain (or loss) for
each trade that month?
82. John is a new trader of stocks. During one month he had gains of $1560 and $720, and
losses of $865, $1890, and $4250. What was his average gain (or loss) for each trade
that month?
83. Are undefined and indeterminate the same thing? Explain your answer.
84. Is the average of a set of integers necessarily an integer? Explain your answer.
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