# Fractions and Division

• Jul 7th 2011, 10:42 AM
MDS1005
Fractions and Division
I seem to be having trouble with the following types of problems.

$\displaystyle \dfrac{3y^8}{9y^1^2}$

and

$\displaystyle \dfrac{6y^7}{10y^2}$

After a bit of thinking I realized that on the first problem, the three and the nine reduce to one and three, giving me an answer of

$\displaystyle \dfrac{1}{3y^4}$

The problem I was having here was completely forgetting that the 3 and the 9 in the numerator and denominator can be reduced.

In the second problem, it took me the longest time wondering why the answer wasn't written:

$\displaystyle \dfrac{3}{5}y^5$

$\displaystyle \dfrac{3y^5}{5}$

Different ways of writing the same thing? Is one the correct way of writing it, and the other wrong? What I'm getting at here is that I feel I have missed something much more basic when it comes to fractions. I'm not sure where I need to go back to and study. Can someone see what I am missing and suggest something for me to review to bring my skills in this area up to par?
• Jul 7th 2011, 10:45 AM
e^(i*pi)
Re: Fractions and Division
Quote:

Originally Posted by MDS1005
I seem to be having trouble with the following types of problems.

$\displaystyle \dfrac{3y^8}{9y^1^2}$

and

$\displaystyle \dfrac{6y^7}{10y^2}$

After a bit of thinking I realized that on the first problem, the three and the nine reduce to one and three, giving me an answer of

$\displaystyle \dfrac{1}{3y^4}$

The problem I was having here was completely forgetting that the 3 and the 9 in the numerator and denominator can be reduced.

Correct, numbers cancel along with variables.

Quote:

In the second problem, it took me the longest time wondering why the answer wasn't written:

$\displaystyle \dfrac{3}{5}y^5$

$\displaystyle \dfrac{3y^5}{5}$

Different ways of writing the same thing? Is one the correct way of writing it, and the other wrong? What I'm getting at here is that I feel I have missed something much more basic when it comes to fractions. I'm not sure where I need to go back to and study. Can someone see what I am missing and suggest something for me to review to bring my skills in this area up to par?
They are one and the same and neither is more correct than the other. When it comes to cancelling down I suggest practice, if it helps I've posed some additional examples if you wish to try them

$\displaystyle \dfrac{12y^3}{24y}$

$\displaystyle \dfrac{35x^3}{5x^3}$

$\displaystyle \dfrac{91z^{10}}{28z^{8}}$

Spoiler:

$\displaystyle \dfrac{y^2}{2} = \dfrac{1}{2}y^2$

$\displaystyle 7$

$\displaystyle \dfrac{13x^2}{4} = \dfrac{13}{4}x^2$
• Jul 7th 2011, 04:31 PM
skoker
Re: Fractions and Division
you need remember the laws of exponents in your head. Exponent Laws -- from Wolfram MathWorld

do this for a bit and you will eventually remember them when working. also remember that a coefficient is not part of the variables in the term but multiplied by them.

so you use this rule, quotient rule: $\displaystyle \frac{d^m}{d^n}=d^{m-n}$

\displaystyle \begin{aligned}\frac{10 x^6 y^4 x^2}{5 x^2 y^4 z^6} &= \frac{10}{5}\times \frac{x^6}{x^2}\times \frac{y^4}{y^4}\times \frac{z^2}{z^6} \\ &= \frac{10}{5}\times x^{6-2} \times y^{4-4} \times z^{2-6} \\ &= \frac{2}{1}\times x^{4} \times y^{0} \times z^{-4} \\ &= \frac{2x^4}{z^4} \end{aligned}
• Jul 7th 2011, 04:38 PM
theloser
Re: Fractions and Division
Quote:

Originally Posted by MDS1005
I seem to be having trouble with the following types of problems.

$\displaystyle \dfrac{3y^8}{9y^1^2}$

and

$\displaystyle \dfrac{6y^7}{10y^2}$

After a bit of thinking I realized that on the first problem, the three and the nine reduce to one and three, giving me an answer of

$\displaystyle \dfrac{1}{3y^4}$

The problem I was having here was completely forgetting that the 3 and the 9 in the numerator and denominator can be reduced.

In the second problem, it took me the longest time wondering why the answer wasn't written:

$\displaystyle \dfrac{3}{5}y^5$

$\displaystyle \dfrac{3y^5}{5}$

Different ways of writing the same thing? Is one the correct way of writing it, and the other wrong? What I'm getting at here is that I feel I have missed something much more basic when it comes to fractions. I'm not sure where I need to go back to and study. Can someone see what I am missing and suggest something for me to review to bring my skills in this area up to par?

