In some textbooks I have seen the archimedean property defined as:

for some positive real $x$, real number $y$, there exists a natural $n$ such that $nx>y$.

In other textbooks the archimedean property is defined as:

for any real $x$, we can find a natural $n$ such that $x \leq n$

I'm guessing I can prove that the two definitions are equivalent, but its just my guess, so if my reasoning is incorrect please correct me. Also if there are any other ways of looking at how the two definitions are equivalent I would love to learn more.

My attempt at proving:

To see how the first definition implies the second, we let $x=1$ in the first definition, then we have $n>y$. But then this means for any real $y$, we can find n such that n satisfies $n \geq y$(since n satisfies $n>y$), which is exactly the second definition.

To see how the the second definition implies the first, for any positive real $x$, real $y$, we pick a real $z$ such that $z=y/x$. Then we can find $n$ such that $n \geq z=y/x$. Since $n+1>n$, we have $n+1>y/x$ . Since $x>0$, we multiply both sides of the inequality and it does not change the order, to get $x(n+1)>y$. Since $n+1$ is a positive integer, this is equivalent to the first definition.


1 Answer 1


You've shown that they are equivalent under the assumption of the properties of the real numbers that you used to prove the equivalence. Luckily, the properties of the order on the real numbers do not depend on the Archimedean property for their proof, so your proof has some value.

In some books I have seen the Archimedean property stated as "for any $x \in \mathbb{R}$, there exists $n \in \mathbb{Z}$ such that $n \leq x < n + 1$". This one is also equivalent to the two you have there. To prove the equivalence, I used the fact that every non-empty subset of $\mathbb{N}$ has a minimum.


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