Every division problem can be written as a fraction and vice-versa. In this lesson, we'll extend this idea to division problems involving negative numbers.
To begin, let's find a fraction that's equivalent to the following division problem:
Notice that we are dividing a negative number by a positive one So the result will be negative:
To write down the fraction that goes in the box, recall that
Therefore, we get
Find a fraction that's equivalent to
We have to divide a positive number by a negative one So the result will be negative:
To write down the number that goes in the box, recall that
Therefore, we get
$(-2) \div 3$ is equivalent to
|
a
|
$-6$ |
|
b
|
$\dfrac{2}{3}$ |
|
c
|
$-\dfrac{2}{3}$ |
|
d
|
$\dfrac{3}{2}$ |
|
e
|
$-\dfrac{3}{2}$ |
$7 \div (-4)$ is equivalent to
|
a
|
$-\dfrac{7}{4}$ |
|
b
|
$\dfrac{4}{7}$ |
|
c
|
$\dfrac{7}{4}$ |
|
d
|
$-\dfrac{4}{7}$ |
|
e
|
$-28$ |
$(-5) \div (-8)$ is equivalent to
|
a
|
$-\dfrac{5}{8}$ |
|
b
|
$\dfrac{5}{8}$ |
|
c
|
$40$ |
|
d
|
$\dfrac{8}{5}$ |
|
e
|
$-40$ |
Find a fraction that's equivalent to
Here, we are dividing a negative number by another negative number So the result will be positive:
Now, to write down the fraction that goes in the box, recall that
Therefore, we get
$(-7) \div 2$ is equivalent to
|
a
|
$-5 \, \dfrac{1}{2}$ |
|
b
|
$-3 \, \dfrac{1}{2}$ |
|
c
|
$2 \, \dfrac{1}{2}$ |
|
d
|
$-2 \, \dfrac{1}{2}$ |
|
e
|
$3 \, \dfrac{1}{2}$ |
$4 \div (-3)$ is equivalent to
|
a
|
$-1 \, \dfrac{1}{3}$ |
|
b
|
$1 \, \dfrac{1}{3}$ |
|
c
|
$2 \, \dfrac{1}{3}$ |
|
d
|
$-2 \, \dfrac{1}{3}$ |
|
e
|
$-1 \, \dfrac{2}{3}$ |
What is the missing part of the expression below?
A negative number can be obtained if we divide two numbers, one of which is positive and the other negative.
So, given a negative fraction, we have the following equivalent notations:
In our case, we see that the denominator on the right-hand side is negative Therefore, the missing numerator must be
What is the missing part of the expression below?
A positive number can be obtained if we divide two numbers that are either both positive or both negative.
So, given a positive fraction, we have the following equivalent notations:
In our case, we see that the numerator on the right-hand side is negative Therefore, the missing denominator must be
What is the missing part of the expression below? \[ -\dfrac{13}{5} = \dfrac{\bbox[1pt, border: 1pt solid black]{\phantom{A}}}{5} \]
|
a
|
$-65$ |
|
b
|
$5$ |
|
c
|
$-13$ |
|
d
|
$65$ |
|
e
|
$13$ |
What is the missing part of the expression below? \[ \dfrac{7}{23} = \dfrac{-7}{\bbox[1pt, border: 1pt solid black]{\phantom{A}}} \]
|
a
|
$-161$ |
|
b
|
$-23$ |
|
c
|
$161$ |
|
d
|
$23$ |
|
e
|
$237$ |