Solving Complex Fraction: Hint for Rule to Follow

[Raizy]

Homework Statement

I am just looking for a hint on which rule to follow first; what would you tell me if I already know how to deal with an ordinary complex fraction?

I tried using x (and multiplied everything) as the LCD and got the wrong answer.

$$3 - \frac {2}{1-\frac{2}{3-\frac{2}{x}}}$$

If it's hard to see the bottom part, it is:

$${1-\frac{2}{3-\frac{2}{x}}}$$

 

[Phrak]

Is x complex? Why don't we call it z; then z=x+iy?


$$3 - \frac {2}{1-\frac{2}{3-\frac{2}{z}}}$$


Multiply the top and bottom of z/2 by the complex conjugate of z.

 

[gabbagabbahey]

Is x complex? Why don't we call it z; then z=x+iy?

Multiply the top and bottom of z/2 by the complex conjugate of z.

I don't think x is a complex number. I'm pretty sure that he is just supposed to simplify this expression as much as possible, these types of embedded fractions are often called a complex fraction, even though it has nothing to do with complex numbers

I tried using x (and multiplied everything) as the LCD and got the wrong answer.

Why don't you show us what you tried, and why you think it is the wrong answer...

 

[Phrak]

I don't think x is a complex number. I'm pretty sure that he is just supposed to simplify this expression as much as possible, these types of embedded fractions are often called a complex fraction, even though it has nothing to do with complex numbers.

Yeah. I hadn't thought of that... It is rather complex as fractions go.

 

[HallsofIvy]

Homework Statement

I am just looking for a hint on which rule to follow first; what would you tell me if I already know how to deal with an ordinary complex fraction?

I tried using x (and multiplied everything) as the LCD and got the wrong answer.

$$3 - \frac {2}{1-\frac{2}{3-\frac{2}{x}}}$$

If it's hard to see the bottom part, it is:

$${1-\frac{2}{3-\frac{2}{x}}}$$

First step: multiply both numerator and denominator of that fraction:
$$\frac{2x}{(3- \frac{2}{x})x}= \frac{2x}{3x- 2}$$
Now subtract:
[tex]1- \frac{2x}{3x-1}[/itex] <br /> and continue.[/tex]

 

[jambaugh]

To OP, did you mean to say "compound fraction"?

I find it helps to put in the grouping symbols implicit in the bar fraction notation and then work from the inside out:

$$3 - \frac{2}{1 - \frac{2}{3-\frac{2}{x}}}$$
$$3 - \frac{2}{\left(1 -\frac{2}{\left(3-\frac{2}{x}\right)}\right)}$$

Now simplify:
$$3 - \frac{2}{x} = \frac{3x - 2}{x}$$
$$3 - \frac{2}{\left(1 - \frac{2}{\frac{3x-2}{x}\right)}\right)}=3 - \frac{2}{\left(1 -\frac{2x}{3x-2}\right)}\right)}$$

and so on...

 

[Raizy]

Thanks all for the advice. I tried both Jambaugh's advice and HallsofIvy's. I preferred Halls method since it skips a step Okay... so, in BEDMAS, when they say deal with brackets from innermost first, does this mean the same thing for fraction bars? Fraction bars are classified as a type of bracket?

The original expression I had to simplify:

$$3-\frac {2}{1-\frac{2}{3-\frac{2}{x}}$$

The book's answer is: $$-\frac{3x+2}{x-2}$$

Attempt #1 (Jambaugh's advice):

=$$3 - \frac{2}{1 -\frac{2}{\left(3-\frac{2}{x}\right)}\right)}$$

=$$3 - \frac{2}{\left(1 - \frac{2}{\frac{3x-2}{x}\right)}\right)}=3 - \frac{2}{\left(1 -\frac{2x}{3x-2}\right)}\right)}$$

=$$3-\frac{6x-4}{3x-2-2x}=3-\frac{6x-4}{x-2}$$

=$$\frac{3x-6-6x+4}{x-2}$$

My final answer: = $$\frac{-3x-2}{x-2}$$
Now this is probably equivalent to: $$-\frac{3x+2}{x-2}$$ ... but I am still wondering why that makes sense...Attempt #2: HallsofIvy's advice:

$$3-\frac{2}{1-\left(\frac{2}{3-\frac{2}{x}}}\right)$$
=$$3-\frac{2}{1-\frac{2x}{3x-2}}=3-\frac{6x-4}{3x-2-2x}$$

My final answer: $$\frac{-3x-2}{x-2}$$

Here is my original attempt, I missed some steps, as follows:

=$$3-\frac{2}{1-\frac{2}{3x-2}}=3-\frac{6x-4}{3x-2-2}=3-\frac{6x-4}{3x-4}$$

=$$\frac{9x-12-6x+4}{3x-4}=\frac{3x-8}{3x-4}$$

There were a few other attempts, all which were wrong. My mind was all over the place; very unorganized, very confusing, very frustrating...

 

[tiny-tim]

Hi Raizy!

Okay... so, in BEDMAS, when they say deal with brackets from innermost first, does this mean the same thing for fraction bars? Fraction bars are classified as a type of bracket?

Yes, when you have a fraction, imagine it's all on one line, like (a + b)/(c - d), so you have to put brackets round the numerator or denominator

for more detail, see http://en.wikipedia.org/wiki/BODMAS#Proper_use_of_parentheses_and_other_grouping_symbols

 

[Mark44]

Thanks all for the advice. I tried both Jambaugh's advice and HallsofIvy's. I preferred Halls method since it skips a step Okay... so, in BEDMAS, when they say deal with brackets from innermost first, does this mean the same thing for fraction bars? Fraction bars are classified as a type of bracket?

