Numbers Battle

Consider

$\small \frac {\boxed{\dfrac {10^{1998}+1}{10^{1999}+1}}}{\dfrac {10^{1999}+1}{10^{2000}+1}} = \frac {\left(10^{1998}+1 \right)\left(10^{2000}+1\right)}{\left(10^{1999}+1\right)^2} = \frac {10^{3998}+10^{1998}+10^{2000}+1}{10^{3998}+2\cdot 10^{1999}+1} = = \frac {10^{3998}+ \blue{101 \cdot 10^{1998}}+1}{10^{3998}+\blue{20\cdot 10^{1998}}+1} > 1 \\ \implies \boxed{\frac {10^{1998}+1}{10^{1999}+1}} > \frac {10^{1999}+1}{10^{2000}+1}$

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Solution using @Barry Leung 's approach: Define $f(x)$ as follows:

$f(x) = \frac {10^x +1}{10^{x+1}+1} = \frac {1+\frac 1{10^x}}{10+\frac 1{10^x}} = 1 - \frac 9{10+\frac 1{10^x}}$

This means that the larger the $x$ , the smaller the $f(x)$ . Therefore $\boxed{f(1998)} > f(1999)$ .

We can use functions to help us solve this problem.

The above values can be represented as $\large f(x) = \frac {10^{x} + 1 }{10^{x+1} + 1}$ , where they are $f(1998)$ and $f(1999)$ respectively.

We can rewrite the function as $\large f(x) = \frac{10^{x} + 1 }{10 \times 10^{x} + 1 + 9 - 9}$ , this expression is equivalent to $\large f(x) = \frac{10^{x} + 1}{10 \times {(10^{x} + 1)} - 9}$ .

And to maximize the value of this function means minimizing the value of its reciprocal, therefore, we have $\large \frac{10 \times {(10^{x} + 1)} - 9}{10^{x} + 1}$ .

We can split the fraction into $\large 10 - \frac {9}{10^{x} + 1}$ .

From here we want $\large \frac{9}{10^{x} + 1}$ to be as big as possible so that $\large 10 - \frac {9}{10^{x} + 1}$ can be as small as possible.

To maximize $\large \frac{9}{10^{x} + 1}$ , we need to minimize the denominator, this means a smaller value of $\large 10^{x} + 1$ and consequently a smaller value of $x$ .

In our problem, we have $f(1998)$ and $f(1999)$ , and since 1998 < 1999, $f(1998)$ is bigger.

So our answer is $\large \frac{10^{1998} + 1}{10^{1999} + 1}$ .

$\dfrac {10^{1998}+1}{10^{1999}+1}-\dfrac {10^{1999}+1}{10^{2000}+1}$

$\dfrac {81\times 10^{1998}}{(10^{1999}+1)(10^{2000}+1)}>0$

So, $\boxed {\dfrac {10^{1998}+1}{10^{1999}+1}>\dfrac {10^{1999}+1}{10^{2000}+1}}$

We can hide most of the complexity (and also obtain a more general solution) by letting $x = 10^{1998}$ . The two fractions become $\frac{x+1}{10x+1}$ and $\frac{10x+1}{100x+1}$ respectively. Then we can ask whether $\frac{x+1}{10x+1} > \frac{10x+1}{100x+1}$ , which simplifies to $x > 0$ . Clearly our x is bigger than 0, so the first fraction is the larger one.

(\frac {1}{1 + \frac {1}{x}}/)

Let $\displaystyle \frac{10^{1999}+1}{10^{2000}+1}$ = $\displaystyle \frac{10^{y-1}+1}{10^y+1}$ = $\displaystyle \frac{10^y+10}{10(10y+1)}$

Similarly $\displaystyle \frac{10^{1998}+1}{10^{1999}+1}$ = $\displaystyle \frac{10^{y-2}+1}{10^{y-1}+1}$ = $\displaystyle \frac{10^y+10^2}{10(10y+1)}$

Clearly $\displaystyle \frac{10^y+10^2}{10(10y+1)}$ > $\displaystyle \frac{10^y+10}{10(10y+1)}$

Hence $\displaystyle \frac{10^{1998}+1}{10^{1999}+1}$ > $\displaystyle \frac{10^{1999}+1}{10^{2000}+1}$