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
Best Answer
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.