# A strange simplification?

#### snowJT

I was copying notes in class down, now that class just ended, I was reading over the notes, and I saw this line that went from...

THIS:
$$\frac{x-y}{2x} = \frac{dy}{dx}$$

To THIS:
$$\frac{1}{2}(1-\frac{y}{x}) = \frac{dy}{dx}$$

But I'm wondering... wouldn't It be this?

$$\frac{1}{2}(x-\frac{y}{x}) = \frac{dy}{dx}$$

if not.. how come?

#### arildno

Homework Helper
Gold Member
Dearly Missed
Do you think that:
$$\frac{7-2}{7}=7-\frac{2}{7}$$??

#### neutrino

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

#### snowJT

thanks, I see it like that; however, when you try and multiply it back....

$$\frac{1}{2}(1-\frac{y}{x})= \frac{1}{2}-\frac{y}{2x}$$

no?

#### arildno

Homework Helper
Gold Member
Dearly Missed
That is correct. So?

#### snowJT

Yes, but you can't multiply it back to get $$\frac{x}{x}$$ to get $$\frac{x-y}{2x}$$

Homework Helper
What......?

#### snowJT

well...

I don't think this makes sense... because how can you get x back when you go from this

$$\frac{1}{2}(1-\frac{y}{x})$$

to this:

$$\frac{1}{2}\left(\frac{x}{x} - \frac{y}{x}\right)$$ <-- x over x appears from 1 over 1 which doesn't make sense

Last edited:

#### matt grime

Homework Helper
So you're saying that 2 divided by 2 is not 1?

#### snowJT

no, I'm saying you can't get x when you go backwards

#### cristo

Staff Emeritus
no, I'm saying you can't get x when you go backwards
I don't know what your mean. This can be simplified as $$\frac{1}{2}(1-\frac{y}{x})=\frac{1}{2}-\frac{y}{2x}=\frac{2x-2y}{4x}=\frac{x-y}{2x}$$

What's wrong with that?

#### snowJT

thanks... that cleared it up well, I see it thanks... there were some other people here beside me at school who wern't getting it too, but now they do too, lol

#### matt grime

Homework Helper
@cristo: what's wrong is that you needlessly introduce extra factors of 2 for no reason.
@snowJT: you were taught in elementary school or primary school, or some time before the age of 11 to put things over a common denominator, i.e. that a/b + c/d = (ad+bc)/bd, so use it, but sensibly so that you don't introduce unnecessary factors that are going to cancel out later. You are just adding together fractions. That is something you have been doing for years.

#### cristo

Staff Emeritus
@cristo: what's wrong is that you needlessly introduce extra factors of 2 for no reason.
How is that wrong? Clearly I know that the factor of two was included unnecessarily, but since the OP could not see the answer, I expanded out the fraction fully, then used the rule that he would be familiar with, namely that a/b+c/d=(ad+bc)/bd.

I understand your point, but sometimes, when someone doesn't see what is going on, it is *simpler* to follow rules that they will know, and then simplify later. Skipping steps, in this case, may have added confusion.

#### matt grime

Homework Helper
"Putting things over a common denominator" should be well known from the age of *insert age when you learn what a fraction is*, and knowing that that if you have denominators 2x and 2 then you need only 2x shuold be clear if you explain that is all you're doing. Of course, multipling the 1/2 in is the real culprit here.

#### Gib Z

Homework Helper
haha I know this is kinda random, but i remember when I was like 3 and my sister said she was learning algebra at school, and asked me what a+a was, i was convinecd it was b :D

it's not b

#### uart

How is that wrong? Clearly I know that the factor of two was included unnecessarily, but since the OP could not see the answer, I expanded out the fraction fully, then used the rule that he would be familiar with, namely that a/b+c/d=(ad+bc)/bd.
I found it interesting that the OP could understand the algebra when cristo included the redundant factor but could not understand it previously.

It seems to be a case of OP not being able to grasp that 1/2 = x/2x but still being able to use the "method of cross multiplication" to add fractions. I think this implies that he was taught how to do things but not why those procedures work. Sad.

#### uart

That's actually true if you're talkling in ASCII. 