$\displaystyle \frac{a}{a}=1$ So therefore you can slash and burn over factors but not terms such as $\displaystyle \frac{a-1}{a}$.
$\displaystyle \frac{a}{b}x^2$ is the same as $\displaystyle \frac{ax^2}{b}$ because it is $\displaystyle \frac{a}{b} * \frac{x^2}{1}$.
Try reviewing the basic forms of fractions again:
$\displaystyle \frac{a}{b}+\frac{c}{b}=\frac{a+c}{b}$

$\displaystyle \frac{a}{b}*\frac{c}{d}=\frac{ac}{bd}$

$\displaystyle \frac{a}{b}*\frac{c}{1}=\frac{ac}{b}$ and etc...
• Jul 8th 2011, 10:31 AM
MDS1005
Re: Fractions and Division
Ok, I get it, thanks for the help everyone.

I'm finding that studying all this on my own with no teacher, and just the aid of a textbook, while frustrating, is a very good exercise in logic. I keep encountering examples that weren't fully explained in the text, and having to work out the reasoning behind the problems myself. The following is an example where there was no answer in the back of the book. I'd like to know if I was correct.

$\displaystyle \dfrac{5x^2}{10x^-^5}$

This is the first time I've encountered a positive exponent divided by a negative exponent. Now remembering (thanks to everyones help) to first reduce the 5 and the 10, I then move onto working out the exponents. It would stand to reason that I'm simply subtracting -5 from 2, obviously giving me 7. Remembering now (again, thanks to everyones help) that x^7 is the same as x^7/1, I came up with the following answer:

$\displaystyle \dfrac{x^7}{2}$

Correct?
• Jul 8th 2011, 10:42 AM
MDS1005
Re: Fractions and Division
Quote:

Originally Posted by theloser
Try reviewing the basic forms of fractions again:
$\displaystyle \frac{a}{b}+\frac{c}{b}=\frac{a+b}{c}$

Is this wrong? Or am I misunderstanding something? Plug in some numbers for the variables such as a=1, b=4, c=2

$\displaystyle \dfrac{1}{4}+\dfrac{2}{4}=\dfrac{3}{4}$

$\displaystyle \dfrac{3}{4}$ Does Not Equal $\displaystyle \dfrac{1+4}{2}$

Should be a/c+b/c=a+b/c correct?
• Jul 8th 2011, 10:58 AM
e^(i*pi)
Re: Fractions and Division
Quote:

Originally Posted by MDS1005
Ok, I get it, thanks for the help everyone.

I'm finding that studying all this on my own with no teacher, and just the aid of a textbook, while frustrating, is a very good exercise in logic. I keep encountering examples that weren't fully explained in the text, and having to work out the reasoning behind the problems myself. The following is an example where there was no answer in the back of the book. I'd like to know if I was correct.

$\displaystyle \dfrac{5x^2}{10x^-^5}$

This is the first time I've encountered a positive exponent divided by a negative exponent. Now remembering (thanks to everyones help) to first reduce the 5 and the 10, I then move onto working out the exponents. It would stand to reason that I'm simply subtracting -5 from 2, obviously giving me 7. Remembering now (again, thanks to everyones help) that x^7 is the same as x^7/1, I came up with the following answer:

$\displaystyle \dfrac{x^7}{2}$

Correct?

Yes, that's correct.

Quote:

Originally Posted by MDS1005
Is this wrong? Or am I misunderstanding something? Plug in some numbers for the variables such as a=1, b=4, c=2

$\displaystyle \dfrac{1}{4}+\dfrac{2}{4}=\dfrac{3}{4}$

$\displaystyle \dfrac{3}{4}$ Does Not Equal $\displaystyle \dfrac{1+4}{2}$

Should be a/c+b/c=(a+b)/c correct?

Yes, it should be $\displaystyle \dfrac{a}{c} + \dfrac{b}{c} = \dfrac{a+b}{c}$. I suspect theloser made a typo
• Jul 8th 2011, 11:30 AM
MDS1005
Re: Fractions and Division
Yes, I figured it was a typo as well. Didn't mean to imply he was flat out wrong. With all the confusion I've had over this lately though, I just wanted to make absolutely sure. Thanks one more time to everyone. I feel I have it figured out now, and am comfortable moving on with my studies.