Reduction of $\frac a{b +1}$ Can we convert $\frac a{b+1}$ into a fraction having only $b$ as a denominator? What are the steps in doing so?
 A: You want
$$\frac{a}{b + 1} = \frac{x}{b}$$
for some $x$, so solve this equation for $x$.  You get $x = \frac{ab}{b + 1}$ so
$$\frac{a}{b + 1} = \frac{\left(\frac{ab}{b + 1}\right)}{b}.$$
Note that if you require your fractions to be one integer divided by another integer then you may be out of luck because $\frac{ab}{b + 1}$ isn't always an integer.
A: In general, no, consider $\frac{2}{2+1}=\frac{2}{3}$. We cannot make a fraction of the form $\frac{n}{2}$ with $n\in\mathbb{Z}$ such that $\frac{n}{2}=\frac{2}{3}$. If we allow fractions in the numerator, then $n=\frac{ba}{b+1}$ will work.
A: Lemma $ $ If $\ q = \dfrac{a}{b\!+\!1}\,$ can also be written with denominator $\,b,\,$ i.e.  $\,q = \dfrac{n}b,\,$ then $\,q\,$ is an integer.
Proof $\ $ Since $\,(b\!+\!1)q\,$ and $\,bq\,$ are integers then so too is their difference $\,(b\!+\!1)q-bq\, =\, q.\ \ $ 
Remark $ $ Generally, a fraction can be written with coprime denominators iff it is an integer.
Proof $\ $ If $\, q = a/b = c/d\,$ for coprime $\,b,d\,$ then $\, \color{#c00}{jb\!+\!kd = 1}\,$ for $\,j,k\in\Bbb Z,\,$ by Bezout, so
$$ bq,dq\in\Bbb Z\,\Rightarrow\, j(bq)+k(dq) = (\color{#c00}{jb\!+\!kd})q = q\in \Bbb Z\qquad\qquad\qquad$$
This is a denominator form of this group theoretic theorem: if $\,g^a = 1 = g^b\,$ for coprime $\,a,b\,$ then $\,g = 1,\,$ since the order of $\,g\,$ divides the coprime $\,a,b\,$ so it can only be $1.\,$ The least denominator of a fraction is simply its order in $\,\Bbb Q/\Bbb Z,\,$ so the above is a special case of this result. See also various posts on denominator ideals and order ideals.
