If $b ∈ \Bbb Z$, then there is no integer a such that $b < a < b + 1$

Question

If $b ∈ \Bbb Z$, then there is no integer $a$ such that $b < a < b + 1.$

I am doing an introductory course for mathematical proofs and this question is an exercise related to inequalities and natural numbers.

I am having trouble articulating a proof for this proposition. I know how to prove that $b < b+1$, but I am not sure how I can prove the statement above by showing that $a$ does not exist.

Should I be doing a proof by contradiction? If I assume $a$ exists, then how do I show that this will result in contradiction? Am I supposed to try to show that $a$ is not a natural number?

I am just looking for suggestions or some instructions on how to write the proof, not an actual full written proof. I would like to be able to figure at least some of it out myself, instead of copying off of a posted answer. So please just provide helpful hints as answers.

Thank you

Answer 1

The book you mentioned in a comment under your post, The Art of Proof¹, introduces the set of natural numbers in a way that is quite different from the usual one that defines the integers out of natural numbers.

To summarize,

• it first defines the integers $(\mathbf{Z},+,\cdot)$ using essentially the "ring" axioms in the first chapter.

• Then it defines the natural numbers $\mathbf{N}$ as a subset of $\mathbf{Z}$ that satisfies some properties; in particular, $0\not\in\mathbf{N}$.

• Then it defines the order on $\mathbf{Z}$ using the natural numbers $\mathbf{N}$.

And then it introduces the axiom of mathematical induction; this is the section where your exercise is in.

Your book (p.20) suggests that you should first prove the following:

Proposition 2.21. There exists no integer $x$ such that $0<x<1$.

Proof.

Suppose there exists an integer $x$ such that $0<x<1$. Proposition 2.20 says that for all $k\in\mathbf{N}$, $k\ge 1$. Therefore we must have $x\not\in\mathbf{N}$.

On the other hand, by the definition of the order of integers (section 2.2, page 15), $$ x=x-0\in\mathbf{N} $$ which contradicts to $x\not\in\mathbf{N}$.

Now we prove

Corollary 2.22 Let $n\in\mathbf{Z}$, There exists no integer $x$ such that $n<x<n+1$.

Let $P(k)$ be the statement that there exists no integer $x$ with $k-1<x<k$. Then $P(1)$ is true by Proposition 2.21. Let $n\in\mathbf{N}$ and assume that $P(n)$ is true: there is no integer $x$ with $n-1<x<n$.

For sake of argument, suppose there exists an integer $x$ with $n<x<n+1$. Then $$ n-1=n+(-1)<x+(-1)<n $$ which contradicts the induction hypothesis. By the induction axiom 2.15, $P(k)$ is true for all $k\in\mathbf{N}$.

So far, we have shown Corollary 2.22 for all integer $n\ge 0$. Now suppose $n<0$. If there exists $x\in\mathbf{Z}$ with $n<x<n+1$, then $$ -(n+1)<x<-n $$ which contradicts what we have shown since $-n\in\mathbf{N}$ and $-(n+1)\in\mathbf{N}$ or $-(n+1)=0$.

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¹ The authors should be Beck and Geoghegan, not Allistair Savage, who is the one wrote his lecture notes based on the book.

Answer 2

Assume that $a \in \mathbb{N}$. Since your inequalities give $a \gt b$, then with $b \in \mathbb{N}$, as you state in your comment, we have $a - b \in \mathbb{N}$. Since you also state in your comment $0$ is not an element of $\mathbb{N}$, then if you know or can assume that $1$ in the smallest natural number, we also get

$$a - b \ge 1 \tag{1}\label{eq1A}$$

However, with your inequality, we would like to try to get something to compare \eqref{eq1A} against, ideally to get a contradiction. With that in mind, try subtracting $b$ from both sides of $a \lt b + 1$ to see what the result is.