# Dividing Notation ?

When cancelling out common factors eg.
2x-1
-e^(-3x) -- -- --
e^(-3x)

with that bar notation -I see to be moving it together in my head

as one fraction (with nothing hanging off the side)

like -e^(-3x)(2x-1)
--------------
e^(-3x)

so I can see it better...

and then cancelling = -(2x-1)

...I cant seem to get the formatting on here for some reason but I mean

-e^(-3x)*(2x-1)/e^(-3x) ...but write with -e^(-3x) hanging off the side of the fraction ...

I do this a lot ...but NOT everytime ..

like if its really easy or if im going really quickly, I dont move it all as one fraction like that

ill just wrte the answer ...

is this normal?

## Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
I have absolutely no idea what you are talking about! What do you mean by "hanging off the side"?

Are you talking about
$$\frac{(2x-1)}{e^{-3x}}e^{-3x}$$
versus
$$\frac{e^{-3x}(2x-1)}{e^{-3x}}$$ ?

And what do you mean by "move it all as one fraction"?

I have absolutely no idea what you are talking about! What do you mean by "hanging off the side"?

Are you talking about
$$\frac{(2x-1)}{e^{-3x}}e^{-3x}$$
versus
$$\frac{e^{-3x}(2x-1)}{e^{-3x}}$$ ?

And what do you mean by "move it all as one fraction"?

Yea, if it were something like $$\frac{2x-1}{e^{-3x}}e^{-3x}$$

then I usually bring it up in my head before I cancel off common factors...

like $$\frac{e^{-3x}(2x-1)}{e^{-3x}}$$

because if I just cancel off from $$\frac{2x-1}{e^{-3x}}e^{-3x}$$, sometimes the

____________ is left over ...and I'd have to put a 1 there for it to make sense ..and ive never seen that in a book or anything

and then the basic rule I thought was a^m/a^n=a^(m-n) ...but then does it have to be

all over a single fraction like that before you cancel, ...

I think i'd be more comfortable bring it all up as multiplication ...so I can see how the rule

applies exactly ...but I know people dont do that and use both

the a^m / a^n = a^(m-n) IS used. e^(-3x) / e^(-3x) = e^(-3x - (-3x)) = e^(-3x + 3x) = e^(0) = 1 which is what you'd expect when dividing a/a.

It does not matter which order you do it in, multiplication for your sake is commutative, associative, etc.

I'm still not sure what your point is though...

the a^m / a^n = a^(m-n) IS used. e^(-3x) / e^(-3x) = e^(-3x - (-3x)) = e^(-3x + 3x) = e^(0) = 1 which is what you'd expect when dividing a/a.

It does not matter which order you do it in, multiplication for your sake is commutative, associative, etc.

I'm still not sure what your point is though...

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

but then if I used a^m/a^n=a^(m-n) right away

then id have
2x-1
- ------------------

- (2x-1) on top of the ---- bar thing

I know that's supposed to be -(2x-1)

but it looks weird ...

I mean technically to use a^m/a^n that rule fully ...

dont I have to write it like

$$\frac{2x-1}{e^{-3x}}e^{-3x}(-1)$$

the bar ....

=-(2x-1)*e^(-3x)/e^(-3x) ...then use the rule ...so it fits exactly

I guess my question is ...how can people not apply the rule exactly ...

like if I had something really long over the ______ bar

and then people just cancel ....

but ___something1*something2*something3________ * something6
something4*somethin5
isnt Exactly the same as something/something

(Even though I know it is by experience)

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

but then if I used a^m/a^n=a^(m-n) right away

then id have
2x-1
- ------------------

- (2x-1) on top of the ---- bar thing

I know that's supposed to be -(2x-1)

but it looks weird ...

so you are saying you'd have -(2x-1)/1 vs. -(2x-1)?

Sorry I'm not getting anywhere in anything you are saying. You need to clearly define what you are thinking, then clearly define what part of that that deviates from the "norm" is making you confused. Typing out your thoughts on here isn't helpful. Maybe it's because you need to formulate it better or maybe it is because I just don't get it. In either case good luck.

okay for eg if I had

-x^3*(3x^2-1)/(-x^3)*6x I can easily see that the -x^3 cancels off...
=(3x^2-1)*6x

but if it's written with the ______________ instead of all close together like that

-x^3 * $$\frac{3x^2-1}{-x^3}$$*6x

then I find it harder to see which thing cancels off ...and I usually ( but not all the time)

write it in the form in my head

-x^3*(3x^2-1)/(-x^3)*6x and I can see what cancels off easily...

is this normal? Or is everyone else doing it differently?

