$a=987654312$ and $b=987654321$

Does it means that $a$ and $b$ altogether has digit from $1$ to $9$ exactly once like $a=12345,b=6789$ ??

DylanN wrote: Together, the two positive integers $a$ and $b$ have $9$ digits and contain each of the digits $1, 2, 3, 4, 5, 6, 7, 8, 9$ exactly once. For which possible values of $a$ and $b$ is the fraction $a/b$ closest to $1$? If $a>b$, then $\frac ab-1>1-\frac ba$ and so $a<b$ So obviously $a$ has four digits while $b$ has five. If leftmost digit of $a$ is not $9$, swapping $9$ and this leftmost digit increases $\frac ab$ So leftmost digit of $a$ is $9$ If leftmost digit of $b$ is not $1$, swapping $1$ and this leftmost digit increases $\frac ab$ So leftmost digit of $b$ is $1$ And repeating this method easily gives the result $\boxed{\frac{9876}{12345}}$

pco wrote: DylanN wrote: Together, the two positive integers $a$ and $b$ have $9$ digits and contain each of the digits $1, 2, 3, 4, 5, 6, 7, 8, 9$ exactly once. For which possible values of $a$ and $b$ is the fraction $a/b$ closest to $1$? If $a>b$, then $\frac ab-1>1-\frac ba$ and so $a<b$ So obviously $a$ has four digits while $b$ has five. If leftmost digit of $a$ is not $9$, swapping $9$ and this leftmost digit increases $\frac ab$ So leftmost digit of $a$ is $9$ If leftmost digit of $b$ is not $1$, swapping $1$ and this leftmost digit increases $\frac ab$ So leftmost digit of $b$ is $1$ And repeating this method easily gives the result $\boxed{\frac{9876}{12345}}$ But if $\frac ab-1>1-\frac ba$ then $\frac {a}{b}+ \frac{b}{a} > 1+1$ $\frac {a^2+b^2}{ab} >2$ $a^2+b^2>2ab$ $a^2+b^2-2ab>0$ $(a-b)^2>0$ $a-b>0$ $a>b$ So it´s not possible to say that $a<b$.

$(a-b)^2 > 0$ does not imply $a-b > 0$

@Mewi What I say is that since trivially $\frac ab-1>1-\frac ba$ $\forall a>b>0$, then fraction $\frac ba$ is closest to $1$ than fraction $\frac ab$ in these conditions. So we just have to look for solutions $\frac ab$ where $0<a<b$