Proving that $\frac{a}{b} = \frac{c}{d}$ if and only if $ad = bc$ I was working on a problem which asked:

Prove that $ \frac{a}{b} = \frac{c}{d} $ if and only if $ad=bc$, provided $c,d \neq 0$.

Is it sufficient to manipulate $ \frac{a}{b} = \frac{c}{d} $ via multiplications to reach $ad=bc$? Or is there something more? My feeling is, yes, it is enough, because making valid algebraic manipulations to both sides of an equation should, I think, yield an equivalent statement.
Insight would be appreciated. Thanks.
 A: $$ab^{-1}=cd^{-1} \iff (ab^{-1})(bd)=(cd^{-1})(bd) \iff a(b^{-1}b)d=(cd^{-1})(db)$$
$$ \iff ad=c(d^{-1}d)b \iff ad=cb \iff ad=bc.$$

Notes: 


*

*I think you meant "provided $b,d \neq 0$".

*The right side took some time to catch up. 

A: It depends what you're doing. If you are at the stage where rational numbers are being introduced and constructed from the integers, the fact that
$$
\frac{a}{b}=\frac{c}{d}
\quad\text{if and only if}\quad
ad=bc
$$
is part of the definition of the rational number $\dfrac{a}{b}$. So there's nothing to prove.
If you are at a different stage, when the quotient $a/b$ has already been introduced and the basic fact that
$$
\frac{a}{b}=ab^{-1}
$$
is known, then you can just manipulate.
Suppose $a/b=c/d$; then we can multiply both sides by $bd$ and the equality is preserved:
$$
\frac{a}{b}bd=\frac{c}{d}bd
$$
that can be rewritten, using commutativity of multiplication, as
$$
ab^{-1}bd=cd^{-1}db
$$
that becomes
$$
ad=cb
$$
Commutativity yields $ad=bc$.
Suppose instead $ad=bc$; then multiply both sides by $b^{-1}d^{-1}$, which you can do because $b\ne0$ and $d\ne0$. So
$$
adb^{-1}d^{-1}=bcb^{-1}d^{-1}
$$
and rearranging terms with commutativity, easily yields
$$
ab^{-1}=cd^{-1}
$$
or
$$
\frac{a}{b}=\frac{c}{d}
$$
A: Let $(a,b),(c,d)\in\mathbb{R}\times\mathbb{R}^*$.
Multiplying by $b$, one has: $$\frac{a}{b}=\frac{c}{d}\Rightarrow a=\frac{bc}{d}\tag{1}.$$
Since $b\neq 0$, multiplying by $\displaystyle\frac{1}{b}$, one has: $$a=\frac{bc}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}.\tag{2}$$
Therefore, according to $(1)$ and $(2)$, one has: $$\frac{a}{b}=\frac{c}{d}\Leftrightarrow a=\frac{bc}{d}\tag{3}.$$
Proceeding the same way with $d\neq 0$, one has: $$a=\frac{bc}{d}\Leftrightarrow ad=bc.\tag{4}$$
Finally, it follows from $(3)$ and $(4)$ that: $$\frac{a}{b}=\frac{c}{d}\Leftrightarrow ad=bc.$$
A: $\frac{a}{b}=\frac{c}{d}$
Multiply by $a^{-1}$ on both sides.(We are using the multiplicative inverse of a, following the axioms of real numbers)
I am only writing the basics.
$\frac{1}{b}=\frac{c}{ad}$ (We use the multiplicative identity on the LHS to obtain the $1$. This is repeated in the other steps also.)
Using multiplicative inverse of $\frac{1}{d}$ on both sides.
$\frac{d}{b}=\frac{c}{a}$
using multiplicative inverse of $\frac{1}{b}$ on both sides.
$d=\frac{cb}{a}$
using multiplicative inverse of $\frac{1}{b}$ on both sides.
$ad=bc$, hence shown.
A: The detailed proof is a bit lengthy to develop here and involves fundamental results on algebraic structures. Given some set $E$ and a law of composition on it $(x,y)\mapsto x.y$, one can define the set of fractions with numerators in $E$ and denominators in some non empty subset $S$ of $E$. This set can be conveniently defined if the law on $E$ commutative, associative, has an identity element $1$; in other words when $E$ is a commutative monoïd. 
