17. Efficient Ways to Add with unlike Denominators
After reading this page, provide answers for the
following questions and/or exercises
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How is the list method used to obtain a least common
denominator = the least common multiple of a pair of
denominators?
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How can the prime number decomposition (also known as
factorization) be used to calculate the LCM and GCD of a pair of
whole numbers?
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How can Euclid's algorithm be used forwards and
backwards to calculate the GCD (the normal result) and then the LCM
of a pair of whole numbers?
-
Employ the M-method below to find the sum of eighteen
21st and nine 14-ths? Say which of the latter is more than the other,
and find how much more.
Two fractions may be added together using any common denominator. For
example, the use of common denominator 12 = 2×6 = 3×4 leads to
15
6
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+
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7
4
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=
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30
12
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+
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21
12
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=
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51
12
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=
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4
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3
12
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=
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4
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1
4
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the use of common denominator 24 = 4×6 = 6×4 leads to
15
6
|
+
|
7
4
|
=
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60
24
|
+
|
42
24
|
=
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102
24
|
=
|
4
|
6
24
|
=
|
4
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1
4
|
and use of common denominator 36 = 6×6 = 9×4 leads to
15
6
|
+
|
7
4
|
=
|
90
36
|
+
|
63
36
|
=
|
153
36
|
=
|
4
|
9
36
|
=
|
4
|
1
4
|
For all three choices of common denominators, the least and other,
conversion to a like denominator, addition and simplification all lead to
the result 4¼ . But the use of smaller common denominators involves
smaller numbers in the computation and hence less simplification work in
the end. The use of the least common denominators usually gives the
most efficient way to add and subtract fractions with unlike
denominators. So try to use the least common denominator.
There is one exception that comes to mind, that occurs when the product
of the original denominators in the addends (the fractions being
added) gives a power of ten, for example 10, 100, 1000, 10000, and so
on. In the latter case, divisibility rules for division by 2, 5 and 10
may lead to easier simplification despite the presence of larger
numbers.
Oops: The following is Algebraic But you should see how much you can
follow alone or with help.
Methods for adding and subtracting efficiently.
Take A = 15, B = 6, C = 7, D = 4 and M = 12 or 24 on first
reading.
Let M can be any common multiple of B and D (or not, as you may later
discover). I
The foregoing implies.
To apply these formulas, remember the lowest common multiple M of left
hand sides denominators B and D usually gives less work in the
simplification of the right hand sides. That being said, any and all
common multiples of the left hand side denominators will suffice with
most likely bigger numbers on the right hand side and hence more work to
do in simplification (reducing terms).
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This algebraic reason or proof for the formulas is optional
reading. One that you should try to follow now (or after
reading an introduction to algebra here or elsewhere.)
|
|
Algebraic Proof of Formulas: Let M denote a multiple
of both B and D. Now M = p B gives M ÷ B = p,
and
A( M ÷ B)
M
|
=
|
Ap
M
|
=
|
Ap
pB
|
=
|
A
B
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Likewise M = q D gives M ÷ D = q and
C( M ÷ D)
M
|
=
|
Cq
M
|
=
|
Cq
qD
|
=
|
C
D
|
The addition and subtraction formulas above are immediate
consequence of the latter expression. Q. E. D
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Easy Consequence of the Proof: The
expressions
A( M ÷ B)
M
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and
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C( M ÷ D)
M
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could be used to compare the fractions A/B and C/D. Do you
understand why?
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First Example Above Revisited with M = 12:
15
6
|
+
|
7
4
|
=
|
15(12 ÷ 6) + 7(12 ÷4)
12
|
=
|
15(2) + 7(3)
12
|
=
|
51
12
|
=
|
4
|
3
12
|
=
|
4
|
1
4
|
First Example Above Revisited with M = 6 × 4 = 24:
15
6
|
+
|
7
4
|
=
|
15(24 ÷ 6) + 7(24 ÷4)
24
|
|
|
|
=
|
15(4) + 7(6)
24
|
|
|
|
=
|
60 + 42
24
|
|
|
|
=
|
102
24
|
|
|
|
=
|
51
12
|
|
|
|
=
|
|
|
|
|
=
|
|
Here the use of a larger common denominators leads to more work.
