Question

Perform the indicated operation. 9/32-1/16
I did the math:
9*16/32*16 - 1*32/32*16 = 144/512 - 32/512 = 112/512
How would I simplify this?

Answer 1

I would recommend that you start like so:
\[\frac{9}{32}-\frac{1}{16}=\]\[\frac{9}{32}-\frac{1*2}{16*2}=\]
\[\frac{9}{32}-\frac{2}{32}= \frac{7}{32}\]

Answer 2

However, \[\frac{112}{512}\]is also correct, and we can simplify it so you know how for another time!

Answer 3

Oh ok, that would be great if you could help with that. Im lost with that big of a number

Answer 4

Okay! The quickest way is to divide by the "greatest common factor".

But then you have to know the greatest common factor.

So instead, I divide top and bottom by prime numbers to get whole numbers on top and bottom until I can't anymore!

Answer 5

I divide top and bottom by two until I get a decimal from one.

Answer 6

So divide them by 2 until the smallest number can't divide down?

Answer 7

\[\frac{112}{512}=\frac{112\div2}{512\div2}=\frac{56}{256}\]

Answer 8

Yep! Keep on dividing down.

Answer 9

Oh, I kept dividing till I got 1/2

Answer 10

But the answers are: 1/4. 5/16. and 7/32

Answer 11

\[\frac{56\div2}{256\div2}=\frac{28}{128}\]

Answer 12

\[\frac{28\div2}{128\div2}=\frac{14}{64}\]

Answer 13

oh so just divide by 2 one more time?

Answer 14

\[\frac{14\div2}{64\div2}=\frac{7}{32}\]

Answer 15

7 is a prime number, so it can't be divided by anything but itself to get a whole number.
So you could divide top and bottom by 7, but 7 doesn't go evenly into 32, so you're left with \[\frac{7}{32}\]

Answer 16

The trick is dividing by prime numbers as long as you can.

Answer 17

Once you hit a prime number, you know it can be divided only by itself.

Once a prime number is on the bottom, you're done (unless you divide the bottom by itself, and then you don't have a fraction).

For the top, just keep trying to divide by prime numbers until there are no prime numbers left that are smaller than the number on top.

Answer 18

Thanks! Take care!
Ask any follow-up questions here and I'll look for them later!
Lists of prime numbers are on the internet!