# Thread: Proof of (a/b) + (c/d) = (ad + bc)/bd

1. ## Proof of (a/b) + (c/d) = (ad + bc)/bd

Hi folks,

Starting with 6 field axioms, I'm proving a sequence of 15 algebraic laws.

Number 13 in my list is: $(ab) + (cd) = (ad + bc)/(bd)$

Using the axioms and prior theorems:

$a\cdot 1/b + c\cdot 1/d$
$a\cdot 1/b\cdot d\cdot 1/d + c\cdot 1/d\cdot b\cdot 1/b$
$ad\cdot 1/b\cdot 1/d + bc\cdot 1/b\cdot 1/d$
$(ad +bc)\cdot 1/b\cdot 1/d$

I'm stuck here. I'm not seeing how to get from $1/b\cdot 1/d$ to $1/bd$ thus allowing the final step achieving the RHS. Unfortunately, the general case $(a/b)(c/d) = (ac)/(bd)$ is the next theorem in the list, and so, not available.

If you can provide a hint, I'd appreciate it.

Thanks,
Scott

2. ## solved it

$a\cdot 1/b + c\cdot 1/d$
$a\cdot 1/b\cdot d\cdot 1/d + c\cdot 1/d\cdot b\cdot 1/b$
$ad\cdot 1/b\cdot 1/d + bc\cdot 1/b\cdot 1/d$
$(ad +bc)\cdot 1/b\cdot 1/d$
Well, while messing around with my next proof, I found I had the same dilemma and ended up solving both proofs.

$(ad +bc)\cdot 1/b\cdot 1/d = (ad + bc)/(bd)$
$(ad +bc)\cdot 1/b\cdot 1/d\cdot b\cdot d = [(ad + bc)/(bd)]\cdot bd$
$(ad +bc)\cdot 1\cdot 1 = [(ad + bc)\cdot 1/(bd) \cdot bd$
$(ad +bc) = (ad + bc) \cdot 1$
$(ad +bc) = (ad + bc)$

Scott

3. Having no idea what your particular set of field axioms is, the best I can offer is what we would do.
Maybe you can modify to suit yours.
Note that we define $\frac{a}{b} = ab^{ - 1}$, so $\frac{a}{b} + \frac{c}{d} = ab^{ - 1} + cd^{ - 1}$.
Then
$\begin{gathered} ab^{ - 1} + cd^{ - 1} = ab^{ - 1} \left( {dd^{ - 1} } \right) + cd^{ - 1} \left( {bb^{ - 1} } \right) \hfill \\ = \left( {b^{ - 1} d^{ - 1} } \right)\left[ {ad + cb} \right] \hfill \\ = \frac{{ad + cb}}{{bd}} \hfill \\
\end{gathered}$
.

There are several gaps to be filled in. But I hope it helps.

4. Originally Posted by Plato
Having no idea what your particular set of field axioms is, the best I can offer is what we would do.
Yes, I thought of that after I posted. Not to mention the additional dozen theorems I had to work with. Sorry about that.

Anyway, your info was helpful, and thanks for taking the time to post.

Scott

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