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### Course: Algebra basics > Unit 2

Lesson 7: Nested fractions# Nested fractions

Sal starts with a simple case of multiplying two fractions and proceeds to simplifying a nested fraction expression.

## Want to join the conversation?

- The exercise is so confusing, and it's definitely way more complex than the ones shown in the video! HELP!!(130 votes)
- I have to say this area has not been explained very well, I have had no issue up until now, then all of a sudden it feels like its just jumping ahead in difficulty, I wish this had been explained better.. great website otherwise(86 votes)
- I'm right there with you. Everything else made sense and now we go from simple fractions to pre-calculus(3 votes)

- At3:411/a - 1/b then becomes b/ba - a/ba why did the a and b get reversed? shouldnt it be a/ba - b/ba still? sorry, this video is the first on the site that has confused me. thanks(40 votes)
- We had:

(1/𝑎) - (1/𝑏)

Multiplying the first fraction by 𝑏/𝑏:

(𝑏/𝑎𝑏) - (1/𝑏)

Multiplying the second fraction by 𝑎/𝑎:

(𝑏/𝑎𝑏) - (𝑎/𝑎𝑏)

Therefore:

(1/𝑎) - (1/𝑏) = (𝑏/𝑎𝑏) - (𝑎/𝑎𝑏)

Comment if you have questions.(21 votes)

- How is the reciprocal of 1/C = 1/C ?(15 votes)
- In the video, he was telling us that we were dividing by "c" (i.e. c/1). So, the reciprocal is 1/c :)(22 votes)

- I have watched this video and the one before it but can't understand what they are trying to teach. Can anyone explain the basic rule of nested fractions?(18 votes)
- The big idea is dividing rational numbers (fractions). Generally speaking I advise my students to rewrite fraction division as the product of the reciprocal.

So if you had (1/2)/(3/4) I would suggest re writing the problem as 1/2 x 4/3 then multiply the fractions as you normally would.

There isn't a basic rule in the sense of step 1 do this, step 2 do this every time. This is where math can be a puzzle or brain teaser, combine what you can using the rules of algebra, until you get everything down to only one fraction/division bar.(11 votes)

- Not a question as I can't even articulate one right now! I was doing pretty well until I got to this. Now I'm completely lost. Much older student here... trying to get math skills brushed up based on my placement exam so I can take my required course. Seems futile at the moment. (going to re-watch the video after reading some of these q's here and the explanations provided, not sure I'll get anywhere but will certainly try.)(14 votes)
- This video is not the best on this topic, and its the only one on this site. I would suggest that you do an internet search on "simplifying complex fractions". You'll find some much better explanations than this one.(13 votes)

- I have been working on nested fractions for an hour so far today, and I am not understanding. I get to the part where I multiply by the reciprocal but when it comes to actually doing the math and writing down what goes where, I get lost. Any suggestions or steps I need to be taking? Thanks!(11 votes)
- I've been at this 3 days and I'm so frustrated. Every time I think I have it, it doesn't make sense.(1 vote)

- Is there a method for retaining all this information? I practice EVERY DAY and 24 hours later, it's all gone. I've been working through the Algebra Basics for FIVE YEARS now and not making any progress. How do I make sense of this?(12 votes)
- This is confusing and the video doesn't help at all. Out of all the helpful videos i've seen so far this is the only unhelpful one.

There are many parts in the video where it doesn't explain what it's doing or rushes through an explanation. Not only that but it seems "Nested fractions" was a term khan academy made up. A more true name would be "Complex fractions."

I also have another problem with this specific video, why solve the problem this way? Other tutorials do not do what he does and often try to find the common denominator first. Here he just flips everything with little explanation and gets variables for nowhere.(9 votes)- You're right - this is not one of their better videos. I've commented before on that myself. However, Sal's steps are valid. He is following order of operations rules. He is not getting variables from no where. The work is all driven by what is in the problem.

Sal's example is a little unusual because he has a complex fraction divided by another fraction. Usual problems deal with just one complex fraction.

In my experience if you don't understand working with basic fractions and/or order of operations, then you are likely to get confused. And, the variables are apt to also confuse some people.

Anyway... To solve using order of operations:

1) Sal starts by subtracting the 2 fractions in the numerator. Subtracting fractions requires a common denominator, which for this problem is "ab". Sal converts each fraction to the common denominator and then combines the results to create the fraction: (b-a)/ab

2) Next, Sal takes this fraction and divides it by "c". To divide fractions, we change the division to multiplication by using the reciprocal of the 2nd fraction. This is where the 1/c comes from.

(b-a)/ab divided by c = (b-a)/ab * 1/c.

