The steps of simplifying a fraction?

Question

So I'm in an Adult Education class for my GED and I'm trying hard to study on my Math which is the only subject I have trouble with. I only have "barely" a 6th grade education to Math so I'm having a lot of trouble totally understanding it all.

I'm trying to learn how to simplify fractions on Khan Academy which is the current task we're learning in the class. I am not getting any of it. I'm trying to answer the question

Simplify to lowest terms.

36/60

I tried to Google the answer to see if I can find an explaination how to do this and found one on Y/A but the question was how to simplify 60/36 instead. I assume it was the same steps.

You're looking to get rid of the common factors, and inspection is probably easiest here.

First try 6 as a factor, so

(60/6) /(36/6) = 10/6

10/6 is even, so divide by 2

(10/2) / (6/2) = 5/3

OK, that's pretty simple, and both sides are prime, so the only thing you could do further is change

5/3 = (3 + 2)/3 = 3/3 + 2/3 = 1 2/3

5/3 or 1 2/3 - take your pick as to the simplest!

I'm still not understanding it.

I understand that:

• Find the number that goes into both 36 and 60
• Multiply the number and count how many times it goes into 36/60
• ???

I'm just not sure what to do after that. What is the next step?

Answer 1

If both the denominator and numerator are divisible by a common number, divide both by that number. Rinse and repeat until the only number that they are both divisible by is 1. I think that's really all there is to it.

In your example, you could divide by 6 then 2. Or you could just directly divide by 12. Or you could divide by 3, then 2, then 2. It'll get you to the same answer of $3/5$ either way. In terms of doing problems quickly, it should be pretty obvious that you'd want to divide by the highest number that they're both divisible by (here 12), so you don't need multiple steps, but it doesn't really matter...

Answer 2

Are you familiarized with the prime number decomposition? You can also decompose the two numbers into prime ones and simply cancel the common ones; in this case,

$36 = 2·2·3·3$

$60 = 2·2·3·5$

hence $\frac{36}{60} = \frac{2·2·3·3}{2·2·3·5} = \frac{3}{5}$

Answer 3

The basic idea is that, if you have a fraction (where the denominator is not zero), you can multiply or divide the numerator and denominator by the same nonzero value without changing the value of the fraction.

In your case you have $\dfrac{36}{60} $. You noticed that $36=6\cdot6$ and $60 = 6\cdot10$, so you can write your fraction as $\dfrac{6\cdot6}{6\cdot10} $. Seeing the $6$ in both the numerator and denominator, you can cancel it (or divide it out) to get $\dfrac{6}{10} $.

Another way to see this is to notice that $\dfrac{6\cdot6}{6\cdot10} =\dfrac{6}{6}\cdot\dfrac{6}{10} =1\cdot\dfrac{6}{10} =\dfrac{6}{10} $.

At this stage, you notice that $6=2\cdot3$ and $10=2\cdot5$, so $\dfrac{6}{10} =\dfrac{2\cdot3}{2\cdot5} =\dfrac{3}{5} $.

Since no integer (greater than one) divides both $3$ and $5$, this fraction is in lowest terms so you are done.

Answer 4

The complete general way is to prime factorize the numerator and denominator. To factorize a number into primes is to find a product of primes that equals the given number. This may seem slightly challenging, but as long as the numbers aren't that large, it is usually not too difficult.

Let me give you an example of a prime factorization of, say, $78$. We first note that it is an even number, so $2$ is a factor, i.e. $78 = 2\cdot 39$. Since $2$ is prime we can not reduce that futher, but $39$ is not. We find the factors of $39 = 3\cdot13$. Therefore $78 = 2\cdot3\cdot13$, where all the factors are now prime.

In the fraction we just cancel each common prime factor, i.e. $$\require{cancel}\frac{36}{60}=\frac{2\cdot 18}{2\cdot 30}=\frac{2\cdot2\cdot9}{2\cdot2\cdot15}= \frac{\cancel{2\cdot2\cdot3}\cdot3}{\cancel{2\cdot2\cdot3}\cdot5}=\frac{3}{5}.$$

Answer 5

your difficulty may relate either to unfamiliarity with prime factorization of integers, or to the fact that fractions ( called rational numbers) are a little more complicated than appears at first sight. i will address the second of these sources of difficulty, and suggest how you may improve your understanding:

THE KEY INSIGHT

suppose $B$ and $D$ are both unequal to zero. then we will have $$ \frac{a}{B} = \frac{c}{D} $$ if and only if it is true that: $$ aD=cB $$ once you begin to feel comfortable with this key insight you may find it useful to start off with a number that you know has quite a few divisors, say $5040=1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7$ begin with the equation (which is obviously true!) $$ 5040 = 5040 $$ and break down each side in different ways to get an equation expressing equality of two (different-looking) fractions. i will work through one example, but you should repeat the process, using these numbers or others of your own choice, until you feel confident $$ ( 1 \times 3 \times 5) \times (2 \times 4 \times 6 \times 7) = (1 \times 4 \times 7) \times (2 \times 3 \times 5 \times 6) $$ i.e. $$ 15 \times 336 = 28 \times 180 $$ so here we may take $a=15, D=336, c=28, B=180$

this gives
$$ \frac{15}{180} = \frac{28}{336} $$ and also (by rewriting the equality as $15 \times 336 = 180 \times 28$) $$ \frac{15}{28} = \frac{180}{336} $$

Answer 6

Just for completeness, one should mention Euclid's algorithm. If you want to reduce $\dfrac{51}{68}$ to lowest terms, it's not hard to write $$ \frac{51}{68} = \frac{3\times17}{4\times17} $$ and then cancel the $17$s, but suppose you have $\dfrac{4953}{2159}$. Ultimately it's $\dfrac{127\times39}{127\times17}$ and you cancel the $127$s, but how do you find the $127$ so that you can do that?

$127$ is the "$\gcd$" of $4953$ and $2159$, i.e., the "greatest common divisor" of $4953$ and $2159$, the largest number that divides both of them.

Euclid's algorithm is a simple and highly efficient way to find that $\gcd$.

Here's how it works: You want to find $\gcd(4953,2159)$ (which will come to $127$).

Divide the larger number by the smaller: $2159$ goes into $4953$ a total of $2$ times and the remainder is $635$. So you put $635$ in place of the larger number, and you have $$ \gcd(4953,2159) = \gcd(635,2159). $$ The $\gcd$ is still the same. You've reduced the problem to one involving smaller numbers. You keep reducing it as many times as it takes.

Next step: $635$ goes into $2159$ a total of $3$ times and leaves a remainder of $254$. So $$ \gcd(4953,2159) = \gcd(635,2159) = \gcd(635,254). $$

Next step: $254$ goes into $635$ a total of $2$ times and leaves a remainder of $127$. $$ \gcd(4953,2159) = \gcd(635,2159) = \gcd(635,254)= \gcd(127,254). $$

Next step: $127$ goes into $254$ a total of $2$ times and leaves a remainder of $0$: $$ \gcd(4953,2159) = \gcd(635,2159) = \gcd(635,254)= \gcd(127,254) = \gcd(127,0). $$

And you're done. The $\gcd$ of $127$ and $0$ is $127$, since $\dfrac{0}{127} = \dfrac{0\times127}{1\times127}$.

Euclid was a Greek who lived in the third century BC in Alexandria (in Egypt). He explained why this works, and this is the oldest algorithm still in standard use today.