Krdytkyu

Consider the expression

$$
10^5 - frac{10^{10}}{1+10^5}.
$$

Using the elementary properties of fractions we can evaluate the expression as

$$
10^5 - frac{10^{10}}{1+10^5} = frac{10^5 + 10^{10} - 10^{10}}{1+10^5} = frac{10^5}{1+10^5}approx 1.
$$

Note that the approximation $10^5+1 approx 10^5$ is used in the last step. Now suppose we use the same approximation, but apply it before we perform the subtraction. We get

$$
10^5 - frac{10^{10}}{1+10^5} approx 10^5 - frac{10^{10}}{10^5} = 0.
$$

The same logic works for

$$
10^p - frac{10^{2p}}{1+10^p}
$$

for arbitrary large $p$, so it cannot be simply an issue with the accuracy of the approximation.

Is there an easy explanation of what's going on here?

It is simply an issue of accuracy of approximation. Let me write $x = 10^p$. Then your expression is
$$ x - frac{x^2}{1+x}$$

Note that $$frac{x^2}{1+x} = frac{x}{1/x + 1} = x (1 - 1/x + O(1/x^2)) = x - 1 + O(1/x)$$
so that
$$ x - frac{x^2}{1+x} = x - (x - 1 + O(1/x)) = 1 + O(1/x)$$

In your second calculation you only evaluated $x^2/(1+x)$ to within $O(1)$, not $O(1/x)$, so naturally you have an error at the end that is $O(1)$.

The first approximation is fine. The second on is not, because, $10^5$ and $dfrac{10^{10}}{1+10^5}$ are large numbers with approximately the same size. You are saying that since $10,001$ is close to $10,000$, then $1$ is close to $0$.

There is no paradox.

When you approximate $$frac{10^5}{1+10^5}=1-0.000099999000cdots$$ with $1$, the error is on the order of $10^{-5}$.

But in the second case, the same error is multiplied by $10^5$, so that it is no more negligible.