# How to prove the following statements are equivalent?

Let $\{a_{n}\}^{\infty}_{n=1}$ be a sequence of real numbers. How to prove that the following statements are equivalent?

$a)$ For every $n_1, n_2\in \Bbb{N}$, if $n_1\gt n_2$, then $a_{n_1}\gt a_{n_2}$

$b)$ For every $n\in \Bbb{N}$, $a_{n+1}\gt a_n$

Here is my try:

$a\Rightarrow b$: Let $n_2+1\ge n_1 \gt n_2$, then $a_{n_2+1}\ge a_{n_1}\gt a_{n_2}$, then $a_{n+1}\gt a_n$.

$a\Leftarrow b$: Let $n_1=n+1$ and $n_2=n$, then it is apparent that $n_1\gt n_2$, and also $a_{n_1}\gt a_{n_2}$, then for every $n_1, n_2\in \Bbb{N}$, if $n_1\gt n_2$, then $a_{n_1}\gt a_{n_2}$

But I think there are lots of flaws in the proof. Could someone fix them?

• a=>b: This is no really nicely argumented. Let $n\in\mathbb N$. Put $n_1 := n+1$ and $n_2 := n$. Then $n_1 > n_2$ and thus $a_{n+1} = a_{n_1} > a_{n_2} = a_n$. For the proof of b=>a you should use induction. Mar 2, 2016 at 2:21

For the $a\Rightarrow b$ part, you can simply take $n_2=n$ and $n_1=n+1$.

For the $a\Leftarrow b$ part, given $n_1$ and $n_2\in\mathbb{N}$ with $n_1>n_2$, let $k\in\mathbb{N}$ be such that $n_1=n_2+k$. Then you have, by applying $b$ $k$ times:

$$a_{n_1}=a_{n_2+k}>a_{n_2+k-1}>\ldots>a_{n_2+k-(k-1)}=a_{n_2+1}>a_{n_2}$$

And since $n_1$ and $n_2$ were arbitrary, the proof follows.

$a \Rightarrow b$ : Assume $a$. Now take any $n \in \Bbb{N}$. $n+1 > n$, so by our assumption, $a_{n+1} > a_n$.

$b \Rightarrow a$ : Assume $b$. Now take any $n_1, n_2 \in \Bbb{N}$ such that $n_1 > n_2$. Say that $n_1 - n_2 = k$. Then $n_1 > n_1 - 1 > n_1 -2 > \cdots > n_1 - k = n_2$, so $a_{n_1} > a_{n_1 - 1} > a_{n_1 -2} > \cdots > a_{n_1 - k} = a_{n_2}$. Thus $a_{n_1} > a_{n_2}$.

This completes the proof.

If for every $n_1,n_2 \in \mathbb{N}$ with $n_1>n_2$ you have $a_{n_1}>a_{n_2}$ then letting $n_1 = n+1$ and $n = n_2$ you get $a_{n+1}>a_n$.

For the converse, if for every $n \in \mathbb{N}$ we have $a_{n+1}>a_n$ then if we suppose $n_1>n_2$ then $n_1 = n_2+k$ for $k \in \mathbb{N}$ and you can finish it from here.

@Larara beat me to it haha.