Chapter A.1, Problem 149AYU

To determine

To find: Is there a positive real number “closest” to 0?

Answer to Problem 149AYU

No.

Explanation of Solution

If that is the intent, the answer is that such does not exist. The set of real numbers is referred to as being densely ordered: Between any two distinct real numbers there exists another real number. (The word “between” means the new number is not equal to either of the two given numbers.) A specific example is that for real numbers $x$ and $y$ with $x<y$, we have $x<\left(x+y\right)/2<y$; in words, the average of two distinct real numbers is always between those two real numbers. As I stated an implication of being densely ordered is that given two distinct real numbers $x$ and $y$ with $x<y$, there is another real number $a$ satisfying $x<a<y$. We can do this again: there is $b$ satisfying $x<b<a$. And again: there is a $c$ satisfying $x<c<b$. This process can go on forever, meaning that, not only is there at least one, but there is an infinite count of, real values between $x$ and $y$.

Let’s let $x=0$. Somebody thinks they have found a value greater than 0 (the same argument can be repeated for values less than 0) that is the closest such value to 0. We will call that $y$. However, based on the above discussion the average of 0 and $y$ is greater than 0 (so not the same as 0) and less than $y$ (so closer to 0 than $y$ is), but $y$ is supposed to be the closest value to 0 and you just found a closer one, so your assertion yields a contradiction. Therefore, your assertion is wrong. No matter what nonzero value you pick for $y$ there is always one (actually an infinite count of them) closer to 0. Therefore, there is no closest nonzero real value to 0. Writing a bit loosely, this is equivalent to saying there is no next point after 0 on the real number line between any two distinct points on a line, there is always another point between them.