(Proof)
For positive numbers \(a,b\) and \(H\),
suppose a real number \(r\) exists such that
\(a=H+\dfrac{a}{r}\) and \(H=b+\dfrac{b}{r}\)
\((A)\)
is satisfied. Then \(a\ne b\) and
\(\dfrac{a-H}{a}= \fbox{\(\;(\alpha)\;\)}\)
\((B)\)
therefore \(H=\fbox{\(\;(\beta)\;\)}\).
Conversely, for positive numbers \(a\) and \(b\) where \(a\ne b\),
if \(H=\fbox{\(\;(\beta)\;\)}\),
then the expression \((B)\) holds and \(\dfrac{a-H}{a}\ne 0\).
From \((B)\), let \(\dfrac{a-H}{a}=\dfrac{1}{r}\).
Then the expression \((A)\) holds.
Therefore, for positive numbers \(a, b\) and \(H\),
‘\(a\ne b\) and \(H=\fbox{\(\;(\beta)\;\)}\)’
is \(\fbox{\(\;(\gamma)\;\)}\) for
a real number \(r\) to exist such that expression \((A)\) holds.
In the proof above, what are appropriate for
\((\alpha), (\beta)\) and \((\gamma)\) in this order?