Let's compare the fractions and using fraction models.
By comparing the shaded areas, we can tell that the fraction on the left is smaller than the fraction on the right. But how can we show this mathematically?
One way to compare the fractions is to put them over a common denominator. In other words, we can compare the fractions by making their denominators the same.
Notice that is a multiple of In fact,
So, to make the denominators the same, we split each part of the left fraction into equal pieces.
The fractions now have the same denominators, so we can compare them by comparing their numerators.
Since the correct comparison statement is
Therefore, we conclude that
Consider the comparison statement above. By putting both fractions over a denominator of this comparison statement becomes
What is the missing number?
The denominators are and and is a multiple of In fact
We can put both fractions over a common denominator of by splitting each of the parts on the left into equal pieces.
Therefore, the missing number is
Consider the comparison statement above. By putting both fractions over a denominator of $8,$ this comparison statement becomes
\[ \dfrac58 < \dfrac{\fbox{$\phantom{0}$}}8. \]
What is the missing number?
|
a
|
$6$ |
|
b
|
$9$ |
|
c
|
$8$ |
|
d
|
$3$ |
|
e
|
$7$ |
Consider the comparison statement above. By putting both fractions over a denominator of $12,$ this comparison statement becomes
\[ \dfrac{\fbox{$\phantom{0}$}}{12} < \dfrac5{12}. \]
What is the missing number?
|
a
|
$4$ |
|
b
|
$2$ |
|
c
|
$3$ |
|
d
|
$1$ |
|
e
|
$6$ |
Consider the comparison statement above. By putting the fractions over a common denominator, we can rewrite this comparison statement as follows:
What is the missing fraction?
The denominators are and and is a multiple of In fact
We can put both fractions over a common denominator of by splitting each of the parts on the left into equal pieces.
The fractions now have the same denominator, , and the comparison statement is
Therefore, the missing fraction is
Consider the comparison statement above. By putting the fractions over a common denominator, we can rewrite this comparison statement as follows:
\[ \dfrac{\fbox{$\phantom{0}$}}{\fbox{$\phantom{0}$}} < \dfrac5{6} \]
What is the missing fraction?
|
a
|
$\dfrac46$ |
|
b
|
$\dfrac58$ |
|
c
|
$\dfrac59$ |
|
d
|
$\dfrac36$ |
|
e
|
$\dfrac56$ |
Consider the comparison statement above. By putting the fractions over a common denominator, we can rewrite this comparison statement as follows:
\[ \dfrac{\fbox{$\phantom{0}$}}{\fbox{$\phantom{0}$}} < \dfrac7{12} \]
What is the missing fraction?
|
a
|
$\dfrac6{12}$ |
|
b
|
$\dfrac4{12}$ |
|
c
|
$\dfrac2{6}$ |
|
d
|
$\dfrac5{12}$ |
|
e
|
$\dfrac4{6}$ |
Let's compare the fractions and using the models shown below.
Note the following:
is not a multiple of
In cases like this, we can find a common denominator by taking the product of the two denominators.
So for the fractions above, we can create a common denominator of
First, we put the left fraction over a denominator of by splitting each of its parts into equal pieces.
Next, we put the right fraction over a denominator of by splitting each of its parts into equal pieces.
The fractions now have the same denominator, , so we can compare them by comparing their numerators.
Since the correct statement is
Therefore, we conclude that
Consider the comparison statement above. By putting both fractions over a denominator of this comparison statement becomes
What is the missing number?
The left and right sides represent fractions with denominators of and respectively. To compare them, we can put both fractions over a common denominator of
We put the left fraction over a denominator of by splitting each of its parts into equal pieces.
We put the right side over a denominator of by splitting each of its parts into equal pieces.
The fractions now have the same denominator, , and the comparison statement is
Therefore, the missing number is
Consider the comparison statement above. By putting both fractions over a denominator of $10,$ this comparison statement becomes
\[ \dfrac{\fbox{$\phantom{0}$}}{10} > \dfrac{5}{10}. \]
What is the missing number?
|
a
|
$4$ |
|
b
|
$7$ |
|
c
|
$8$ |
|
d
|
$3$ |
|
e
|
$6$ |
Consider the comparison statement above. By putting both fractions over a denominator of $10,$ this comparison statement becomes \[ \dfrac{\fbox{$\phantom{0}$}}{10} < \dfrac{8}{10}. \]
What is the missing number?
|
a
|
$6$ |
|
b
|
$2$ |
|
c
|
$5$ |
|
d
|
$4$ |
|
e
|
$1$ |
Consider the fraction comparison above. By putting the fractions over a common denominator, we can rewrite this comparison statement as follows:
What number is missing from both boxes?
The left and right sides represent fractions with denominators of and respectively. To compare them, we can put both fractions over a common denominator of
We put the left fraction over a denominator of by splitting each of its parts into equal pieces.
We put the right side over a denominator of by splitting each of its parts into equal pieces.
The fractions now have the same denominator, , and the comparison statement is
Therefore, the number missing from both boxes is
Consider the fraction comparison above. By putting the fractions over a common denominator, we can rewrite this comparison statement as follows:
\[ \dfrac{3}{\fbox{$\phantom{0}$}} > \dfrac{2}{\fbox{$\phantom{0}$}} \]
What number is missing from both boxes?
|
a
|
$4$ |
|
b
|
$6$ |
|
c
|
$12$ |
|
d
|
$9$ |
|
e
|
$8$ |
Consider the fraction comparison above. By putting the fractions over a common denominator, we can rewrite this comparison statement as follows:
\[ \dfrac{9}{\fbox{$\phantom{0}$}} > \dfrac{8}{\fbox{$\phantom{0}$}} \]
What number is missing from both boxes?
|
a
|
$10$ |
|
b
|
$9$ |
|
c
|
$11$ |
|
d
|
$14$ |
|
e
|
$12$ |