I'm confused by what's confusing you, the diagonal line (/) vs. the horizontal line (----)? You should be able to look at a * b/c * d as a * b * 1/c * d, if c = a or b or d, then you should immediately see that those 2 cancel. If you don't see it right away, you will with practice.

I do but

with the hoirzontal line ----

I see what factors cancel right away ....

but then I have a tendency in my head to move it up on top of the horizontal line

so it's written as one fraction and then cancel

but is that wrong to do ...or just uncessary ...

sometimes I find it easier

It's not wrong, it's just notation. If you like it, go nuts. As far as unnecessary? If you can't see which factors cancel right away, write them whatever way makes sense to you as long as you get the right answer.

Ok thanks...

I think the problem im having is ....I dont like it when I cancel the only factor in the denominator and then i have to put a bracket...because when I wrote it with the horizontal line _____ there was no bracket to begin with ...
I dont like that for some reason...I like to have it with a bracket and then cancel
so it looks better

I think I found a way to explain it better:

Eg.

-e^(x)*(2x-1)/e^(x) =-(2x-1)

but instead write it with the horizontal line ________

as -e^(x)*$$\frac{2x-1}{e^(x)}$$

Now cancel out the e^(x) right away in your head then it looks like

$$\frac{2x-1}{}$$*(-1)(looks weird to me ...)

=-(2x-1) ...

but if I brought it over one fraction and bracketed first i'd be nicer...

so I'm confused as to when to just do it like this ...
and get weird things with the horizonal line left over (even though I know there's a 1 there )

or when to just bring it all up over the fraction so this doesnt happen ever.

I know I could just skip ahead and do it all in my head at once...and not even consciously look at the intermediate steps...

but sometimes I dont want to

The horizontal line isn't "left over" you have (-1)*(2x-1)/1 = -(2x-1)

The horizontal line isn't "left over" you have (-1)*(2x-1)/1 = -(2x-1)

so is this how you do it ..

and see things with something over the bar and then a 1 and then =-(2x-1)

or do you bring it up to begin with and then cancel out..

idk which is better

There is no better or worse, whatever works for you but -a/a = -1/1 = -1

sorry, im still kind of confused...dont mean to be irritating but

I know how to do it ...Im in university ...im just asking because Im wondering how most people would cancel off ...and want to get in good habit and make sure im doing it the way most people would

if one of the factors is written off to the side like that as in

(-1)*$$\frac{2x-1}{e^{-3x}}e^{-3x}$$

versus

$$\frac{e^{-3x}(2x-1)}{-e^{-3x}}$$

like HallsofIvy said ...

in the first one,

would you normally bring it up and write it like the second one, and then cancel...

or would you normally just do the cancellation (imagine slashes in your head like /)

or do you do a mix of both

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hmm having some difficulty typing it out with the negative, but

without the bracket in the first one ...like how youd normally write it with the horitonzal bar..on paper

http://img174.imageshack.us/img174/2265/divlz1.jpg [Broken]

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in the 2nd one , I have no problem

in the first one, I can cancel off right away, but I often write it like I have in the second one in my head and then cancel

and idk which people usually do ...or if they do a mix of both

Mentallic
Homework Helper
Ahh so you just want to know what the more common approach to this cancelling is.

For the first: $$-e^{-x}.\frac{x^2+1}{e^{-x}}$$ this is simple enough for me to see what cancelling needs to be done with 'moving' the multiple to the numerator as such:
$$\frac{-e^{-x}(x^2+1)}{e^{-x}}$$

I'm in my last year of highschool so I'm thorough enough with these computations so as not to doubt my cancellation was correct. However, if you don't feel as comfortable 'jumping steps' there is no problem quickly re-organising the equation in a way that you please.

Possible steps:

$$-e^{-x}\frac{x^2+1}{e^{-x}}$$

$$\frac{-e^{-x}(x^2+1)}{e^{-x}}$$

$$\frac{-1(x^2+1)}{1}$$

$$-(x^2+1)$$

Ok good, so im not the only one who moves the multiple to the numerator sometimes,