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Let $n$ be a positive integer. Prove that the largest integer which is less than $n$ is $n-1$.

Attempt of a solution: Since $n$ is a positive integer $n>0$. I think I have to use the well-ordering Principle which states: Every non-empty subset of the positive integers contains a smallest member.

Any hints or help would be greatly appreciated.

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1  
Hint: One way is to take the set $\{i\}_{i=0}^n$ and use a contradiction. –  o0BlueBeast0o Jul 13 at 22:52

4 Answers 4

Consider the set of differences:

$$S=\{a: a\in \Bbb N, a<n\}$$

which is all positive integers less than $n$ by definition of $>$. Then this set has a least element, but since $1<k$ for every $k\in\Bbb N$ and $n-a>0\iff n>a$ is a definition, we see that $1\in S$ and so $1$ is the least element, making $n-1$ maximal among all differences $\{n-a: a\in S\}$, for $n-b>n-a\iff a-b>0$ i.e. $a>b$.

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Usually we try by contradiction, and the well-ordering principle.

Lemma: There is no integer between $0$ and $1$.

  • Proof: Suppose there is a integer $n$ between $0$ and $1$. So the set $A = \{ n \in \Bbb Z : 0 < n < 1 \}$ is non-empty, hence exists a minimum element $m = \min~A$. Then $0 < m < 1$ implies $0 < m^2 < m < 1$, and so $m^2 \in A$, contradicting that $m = \min~A$

Now, let's prove that there is no integer between $n-1$ and $n$.

  • Proof: Suppose there exists an integer $r$ between $n-1$ and $n$. So we have $n-1 < r < n$, which implies $0 < r - n + 1 < 1$, which is a contradiction, by the Lemma. $\square$
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Here's a simple proof that does not (directly) appeal to well ordering. First, we cite the fact that for any positive integer $n$, $n^2\geq n$. This can be shown easy by induction on $n$ and I'll leave this to you.

We show by induction that $n-1$ is the first integer less than $n$. Let $n=1$ and suppose that there exists an integer $t$ such that $(1-1)<t<1 \implies 0<t<1 \implies 0<t^2<t\implies t$ is not a positive integer. Since the first integer (which is not a positive integer) less than $1$ is $0$, we have our base case.

Induction Step: Suppose that for all $1\leq k < n$, $k-1$ is the first integer less than $k$. We now show that $n-1$ is the first integer less than $n$. Suppose, for contradiction, that there exists an $x$ such that $x$ is an integer and that $n-1<x<n$. Since $x$ is less than $n$, we may apply our induction hypothesis. Then $x-1$ is the first integer less than $x$. However, $x-1<n-1$ and since $n-1$ is an integer, we have a contradiction.

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In other words, there is no integer $x \in \mathbb{N}$ such that $n-1<x<n$.

We set $A=\{n-1 \in \mathbb{N}: \text{ if } x \in \mathbb{R} \text{ with } n-1<x<n \text{ then } x \notin \mathbb{N} \}$.

We want to show that $A$ is inductive .

  • $1 \in A$

We define $B=1 \cup \{x \in \mathbb{R}: x \geq 2\}$

$B$ is obviously inductive.

So if $1<x<2$ then $x \notin B \Rightarrow x \notin \mathbb{N}$

  • Let $n-1 \in A$. We want to show that $n \in A$, that means that we want to show that if $\exists x \in \mathbb{R}$ with $n<x<n+1$, then $x \notin \mathbb{N}$.

Since $x>n$ then $x>1$.

Let $x \in \mathbb{N}$.

From the sentence:

$\text{ If } n \in \mathbb{N} \text{ and } n \neq 1, \text{ then } n-1 \in \mathbb{N}$

we have that $x-1 \in \mathbb{N}$.

But $n-1<x-1<n$.

That means that $n-1 \notin A$. That is a contradiction.

So, $x \notin \mathbb{N}$.

So, $n \in A$.

$$$$

Therefore, $A$ is inductive.

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