How to put a fraction in simplest form, such as $140/255$?

Question

Given the fraction $$\dfrac{140}{255}$$

How do I find a common factor so it can be easily simplified? I have already tried $2$, $3$ and $4$.

Answer 1

You have the right idea. Let's see if we can get you to the answer.

You've tried $2, 3, 4$. It appears that you're trying to get the fraction in lowest terms.

Some basic divisibility rules might help. In this way you can inspect the numbers (fairly) quickly to see whether or not they evenly divide the number:

• A number is divisible by $2$ if the rightmost digit is even ($2,4,6,8,0$). So, the numerator is divisible by $2$, but the denominator isn't.
• A number is divisible by $3$ if the sum of its digits is divisible by $3$. So, $140$ is not, because $1+4+0 = 5$ is not divisible by $3$, but the denominator is, because $2+5+5 = 12$ is divisible by $3$.
• A number is divisible by $4$ if it is twice divisible by $2$. The numerator is, but the denominator is not.
• A number is divisible by $5$ if its rightmost digit is $5$ or $0$.

AHA! This looks promising:

$$\frac{140}{255} = \frac{5 \cdot 28}{5 \cdot 51} = \frac{28}{51}.$$

Are we done? Well, we can start again with this fraction. If you run through the rules above, we find that none of $2,3,4,5$ divide both the numerator and denominator.

And with that, we're done. Why? Because you don't need to try any numbers that are bigger than either the square root of the numerator, or the square root of the denominator. Since $\sqrt{28} < 6$, we didn't need to test any numbers larger than $5$.

I'll also point out that we didn't need to test the number $4$. Why not? Because $2$ divides $4$. Since the denominator wasn't divisible by $2$, it won't be divisible by $4$, either, so there's no need to test it.

We can extend this a bit and say that we only need to test numbers that have exactly two factors: itself and $1$. These are called prime numbers.

Any natural number greater than $2$ can be written as the product of one or more prime numbers. The so-called prime factorization of the original numerator is $140 = 2 \cdot 2 \cdot 5 \cdot 7$; for the denominator, it's $255 = 3 \cdot 5 \cdot 17$. Notice that the only common factor is one $5$. This is the greatest common factor. Dividing both numerator and denominator by the greatest common factor will get the fraction in lowest terms. This is why we were done with this problem after we divided by $5$.

Hope this helps!

Answer 2

To put a fraction $\frac{a}{b}$ in simplest form, the most straightforward way usually comes from knowing $\gcd(a, b)$. In the case of $\frac{140}{255}$, we have $\gcd(140, 255) = 5$. Divide the numerator by 5 and divide the denominator by 5 to obtain $$\frac{28}{51}.$$ To verify the answer, check that $\gcd(a, b) = 1$.

Answer 3

Like the man who invented 5-Up and 6-Up, you stopped one number short of success. Of $140$ and $255$, only the former is divisible by $2$ and by $4$, and only the latter is divisible by $3$, but both are divisible by $5$. This should immediately lead to the answer: $\frac{28}{51}$.

How can you find a common factor? If none immediately suggest themselves, you can try getting the prime factorization of the numerator and the denominator, and this works reasonably fine for small numbers like $140$ and $255$.

For anything much bigger than that, you might need the Euclidean GCD algorithm. Take for example $$\frac{279244}{508623}.$$ Your computer can immediately see the common factor, but without a computer (e.g., on a test for which you're only allowed a basic calculator), you don't want to spend too much time on trial division. So:

• $508623 = 1 \times 279244 + 229379$
• $279244 = 1 \times 229379 + 49865$
• $229379 = 4 \times 49865 + 29919$
• $49865 = 1 \times 29919 + 19946$
• $29919 = 1 \times 19946 + 9973$
• $19946 = 2 \times 9973 + 0$

On second thought, I hope you don't get a test problem where the GDC is a $4$-digit prime.