rocky heron
Let ( A ) be the set defined as
[ A = \left{ \frac{m}{n} \mid m, n \in \mathbb{N}, m + n \leq 15 \right}. ]
Find ( \sup(A) ). I conjectured that ( \sup(A) = 14 ), because when ( m = 14 )
and ( n = 1 ), we have 14.
Now, to prove that it is an upper bound,
[ m + n \leq 15 ]
[ \frac{m}{n} \leq \frac{15 - n}{n} ]
[ \frac{m}{n} \leq \frac{15}{n} - 1 ]
if ( n = 1 ), we would have,
[ \frac{m}{n} \leq 15 - 1 = 14 ]
Now, using the supremum approximation property, we must have
[ 14 - \varepsilon < c ]
where ( c ) belongs to the set ( A ), that is, ( c ) is of the form ( \frac{p}{q} ),
where ( p + q = 15 ). Then we would have
[ 14 - \varepsilon < \frac{p}{q} ]
but, ( \frac{p}{q} \leq \frac{15 - q}{q} ), so we would have
[ 14 - \varepsilon \leq \frac{15 - q}{q} ]
arranging we have
[ \varepsilon > \frac{15}{q} - 15 ]
This is where I'm stuck; I can't think of a way to use the Archimedean property to find
a transition and conclude that
[ \varepsilon > \frac{15}{q} - 15 ]
sudden vigil