simplification?

• Nov 3rd 2009, 12:56 PM
kais
simplification?
a-a
_
a - a
_
b

not sure if that makes sense like that but im not entirely sure how to write it, basically a take away a over a take away a over b, simplified.

Would I be right to make the last part

a a
_-_
1 b

so it turns into 1-b because the top parts cancel each other out, so I'm left with a-a
_
1-b?

Or is that completely on the wrong track?
• Nov 3rd 2009, 01:31 PM
Defunkt
Did you mean $\frac{a-a}{\frac{a-a}{b}}$?
• Nov 3rd 2009, 02:07 PM
kais
Nope. Hmm, how to explain it.

a-z
_
q - y
_
b

a=z=q=y, but hopefully it gives more clarification, its a minus z over q minus y over b.

edit: nope, cant get the underscores to stay spaced out how id like ><
• Nov 3rd 2009, 02:51 PM
umbrella
I think you mean:

$\frac{(\frac{a - z}{q - y})}{b}$, well if this is the case let's simplify it:

$\frac{a-z}{q-y} \cdot \frac{1}{b}$,

$\frac{a - z}{(q - y) \cdot b}$

There isn't any possible way to simplify this, if I apply this to your question:

$\frac{\frac{a-a}{a-a}}{b}$,

$\frac{a-a}{a-a} \cdot \frac{1}{b}$

$1 \cdot \frac{1}{b} = \frac{1}{b}$

It can't be simplified any further.

In case you have the thing that Defunkt wrote:

$\frac{a-a}{\frac{a-a}{b}}$
$(a - a) \cdot \frac{1}{\frac{a-a}{b}}$
$(a - a) \cdot \frac{b}{a-a} = 0 \cdot \frac{b}{0}$

not possible because we divide by $a-a = 0$

We could've also seen this by:

$\frac{0}{\frac{0}{b}} = \frac{0}{0}$ (this is undefined)