TITLE: What does it mean to show that something is well defined?
QUESTION [14 upvotes]: This is coming from my first course in undergraduate analysis, and it's confusing to me how to show that some operation is "well-defined". For example, my professor left as something for us to figure out on our own, not homework, to show ourselves that if $a,b,c$ and $d$ are integers, and ($b,d\not=0$) that $$\left[\left(\frac{a}{b}\right)\right]+\left[\left(\frac{c}{d}\right)\right]=\left[\left(\frac{ad+bc}{bd}\right)\right]$$ is well defined. He then made an example that said: If $\frac{a'}{b'}\sim\frac{a}{b}$ and $\frac{c'}{d'}\sim \frac{c}{d}$, then $\frac{a'c'}{b'd'}\sim \frac{ac}{bd}$.
Also above in the brackets are supposed to be $2$ classes of element's.

REPLY [3 votes]: Given two fractions $\frac{a}{b}$ and $\frac{c}{d}$, we need to show that the addition operation as defined by $\frac{a}{b} + \frac{c}{d} = \frac{ad+bc}{bd}$ is well defined.
As mentioned above, in order to do this, we need to show that if $\frac{a}{b}=\frac{a'}{b'}$ and $\frac{c}{d} =\frac{c'}{d'}$ then $\frac{a}{b}+\frac{c}{d}=\frac{a'}{b'}+\frac{c'}{d'}$
Which is equivalent to showing that
$\frac{ad+bc}{bd}=\frac{a'd'+b'c'}{b'd'}$ or equivalent to showing that
[1] $(ad+bc)(b'd')=(a'd'+b'c')(bd)$
We begin with what we are given, that  $\frac{a}{b}=\frac{a'}{b'}$ and $\frac{c}{d} =\frac{c'}{d'}$. This implies that 
[2] $ab'=ba'$
[3] $cd'=dc'$
We multiply equation [2] by $dd'$ on both sides and multiply equation [3] by $bb'$on both sides and get:
[2] $(ab')(dd')=(ba')(dd')$
[3] $(cd')(bb')=(dc')(bb')$
We now add both equations and get:
$(ab')(dd')+(cd')(bb')=(ba')(dd')+(dc')(bb')$
We rearrange terms on the right side and the left side of the equation and get:
$(ad)(b'd')+(bc)(b'd') = (bd)(a'd')+(bd)(b'c')$
We pull out the common terms on each side and get:
$(ad+bc)(b'd') = (a'd'+b'c')(bd)$
This is equivalent to equation [1] which we wanted to show is true. This finishes our proof.