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It can’t be $5$. And it can’t be $4.\overline{9}$ because that equals $5$. It looks like there is no solution… but surely there must be?

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There isn’t one. Suppose there were; let’s call it $y$, where $y<5$.

Let $\epsilon = 5 -y$, the difference between $y$ and 5. $\epsilon$ is positive, and so $0 < \frac\epsilon2 < \epsilon$, and then $y < y+\frac\epsilon2 < y+\epsilon = 5$, which shows that $y+\frac\epsilon2$ is even closer to 5 than $y$ was.

So there is no number that is closest to 5. Whatever $y$ you pick, however close it is, there is another number that is even closer.

Consider the analogous question: “$x < \infty$; what is the greatest value of $x$?” There is no such $x$.

The answer is $4$, assuming the domain of $x$ is $\Bbb Z$.

If $x<5$ then $2x<x+5$ so $x<\frac{x+5}{2}$. Similarly, $x<5$ means $x+5<5+5=10$ or $\frac{x+5}{2}<5$. So if $x<5$ we have $x<\frac{x+5}{2}<5$, and therefore there is a larger number, $\frac{x+5}{2}$ less than $5$.

Basically, the average of two different numbers must be strictly between those two numbers.

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