View Full Version : How to reduce fractions to the lowest terms?
alternap18
Sep 21, 2010, 10:04 AM
like for example this one
3
_ = _
7 21
I tried my best to put it up
donf
Sep 21, 2010, 10:10 AM
3/7 is at it's lowest terms.
try this to see if it will help you.
8/32 = 4/16 = 2/8 =1/4.
It looks like you were trying to find the lowest common denominator. Is that what you want? You need to fractions to do that.
wowwy
Sep 21, 2010, 10:26 AM
Simplifying (or reducing) fractions means finding out what can be evenly divided into both top and bottom.
In your example, you are actually doing the reverse... so.. if you multiply 7 x 3 you get 21. You have to multiply the top by 3 also. Your answer would be 9 over 21.
galactus
Sep 22, 2010, 04:59 AM
Finding the prime factorization may help reduce a fraction when it is tough to reduce compared to, say, 8/12.
Suppose we were asked to reduce
\frac{3913}{6149}
3913=43\cdot 7\cdot 13
6149=43\cdot 11\cdot 13
\frac{\not{43}\cdot 7\cdot \not{13}}{\not{43}\cdot 11\cdot \not{13}}
The cancellations reduce it down to \frac{7}{11}
donf
Sep 22, 2010, 05:19 AM
Galactus,
I have never seen that process, not that that surprises me all that much.
Can you take a moment and "PM" with an explanation of how it works.
Unknown008
Sep 22, 2010, 05:46 AM
It's in fact the same process as you did earlier donf.
Take your fraction:
\frac{8}{32}
You divided both numerator and denominator by two successively to get 1/4.
It's like you did it this way:
\frac{8}{32} = \frac{2.2.2}{2.2.2.2.2} = \frac{2.2}{2.2.2.2} = \frac{2}{2.2.2} = \frac{1}{2.2} = \frac{1}{4}
Now, if you get multiples of prime numbers of other factors instead of strictly prime numbers as factors, they still work.
\frac{150}{875} = \frac{\cancel{25}.6}{\cancel{25}.35} = \frac{6}{35}
donf
Sep 22, 2010, 06:22 AM
Hey, thanks,
Now I recognize the process and yes I've used it in the past.
Thanks for helping me out.
Tensyk
Sep 22, 2011, 07:34 AM
Reduce 7/21 to lowest terms
Unknown008
Sep 22, 2011, 08:16 AM
Divide the numerator and the denominator by 7.