Tessellate S.T.E.M.S (2019) - Mathematics - Category C - Set 1 - Objective Problem 3

Let $m = 2019^x + 1, a = x, n = 1, b = 1.$ Then $m^{1/a} - n^{1/b} = (2019^x+1)^{1/x} - 1$ , which approaches $2018$ from above as $x \to\infty.$ So for any $d>0,$ there is some large enough $x$ such that $(2018, 2018+d) \cap S_{x,1}$ is nonempty. Hence the answer is $\fbox{0}.$

Let $A$ be the set of real numbers $d$ as defined in our question

Let $\delta > 0$ be an arbitrary real number no matter how small or how large . Our aim is to represent $m$ and $n$ as functions of $\delta$ .

So let the open interval $(2018,2018+\delta) \cap S_{a,b} = \phi$ .

Now let $m$ and $n$ vary and let us fix $a$ and $b$ such that $a=b=2$ .

Let $m =floor((2.(2018+\delta))^2)$ and $n =floor((2018+\delta)^2) + 1$

So we have $2018 < m^{\frac{1}{a}} - n^{\frac{1}{b}} <2018 +\delta$ . So by contradiction $\delta$ is not an upper bound of $A$ .

But since $\delta$ was arbitrarily small it implies that any $\delta > 0$ cannot be an upper bound.

Hence the $sup A = 0$ .

Now as to how I came to the choice of $m$ and $n$ .

Let $x^{\frac{1}{2}} = 4036 + \delta$

solving for $x$ we see that $x = (4036 + 2\delta)^2$ . Since $m$ is an integer we take $m$ to be $floor(x)$

Let $y^{\frac{1}{2}} = 2018 + \delta$

solving for $y$ we have

$y = (2018 +\delta)^2$ . But we take $n = floor(y) +1$ to account for the upper bound of the open interval. And we can easily see that for such choices of

$m$ , $a$ , $n$ and $b$ , we can find a member which lies in the intersection of the open interval and $S_{a,b}$