Second Example of Addition:
In this example, M = 24 = the least common multiple of the the two
denominators 8 and D = 12 while A = 5 and C = 7. So M/A = 24/8 = 3 and
M/D = 24/12 = 2.
[Play
Video] 3 minutes Another example of how to add fractions
with and without the least common denominators with an explanation that
not using the LCD (least common denominator) leads to ratios that
can be simplified. So use of LCDs is advised.
How to Choose a common denominator M:
Method 1 - List Methods: List the first B multiples of D, and list
the first D multiples of B. The number B x D = D x B is the last
number in each list. Let M < B x D be the smallest number
in both lists. That number will be the smallest common multiple of B and
D.
Subexample: Let B = 8 and D = 12 as above.
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1
|
2
|
3
|
4
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5
|
6
|
7
|
8
|
9
|
10
|
11
|
12
|
|
8
|
16
|
24
|
32
|
40
|
48
|
56
|
64
|
72
|
80
|
88
|
96
|
|
12
|
24
|
36
|
48
|
60
|
72
|
84
|
96
|
|
|
|
|
The number 24 is the smallest in both lists. So 24 = the least common
multiple of 8 and 12. Shortcut: (i) Calculate the multiples
of the largest denominator D first. Then Calculate the first D
multiples of the smallest number B until a multiple of D appears.
The list method is awkward for large numbers. But for small denominators
in the range 2 to 12, you should be able to apply it quickly.
Practice will lead to a knowledge or memory of smallest common
denominators, so that the list method need not be applied.
Method 2 - Prime Decomposition Method: From the prime
factorizations of B and D form a product of primes where each prime in
the product appears to the greatest power that occurs in the prime
decomposition of B and D.
Subexample: B = 8 = 23 and D = 12 =
3×22. Then M = 3×233 = 8 × 4 = 24 as
before.
The
site account of prime decomposition of whole numbers ends with a
quick method for obtaining the decomposition (or determining whether or
not a whole number is prime).
A whole number is prime when and only when it is not a
whole number multiple of any prime less than its square root. You
can calculate the square root with a calculator. Then you start
checking (smallest primes first) whether or not the whole number
is a multiple of any prime less than its square root. With a list of
all primes less than 50, the foregoing route provides a quick method
for discovering whether or not a whole number < 2500 =
502 is prime, and if not a quick method for obtaining its
prime factorization or decomposition. The work here for whole number
less that 2500 can be done with the aid of a calculator provided the
display displays at least three digits after the decimal
point.
Method 3 - Find Greatest Common Divisor using Euclid Algorithm, and
use it to calculate a M.
12 = 1 × 8 + 4
8 = 2×4 = ad
Therefore 12 = 1 × 8 + 4 = 2×4+4 = 3×4 = cd
Now take M = abc = 2×4×3 = 8×3 = 24. The form
of M = adc implies M is c times 8 = ad and a times 12 = bd. Here
M = abc will be the least common multiple of 12 and 8 (why?).
Method 1 works best with pairs of numbers < 15. Each list
is then < 15 numbers long. Method 2 works best if you
know how to obtain the prime factorization of a whole number
quickly. Method 3 works if you know how to divide - a calculator
could be useful tool for doing this exactly.
Real Player Videos
-
[Play
Video] 5 minutes. How to add fractions using common
denominators. Here the common dominators is the lowest or least
common denominator (LCD) and its given by the least common multiple
(LCM) of the denominators in the fractions added together. Here
the listing multiples method is used to compute the LCM. The
alternative of not using the LCD for the fractions is explored to show
what happens when the LCD is not used.
-
[Play
Video] 3 minutes - Another example of the listing multiples
method to find the LCM and thus the LCD for the sum of two
fractions.
-
[Play
Video] 4 minutes - Factorization method to obtain a
common denominator, here the LCM and thus the LCD for the sum of two
fractions. See if you can recognize the GCD of the denominators here.
It is not mentioned here. In this example, the LCD is given by a
product that does not have to be evaluated explicity due to
cancellation of common terms after addition of fractions.
-
[Play
Video] 5 minutes - How to use Prime Factorization or
Decomposition for LCM and LCD for a pair of denominators, an
example.
-
[Play
Video] 6½ minutes. Euclid Algorithm computes GCDs not using
Prime Factorization.
-
[Play
Video] 3 minutes. Another Euclid Algorithm GCD
example with result confirmed using Prime Decomposition.
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