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arun October 7th, 2007 07:19 AM

Any method to compare fractions??
 
Hi friends,

A long time since I posted here.

My query is

* Is there any method/shortcut other than reducing them to decimals to compare 2 fractions say, 25/27 and 65/68 and say which is greater.

It is clear that (a+x)/(b+x) is greater than a/b. Using tis in our example, we can say that 25/27 is smaller than (25+40)/(27+40) or 65/67. But it seems nothin can b said abt 25/27 and 65/68.....


Can someone help..

johnny October 7th, 2007 08:55 AM

Quote:

Hi friends,

A long time since I posted here.
Welcome back. We're glad to have you back in mymathforum.com.

Quote:

My query is

* Is there any method/shortcut other than reducing them to decimals to compare 2 fractions say, 25/27 and 65/68 and say which is greater.

It is clear that (a+x)/(b+x) is greater than a/b. Using tis in our example, we can say that 25/27 is smaller than (25+40)/(27+40) or 65/67. But it seems nothin can b said abt 25/27 and 65/68.....


Can someone help..
Let's say that we want to compare the fractions a/b and c/d. Let's say that there are three possible solutions as shown below.

i. a/b>c/d
ii. a/b=c/d
iii. a/b<c/d

Multiply by bd on both sides, and we get

i. ad>bc
ii. ad=bc
iii. ad<bc

Do some multiplication arithmetic for those, and you can determine if it's equal, or greater, or less of between those two fraction values without doing arithmetic decimal values.

For example, 1/2 and 2/3
Multiply (2)(3) on both fractions
(1)(3) and (2)(2)
3 and 4
Therefore, 2/3 is greater than 1/2.

johnny October 7th, 2007 08:58 AM

Other way to solve is to find the LCM of the two denominators of a/b and c/d, and let lcm(b, d)=m, and therefore, p/m and q/m (p and q are the new numerator values, so you will need to find p and q), and p/m=a/b and q/m=c/d, so therefore you will need to determine whether

i. p>q
ii. p=q
iii. p<q

to find which fraction is greater, less, or equal.

Infinity October 7th, 2007 10:27 AM

Hey, it's good to see you again, 'arun'!

It's easy to tell if two fractions are equal, because all you have to do is flip one of the them and then multiply it by the other. It the result is 1, then they are equal.

I'm tempted to say this also works, but I have no proof for it:

Flip the second of your two fractions and multiply it by the first. If the result is larger than one, then the second fraction is less than the first. Else if the result is less that one, then the second fraction is greater than the first.

Can anyone prove this?

arun October 8th, 2007 02:12 AM

Quote:

Originally Posted by Infinity
Hey, it's good to see you again, 'arun'!

Its good to hear from u too infinity
Quote:

Originally Posted by Infinity
It's easy to tell if two fractions are equal, because all you have to do is flip one of the them and then multiply it by the other. It the result is 1, then they are equal.

I'm tempted to say this also works, but I have no proof for it:

Flip the second of your two fractions and multiply it by the first. If the result is larger than one, then the second fraction is less than the first. Else if the result is less that one, then the second fraction is greater than the first.

Can anyone prove this?

Lets say the two fractions are X=(a/b) and Y=(c/d) .

Now wen the second fraction is flipped, it is 1/Y .

Multiply wit X to get X/Y. If this is greater than 1, it implies X> Y else Y>X...

While flippin and multiplyin, wat we indirectly do is multiply the fractions wit the LCM of the denominators(b,d) and compare them

ie., (a/b) * bd and (c/d)*bd which is ad and bc
( comparing this "ad" and "bc" is nothing but flippin the second fraction Y and multiplyin with the first fraction X.....):roll: :roll: :roll: :roll: { Remember this emoticon infinity :wink: }


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