The original expression I had to simplify:

$$3-\frac {2}{1-\frac{2}{3-\frac{2}{x}}$$

The book's answer is: $$-\frac{3x+2}{x-2}$$

Attempt #1 (Jambaugh's advice):

=$$3 - \frac{2}{1 -\frac{2}{\left(3-\frac{2}{x}\right)}\right)}$$

=$$3 - \frac{2}{\left(1 - \frac{2}{\frac{3x-2}{x}\right)}\right)}=3 - \frac{2}{\left(1 -\frac{2x}{3x-2}\right)}\right)}$$

=$$3-\frac{6x-4}{3x-2-2x}=3-\frac{6x-4}{x-2}$$

=$$\frac{3x-6-6x+4}{x-2}$$

My final answer: = $$\frac{-3x-2}{x-2}$$
Now this is probably equivalent to: $$-\frac{3x+2}{x-2}$$ ... but I am still wondering why that makes sense...

The two results above are equal. The numerator of your answer is -3x - 2. That is equal to -(3x + 2). If you have a rational expression with a negative sign, the sign can appear in the numerator, or the denomintor, or to the front of the fraction. For example, -1/2 = 1/(-2) = -(1/2).

Attempt #2: HallsofIvy's advice:

$$3-\frac{2}{1-\left(\frac{2}{3-\frac{2}{x}}}\right)$$
=$$3-\frac{2}{1-\frac{2x}{3x-2}}=3-\frac{6x-4}{3x-2-2x}$$

My final answer: $$\frac{-3x+2}{x-2}$$

You did something wrong. This answer is not equal to the one you got before, which was the same as the book's answer.

Here is my original attempt, I missed some steps, as follows:

=$$3-\frac{2}{1-\frac{2}{3x-2}}=3-\frac{6x-4}{3x-2-2}=3-\frac{6x-4}{3x-4}$$

=$$\frac{9x-12-6x+4}{3x-4}=\frac{3x-8}{3x-4}$$

This is looks way off.

There were a few other attempts, all which were wrong. My mind was all over the place; very unorganized, very confusing, very frustrating...

 

[Mentallic]

Here is my original attempt, I missed some steps, as follows:

=$$3-\frac{2}{1-\frac{2}{3x-2}}=3-\frac{6x-4}{3x-2-2}=3-\frac{6x-4}{3x-4}$$

=$$\frac{9x-12-6x+4}{3x-4}=\frac{3x-8}{3x-4}$$

There were a few other attempts, all which were wrong. My mind was all over the place; very unorganized, very confusing, very frustrating...

Everything was right except for the first line.

$$3 - \frac {2}{1-\frac{2}{3-\frac{2}{x}}} \neq 3-\frac{2}{1-\frac{2}{3x-2}}$$

Focusing strictly on that smallest fraction part:

$$\frac{2}{3-\frac{2}{x}} = \frac{2}{\left( \frac{3x-2}{x}\right )} =\frac{2x}{3x-2}$$

You have been able to follow these steps correctly for all the other fractions, so I think you just made a little slip-up there.

 

[Raizy]

You did something wrong. This answer is not equal to the one you got before, which was the same as the book's answer.

Ahh, for attempt number 2 I just realized that, it was a copying error. I've edited it so I guess the answer is equivalently right.

I've been practicing these "compound" fractions for about 2 days now. I bet I'll forget everything I did by tomorrow, even if I practice before sleeping. lol. There is just so many details going on in my head it's as if they keep slipping through.

*sigh*... how long until my brain puts this off into long-term memory

 

[Mentallic]

*sigh*... how long until my brain puts this off into long-term memory

How about trying a different approach? Rather than just answering more of these problems, take a step back and work with more simple fractions. Once you understand the concept of how to work with them and manipulate them with relative ease, move your way up the ladder.
Better this way than taking multiple stabs in the dark.

The fractions all work the same way, and these compound fractions are just like simplifying 2+ fractions one after the other in the same problem.


Maybe I'll get you started if you like:

Say you are given the fraction: $$a-\frac{b+c}{d}$$ and asked to simplify it as much as possible.

Find a lowest common denominator (LCD) between the 2 'fractions'.

This would be 'd' so the part of the fraction 'a' needs to be manipulated to have a denominator of d, without changing the fraction. In other words, multiply both numerator and denominator by d.

$$\frac{ad}{d}-\frac{b+c}{d}$$

Now since the 2 fractions have the same denominator, their numerators can be added together:

$$\frac{ad-(b+c)}{d}$$

now simplify by expanding: $$\frac{ad-b-c}{d}$$ and this is the simplest it can be.

If this were a part of a compound fraction, well then now we can move onto the larger outside fraction. such as:

$$\frac{e}{a-\frac{b+c}{d}} = \frac{e}{\left(\frac{ad-b-c}{d}\right)}$$

A fraction on a fraction can of course be simplified. Multiply both numerator and denominator by the denominator in the smaller fraction:

$$\frac{d(e)}{d\left(\frac{ad-b-c}{d}\right)}$$

simplify: $$\frac{de}{ad-b-c}$$

and keep following this process (or any other you choose) for the entire compound fraction.

 

[Raizy]

Thanks Mentallic for that quick lesson. That was a nice review this afternoon I can manage those simpler ones no problem. I guess my problem was realizing I had to deal with the inner most fraction group and work my way outward if there was a bunch of fractions clumped up.