Fractions are defined as elements of the quotient set $(E\times \overline {S})/R$ i.e. as classes of equivalence modulo R where  $\overline S$ is the stable subset (for the considered law) of $E$ generated by $S$ and $R$ is an equivalence on the set $E\times \overline{S}$ defined by $R(x,y):\exists a\in E,\exists c\in E, \exists b\in \overline{S},\exists d\in \overline{S},\exists s\in \overline{S}$ such that $x=(a,b), y=(c,d) \wedge a.d.s= b.c.s$
In $(E\times \overline {S})/R$ the class of equivalence represented by pair $(a,b)$ i.e. $[(a,b)]_R$ is noted $a/b$. It is the fraction of 
numerator $a\in E$ and denominator $b\in \overline S$. One can induce on this quotient set, a law defining it by $(a/b).(a'/b') = (a.a' / b.b')$ which is commutative, associative, has an identity element the fraction $1/1$: one can briefly and refer to $(E\times \overline {S})/R$ as to  the (commutative) monoid of fractions with numerators in $E$ and denominators in $S$. It results from equivalence $R$ that for any two fractions:
$a/b=c/d \iff (\exists s\in \overline{S}) s.a.d=s.b.c$ (0) 
and also that every fraction $a/b$ is  equal to $s.a/s.b$ for any $s,b\in \overline S$ and $a\in E$. In particular $s/s=1/1$ $\forall s\in \overline{S}$. Since $(s/1).(1/s)=s/s$, every $s/1$ is invertible $\forall s\in \overline{S}$ having fraction $1/s$ as an inverse. One can write  $(s/1)^{-1}=1/s$.
One can further consider the mapping from $E$ into  $(E\times \overline {S})/R$ defined by $h(a)=a/1$. It is a unital morphism of monoids since $h(a.b)=(a.b)/1= (a/1).(b/1)=h(a).h(b)$ and $h(1)=1/1$. Every fraction $a/b$ ($a\in E,b\in \overline S$) can be written in terms of $h$ under the form $h(a).{h(b)}^{-1}$. From (0) one gets:
$h(a)=h(c) \iff (\exists s\in \overline{S}) s.a.=s.c$ (1)
Let $C_E$ be the set of all cancellable elements of $E$. ($C_E\neq \emptyset$ since $1\in C_E$).
One can show it is sub-monoid  of E in particular a stable subset $\overline{C_E}=C_E$ of $E$. (if $a$ and $b$ are cancellable so is compound $a.b$)
Reminder: 
$a$ is called left-cancellable (right-cancellable) if $a.x=a.y\rightarrow x=y$ ($x.a=y.a\rightarrow x=y$) $\forall x,y\in E$. It is called 
cancellable when both right-and-left cancellable. If the law to be considered is commutative a right-cancellable element is also left-cancellable and vice-versa, hence cancellable.
From (1): Morphism $h$ is injective iff $\overline S\subset C_E$ 
$\overline{S}\subset C_E\rightarrow S\subset  C_E$. Conversely $S\subset  C_E \rightarrow \overline S\subset \overline{C_E}=C_E$. In other words $S\subset C_E\iff \overline S\subset C_E$. Hence:
Morphism $h$ is injective iff  $S\subset C_E$
Let us suppose $h$ injective now (i.e. $S\subset C_E$) 
$a/b=c/d \iff (\exists s\in S) (s.a.d=s.b.c)$. But now $s.a.d = s.b.c\iff a.d.=b.c.$ (due to the cancellability of every $s\in S$). 
Then one gets,
$a/b=c/d \iff  a.d=b.c$

Conclusion
Let $E$ be a commutative monoid. 
$a/b=c/d \iff  a.d=b.c$  holds between any two elements of the 
commutative monoid of fractions of numerators in $E$ and denominators 
in some $S(\subset E)$ if and only if, all the denominators are cancellable elements for the law of composition on $E$ (i.e. $S\subset C_E$). 
You can apply this result to the case $E=\mathbb R$ and $S=\mathbb R^*$.
Hope this helps
