# Natural number between two real numbers $y,x$ such that $y-x > 1$

Prove that for $$x,y \in \mathbb{R}$$ such that $$y-x>1$$ there is a natural number $$n$$ such that $$x < n < y$$.

Consider the following set:

$$S := \{n \in \mathbb{N} | x < n < y\}$$

Suppose $$S = \emptyset$$, therefore $$\forall n \in \mathbb{N}$$ $$n \leq x$$ or $$n \geq y$$. Let $$n_1$$ be the first natural number before $$x$$. Therefore $$n_1 < x < n_1 + 1$$.

Now we'll show that $$n_1+1 < y$$. Suppose not. Hence $$n_1. Which implies that

$$1 = n_1 + 1 - n_1 > y-x > 1$$

Which is a contradiction, therefore, $$x.

Here's what I'm thinking about my proof: I didn't use the fact that $$x,y$$ are real numbers... Maybe I wasn't supposed to prove it that way? Can someone please point out another proof if mine is not correct for that case? I don't know, it just feels weird to prove something and don't use all our premises.

Thanks!

• I think the premise is missing that $x, y > 0$ (you can't find $n_1$ if $x < 0$). Otherwise, I agree that $\mathbb{R}$ is not special. The same conclusion would hold for $\mathbb{N}$ or $\mathbb{Q}$.
– Dubs
Commented Apr 25, 2019 at 20:02
• It may not be true that $n_1<x<n_1+1$, because $x$ may be an integer. What you can say is $n_1<x\le n_1+1$. Similarly, you should have $n_1<x<y\le n_1+1$. (Your proof still works, by the way.) Commented Apr 26, 2019 at 10:14
• You DID use the fact that $x\in \Bbb R$ when you declared the existence of $n_1\in \Bbb N$ such $x<n_1+1$. It is possible to extend $\Bbb R$ to a larger arithmetic system $R^*$ that has members that are larger than all member of $\Bbb N.$ Commented Apr 26, 2019 at 10:43

## 2 Answers

Your proof is correct, and you implicitly use a two important properties about $$\mathbb{R}$$ and $$\mathbb{N}$$. Although, to make it more explicit that these properties are being used, I would present the proof slightly differently by taking $$n_1$$ to be the least $$n$$ greater than $$y$$. Let's see what that would look like, but before diving into that I want to reiterate that your proof is fully correct. I will just be very precise here to show where you implicitly use certain properties, that you might (rightfully) consider obvious.

The first property I am talking about is the Archimedean property of the reals (as an ordered field). This essentially says that for every real number $$x$$ there is some natural number $$n > x$$. The other property is that the natural numbers are well-ordered. That means that every non-empty subset of the natural numbers has a least element.

You implicitly use the well-ordering of the natural numbers in finding a maximal $$n_1$$ such that $$n_1 \leq x$$ and then use the Archimedean property to conclude that $$n_1 + 1 > x$$. It may not be directly clear how these properties are used here, because we cannot directly use their definitions in this formulation. Let me slightly change your proof to make more clear where we use these properties (also note that there is no reason for a proof by contradiction, we can actually construct the required natural number).

First of all, as was mentioned in the comments already, we clearly need to assume $$0 \leq x$$, because otherwise the statement is simply not true. Let $$y-x > 1$$, then $$y > x + 1 > x \geq 0$$. By the Archimedean property, there is some $$n \geq y$$. So the set $$A = \{n \in \mathbb{N} : n \geq y\}$$ of all natural numbers above (or equal to) $$y$$ is non-empty. Then by the well-ordering of the natural numbers, there must be a least element $$n_1 \in A$$. Since $$y > 1$$, we have that $$n_1 > 1$$ and so $$n_1 - 1 > 0$$ is a natural number. We must have $$y > n_1 - 1$$, since otherwise $$n_1 - 1 \geq y$$ and thus $$n_1 - 1 \in A$$, but $$n_1$$ was the least element of $$A$$. Furthermore, since $$n_1 \geq y > x + 1$$, we have that $$n_1 - 1 > x$$, so we have indeed that $$y > n_1 - 1 > x$$.

Besides the fact that the reals are an ordered field with the Archimedean property, we do not use any other specific properties. That means that this proof would go through for any other ordered field with the Archimedean property. An example of this would be $$\mathbb{Q}$$ (as mentioned in the comments).

• Where was the Archimedean property used? Commented Apr 26, 2019 at 10:13
• For the fact that there is $n \geq y$, so that $A$ is non-empty. I have made the use of these properties boldfaced in my proof, so it is more clear that we actually use them and where we use them. Commented Apr 26, 2019 at 10:14
• If $x<1$ then $n_1$ doesn't exist. Commented Apr 26, 2019 at 10:46
• @DanielWainfleet Why would that be? We take $n_1$ to be the least natural number $\geq y$. This definitely exists. Commented Apr 26, 2019 at 10:48

To supplement other answers, I can offer you a different proof (also i think you assume $$x,y>0$$ in your question, but no need to if you replace natural number with integer). Partition the positive real line as follows: $$[0,\infty ) = \bigcup_{n=0}^{\infty} [n,n+1)$$

Then to have $$y-x>1$$ means that they necessarily have to be in different bins. In other words, for natural numbers $$m , we must have $$x \in [m,m+1) \quad and \quad y \in [n,n+1)$$

But since $$y-x>1$$, we have that $$y>m+1$$. Namely $$x \in [m,m+1) \quad and \quad y \in (m+1,n+1)$$ $$\implies x

(As mentioned in other answers, there's nothing special about the fact that Y and X are real. It would have worked for rationals are natural numbers)