3) And, the problem includes division by 1/d. So, this also gets changed to multiplication by using its reciprocal.

(b-a)/ab * 1/c divided by 1/d = (b-a)/ab * 1/c * d/1

4) Now, the only step left is to complete the multiplication.

Hope this helps.(6 votes)

- At the end when we have db -da/abc wouldn't the answer be zero?

My reasoning on this is the b's and a's would cancel out. That would

leaved the d's and d-d = 0 and 0/c would equal zero right?(6 votes)- No, you can't cancel that way. The numerator is a sum and the denominator is a product. in that situation there isn't anything you can do in the way of cancellation, unless the entire numerator is a factor of the denominator (and often when you get that result it means you've done the problem in a more complicated way than was necessary). You can prove to yourself that it wouldn't work by plugging in numbers for the variables. For example if a=1, b=2, c=3, and d=4, then d(b-a)/abc simplifies to 2/3. If your method was correct the answer would be zero regardless of the actual value of the variables.

One issue I have with Khan Academy (at least for math) is that although most of the concepts are explained well, there aren't many videos showing what NOT to do. I mention this because your question is a common one. Many students make that exact mistake, and most teachers would address it in the classroom. Check out youtube for some variety in teaching techniques, there is a lot of math content.(11 votes)

## Video transcript

- Let's deal with some
algebraic expressions that involve multiplying fractions. So let's say that I had a over b, a over b times c over d. What is this going to be? I encourage you to pause the video and try to figure it out on your own. Well, when you multiply fractions, you just multiply the numerators and multiply the denominators. So the numerators here, a, c, you're just going to multiply those out. It's going to a times c, which we can just write as ac, that just a times c, all of that over the denominator, b times d, b times d. Instead of multiplying,
what would of happened if we were dividing? So if we had a over b, a over b divided by, divided by c over d, what would this be? And once again, encourage you, encourage you to pause the video and figure it out on your own. Well, when you divide by a fraction, it is equivalent to multiplying by it's, by it's reciprocal. So this is going to be the same thing as a over b, a over b times, times the reciprocal of this. So times d over, I'm going to use the same color just so
I don't confuse you, that d was purple, times d over c, times d over c and then it
reduces to a problem like this. You know, and I shouldn't even use this multiplication symbol
now that we're in algebra because you might confuse that with an x, so let me write that as times, times d, d over c, times d over c, and what are you left with? Well the numerator you're
going to have a times d, so it's ad, a, d, over, over bc, over b times c. Now let's do one that's maybe
a little bit more involved and see if you can pull it off. So let's say that I had,
let's say that I had, I don't know, let me write it as 1 over a, minus 1 over b, all of that over, all of that over c, and let's say, let's also divide that by 1 over d. So this is a more involved expression then what we've seen so
far but I think we have all the tools to tackle
it so I encourage you to pause the video and see
if you can simplify this, if you can actually carry
out these operations and come up with a one
fraction that represents this. All right so let's work
through it step by step, so 1 over a minus 1 over b, let me work through just
that part by itself. So 1 over a minus 1 over b, we know how to tackle that, we can find a common denominator, let me write it up here. So 1 over a minus 1 over b, is going to be equal to, we can multiply 1 over a times b over b, so it's going to be b over ba, notice I haven't changed it's value, I just multiplied it times 1, b over b, minus, well I'm going to multiply the numerator and denominator here by a, - a over ab, or I could write that as ba. And the whole reason why I did that is to have the same denominator. So this is going to be equal to b-a, over, I could write is a ba or ab. So this is going to be equal to, this is going to be equal to
this numerator right over here, b-a, over ab, and then, if I'm dividing it by c, that's the same thing as multiplying by the reciprocal of c. So if I'm dividing it by c, that's the same thing as multiplying, that's the same thing as
multiplying times 1 over c. And if I am, and I'll
just keep going here, if I'm dividing by 1 over d, if I'm dividing, notice
this is the same thing as division right over here. If I'm dividing by c,
that's the same thing as multiplying by the reciprocal of c. And then finally, I'm
dividing by 1 over d, that's the same thing as multiplying by the reciprocal of 1 over d. So the reciprocal of 1 over d is d over, d over 1. And so what does this result with? Well in the numerator I
have b-a times 1 times d. So we can write this as d times (b-a), times (b-a) and then in the denominator I have abc, ab and c. And then finally, we can use
the distributive property here, we can distribute this d, and we're going to be left with, we deserve a minor drum
roll at this point, we can write this as d times b, d times b minus d,
woops, I want to do that in the same green color so you really see how it got distributed, minus d times a, all of that over ab, abc. And we are done.