B Reducing Fractions: Learn the Basics and Get Answers

[ophion]

Hey there,

my question is:
do i always have to compare the answer of the multiplied prime numbers by dividing the numerator and denominator with the given answers to see which one is correct. or is there a way to to always see the accurate answer without having to compare them.

here is my way of reducing fractions, feedback would be appreciated I'm pretty new to this.

lets take 108/162

108:2=52:2=27:3=9:3=3
2x2x3x3 = 36
162:2=81:3=27:3=9:3=3
2x3x3x3 = 54

108:36 = 3
162:36 = 4.5

108:54 = 2
162:54 = 3

I feel like I'm doing something incorrect, as i seem to stumble up on some issues with certain fractions like 78/91
78:2=39:3=13
91:7= 13

i don't see why i am supposed to divide this one by 13, if the formula used above are for other fractions.

 

[Eucliddo]

there are some techniques on how to deal with fractions like this. If a number is divisible by 9, the sum of its digits should also be divisible by 9.

Ex. 108 : 1+0+8=9
162 : 1+6+2=9
both are divisible by 9 so you can start dividing both of them by 9.
108/162 = 12/18 then think of a number that can still be divided by both of them.
in this case, it's 6.
12/18 = 2/3.

You can just check your answer by reversed computation. No need to make long calculations just for simplifying fractions.

 

[ophion]

there are some techniques on how to deal with fractions like this. If a number is divisible by 9, the sum of its digits should also be divisible by 9.

Ex. 108 : 1+0+8=9
162 : 1+6+2=9
both are divisible by 9 so you can start dividing both of them by 9.
108/162 = 12/18 then think of a number that can still be divided by both of them.
in this case, it's 6.
12/18 = 2/3.

You can just check your answer by reversed computation. No need to make long calculations just for simplifying fractions.

i feel like reversed computation is quite a hassle. especially when we're using uneven and uncommon fractions like 481/643 for example which is unable to reduce, how would i know this as quickly as possible, without having to deal with all the hassle.

 

[HallsofIvy]

Hey there,

my question is:
do i always have to compare the answer of the multiplied prime numbers by dividing the numerator and denominator with the given answers to see which one is correct. or is there a way to to always see the accurate answer without having to compare them.

here is my way of reducing fractions, feedback would be appreciated I'm pretty new to this.

lets take 108/162

108:2=52:2=27:3=9:3=3
2x2x3x3 = 36

The notation is a bit odd but I presume you are saying that "108 divided by 2 is 52", "52 divided by 2 is 27", "27 divided by 3 is 9", and "9 divided by 3 is 9" so the "prime factorization" of 108 is 2x2x3x3x3. (I don't know what "2x2x3x3= 36" is intended to mean.)

162:2=81:3=27:3=9:3=3

"162 divided by 2 is 81", "81 divided by 3 is 9", "9 divided by 3 is 3" so 162= 2x3x3x3x3

2x3x3x3 = 54

108:36 = 3
162:36 = 4.5

What? A fractional division doesn't help!

108:54 = 2
162:54 = 3

Having seen from the prime factorizations that the numerator is 2x2x3x3x3 and the denominator is 2x3x3x3x3 we see that we can cancel the 2 in the denominator with one of the 2s in the numerator and we can cancel all three 3s in numerator with three of the four in the denominator:
\frac{108}{162}= \frac{2(2)(3)(3)(3)}{2(3)(3)(3)(3)}= \frac{2}{3}.

I feel like I'm doing something incorrect, as i seem to stumble up on some issues with certain fractions like 78/91
78:2=39:3=13
91:7= 13

Okay, 78 divided by 2 is 39 and 39 divided by 3 is 13. 78= 2(3)(13). And 91 is not divisible by 2, 3, or 5, but 91 divided by 7 is 13. 91= 7(13) so that \frac{78}{91}= \frac{2(3)(13)}{7(13)}
The 13s in numerator and denominator cancel leaving
\frac{2(3)}{7}= \frac{6}{7}

i don't see why i am supposed to divide this one by 13, if the formula used above are for other fractions.

You don't "divide by 13". You cancel the 13 in the numerator by the 13 in the denominator.
\frac{13}{13}=1.

 

[Eucliddo]

i feel like reversed computation is quite a hassle. especially when we're using uneven and uncommon fractions like 481/643 for example which is unable to reduce, how would i know this as quickly as possible, without having to deal with all the hassle.

some fractions cannot be reduced because they don't have common factors.. such fractions remain as is.

Reducing fractions require practice and familiarization with numbers. If you do that, you can simplify fractions even by mere inspection.

 

[Mark44]

Hey there,

my question is:
do i always have to compare the answer of the multiplied prime numbers by dividing the numerator and denominator with the given answers to see which one is correct. or is there a way to to always see the accurate answer without having to compare them.

here is my way of reducing fractions, feedback would be appreciated I'm pretty new to this.

lets take 108/162

108:2=52:2=27:3=9:3=3

I get what you're trying to say, but it's terrible notation (and you have a mistake, probably a typo), as you are saying that 108:2 = 3. An improvement would be this:
108 = 2 * 54 = 2 * 2 * 27 = 2 * 2 * 3 * 3 * 3, or $2^2\cdot 3^3$.

2x2x3x3 = 36
162:2=81:3=27:3=9:3=3

162 = 2 * 81 = 2 * 9 * 9 = 2 * 3 * 3 * 3 * 3, or $2 \cdot 3^4$.

2x3x3x3 = 54

108:36 = 3
162:36 = 4.5

108:54 = 2
162:54 = 3

I feel like I'm doing something incorrect, as i seem to stumble up on some issues with certain fractions like 78/91
78:2=39:3=13
91:7= 13

i don't see why i am supposed to divide this one by 13, if the formula used above are for other fractions.

 

[micromass]

i feel like reversed computation is quite a hassle. especially when we're using uneven and uncommon fractions like 481/643 for example which is unable to reduce, how would i know this as quickly as possible, without having to deal with all the hassle.

You know this by computing the greatest common divisor of numerator and denominator. The Euclidean algorithm is a very ancient and very efficient way of doing this. If the greatest common divisor is $1$, then you can't reduce. If it is any other number, then you can divide the numerator and denominator by that number.