Adding Efficiently with unlike Denominators
After reading this page, provide answers for the following questions and/or exercises
- How is the list method used to obtain a least common denominator = the least common multiple of a pair of denominators?
- How can the prime number decomposition (also known as factorization) be used to calculate the LCM and GCD of a pair of whole numbers?
- 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 |
+ | 7 4 |
= | 30 12 |
+ | 21 12 |
= | 51 12 |
= | 4 | 3 12 |
= | 4 | 1 4 |
the use of common denominator 24 = 4×6 = 6×4 leads to
| 15 6 |
+ | 7 4 |
= | 60 24 |
+ | 42 24 |
= | 102 24 |
= | 4 | 6 24 |
= | 4 | 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).
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
BLikewise M = q D gives M ÷ D = q and
C( M ÷ D)
M= Cq
M= Cq
qD= C
DThe addition and subtraction formulas above are immediate consequence of the latter expression. Q. E. D
Easy Consequence of the Proof: The expressions
A( M ÷ B)
Mand
C( M ÷ D)
Mcould be used to compare the fractions A/B and C/D. Do you understand why?
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 |
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| = | 15(4) + 7(6) 24 |
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| = | 60 + 42 24 |
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| = | 102 24 |
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| = | 51 12 |
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| = |
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| = |
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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.
| 1 | 2 | 3 | 4 | 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 = adTherefore 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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Four Topics
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