Why can we include equality in the Archimedean property?

The Archimedean property states that

Let $x$ be a real number. Then there exits $n\in\mathbb{N}$ such that $x<n$.

Sometimes, equality is used in the Archimedean property; that is

Let $x$ be a real number. Then there exits $n\in\mathbb{N}$ such that $x\leq n$.

For example, consider the following problem

Prove that for each $x\in\mathbb{R}$ there exists $n\in\mathbb{Z}$ such that $n\leq x$.

The solution is as follows

By the Archimedean property, for each $-x\in\mathbb{R}$, there exists a corresponding $n\in\mathbb{N}$ such that $-x\leq n$. Multiply both sides of this inequality by $-1$ to get $-n\leq x$. Let $k=-n$, where $k\in\mathbb{Z}$ so that $k\leq x$.

So my question is why in the solution the equality was also used?

• Definitions are equivalent. First obviously implies the second. Second implies the first by considering $n+1$. – MathematicsStudent1122 Mar 12 '17 at 7:44
• @MathematicsStudent1122: $x$ can also be a natural number, is that why first implies the second? if $x\leq n<n+1$ and so $x<n+1$ so if we replace $n+1$ by $n$, we get $x<n$; is this how second implies the first? – user415849 Mar 12 '17 at 8:15
• @MathematicsStudent1122: If that is the case, why do we even bother with two versions of the Archimedean property? we could only give version 2 of the Archimedean property. – user415849 Mar 12 '17 at 8:16
• First implies the second because if a number is less than another number, it is also less than or equal to that number. And yes, that is correct for why the second implies the first. Why do we bother with both? I don't know. It happens a lot in mathematics that we have various equivalent definitions for the same thing. I suggest that you just go with whichever definition your professor is using. – MathematicsStudent1122 Mar 12 '17 at 9:08
• @MathematicsStudent1122: Thanks. By the way, from your profile I can see that we both study at the same university. – user415849 Mar 12 '17 at 9:16