Transcript Reducing Numeric Fractions

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Reducing Numeric Fractions
Suggestion:
140
?
875
Work with
scratch paper
and pencil as you
go through this
presentation.
Turn Mystery into Mastery
Click to Advance
C2006 – DW Vandewater
What if we could look at a
simplified form of both numbers?
140 (2)(2)(5)(7) (2)(2) 4
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875 (5)(5)(5)(7) (5)(5) 25
1.
2.
3.
4.
5.

Figure out the prime factors of both.
Do any factors cancel out? Cancel them.
Write the remaining factors.
Multiply the tops.
Multiply the bottoms.
That’s the simplest form of the fraction!
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What is the Key Skill?
Prime Factorization!
(A big name for a simple process …)
Finding out how to write a number as
the product of it’s prime factors.
Examples:
6 = (2)(3)
70 = (2)(5)(7)
24 = (2)3(3) or (2)(2)(2)(3)
11=(11) because 11 is prime
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Recognizing the Primes between 1 and 20
1 is not considered a prime number
2 is the only even prime number
3, 5, 7 are primes
(3)(3)=9, so 9 is not prime
11, 13, 17, and 19 are prime
There are infinitely many primes above
20.
How can you tell if a large number is prime?
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Is it Prime? You can Find Out!
Use repeated division:
– Start by finding the smallest prime number that divides
evenly into the original number.
– If you can find one, this division yields 2 factors:
one CERTAIN prime and one that MAY be prime
Examples:
36 divided by 2 is 18.
– Therefore, 36=(2)(18)
175 divided by 5 is 35.
– Therefore, 175=(5)(35)
147 divided by 3 is 49.
– Therefore, 147=(3)(49)
18
2 36
49
3 147
35
5 175
11
3 33
33 divided by 3 is 11.
– Therefore, 33=(3)(11)
You still have to check the second factor for more primes
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Is it Prime?
Tricks for recognizing factors 2, 3 and 5
ANY even number can always be divided by 2
– Yes: 3418, 70, 122
No: 37, 120001
Numbers ending in 5 or 0 can always be
divided by 5
– Yes: 2345, 70, 41415
No: 37, 120001
If the sum of a number’s digits divides evenly
by 3, then the number always divides by 3
– Yes: 39, 120, 567
No: 43, 568
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Finding all prime factors:
The “Tree Root” Method
Write down a number
Break it into a pair of
factors (use the smallest prime)
Try to break each
new factor into pairs
Repeat until every
number is prime
Collect the “dangling”
primes
Click to Advance
198
2
99
3
33
3
11
198=(2)(3)(3)(11)
The mechanics of
The “Tree Root” Method
Find the smallest prime
number first
To get the other factor,
divide it into the original
number
Since 66 is positive, 2 must
be a factor
Divide 2 into 66 to get 33
Since 33’s digits add up to 6,
3 must be a factor
Divide 3 into 33 to get 11
All the “dangling” numbers
are prime, so we are done
Click to Advance
66
2
33
3
11
66=(2)(3)(11)
You can also use a linear approach
84=(2)(42)
=(2)(2)(21)
=(2)(2)(3)(7)
=22·3·7
(simpler notation)
Suggestion:
Do your divisions
in a work area to
the right of the
linear factorization
steps.
216=(2)(108)
108
54
2 216
=(2)(2)(54)
2 108
=(2)(2)(2)(27)
9
3
=(2)(2)(2)(3)(9)
3 27
39
=(2)(2)(2)(3)(3)(3)
=23·33
(simpler notation)
Click to Advance
27
2 54
Is a large number prime?
What smaller primes do you have to check?
Here is a useful table of the squares of some small primes:
42=16 52=25 72=49 112=121 132=169 172=289 192=361 232=529
See where the number fits in the table above
Let’s use 151 as an example:
151 is between the squares of 11 and 13
Check all primes before 13: 2, 3, 5, 7, 11
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–
–
–
–
2 won’t work … 151 is not an even number
3 won’t work … 151’s digits sum to 7, which isn’t divisible by 3
5 won’t work … 151 does not end in 5 or 0
7 won’t work … 151/7 has a remainder
21
13
7 151 11 151
11 won’t work … 151/11 has a remainder
So … 151 must be prime
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r4
r 8
Practice:
Let’s Reduce a Fraction
Here’s the problem ->
–
–
–
–
Find the factors of 1848
Find the factors of 990
Rewrite the fraction
Cancel matching factors
– Rewrite the fraction
– Multiply top & bottom
That is the simplest form
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1848
990
1848
2  924
2  2  462
2  2  2  231
2  2  2  3  77
2  2  2  3  7 11
2  2  2  3  7 11
2  3  3  5 11
227
35
28
15
990
2  495
2  3 165
2  3  3  55
2  3  3  5 11
More Practice
See if you can find the simplest forms.
Do the work on paper and click to see the answer
252 2  2  3  3  7 2  3  3 18
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70
257
5
5
2145 3  5 11 13 5 11
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 55
39
3 13
1
91
7 13
13
13
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462 2  3  7 11 2  3 11 66
59
59

both prime
101 101
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Thank You
For Learning about
– Prime Factorization
– Reducing Numeric fractions