0
$\begingroup$

Is $$n \ge 2 \Leftrightarrow n \in \mathbb{N} \backslash \{1\}?$$

This is in a scenario where $n \in \mathbb{N}$ is already implied.

  • 0
    That depends on whether you think $0 \in \mathbb{N}$. Most times I see it, $0$ is not included. But some people take $\mathbb{N}$ to include $0$. – 2017-01-20
  • 0
    @BrianTung yeah, assuming $0 \notin \mathbb{N}. $ – 2017-01-20
  • 0
    This seems to be a trivial question: what are your doubts about this fact? What do you think? – 2017-01-20
  • 2
    My experience has been the reverse: most times I see $\mathbb{N}$ it includes 0. But then I work in algebra where it's convenient to use this as notation for the obvious additive monoid. – 2017-01-20
  • 0
    @Crostul just wasn't very sure, I'm not a very confident mathematician, hehe. – 2017-01-20

1 Answers 1

0

You can prove one implication at time.

If $n\in\Bbb N$ such that $n\geq 2$, then $n\neq 1$ which implies that $n\in\Bbb N\setminus\{1\}$.

To prove the other direction, we will use contrapositive. Let $n\in\Bbb N$ such that $n<2$. But there is only one natural $n$ smaller than $2$: $n = 1$. This means that $n\not\in\Bbb N\setminus\{1\}$.

We conclude that for $n\in\Bbb N$, $n\geq 2$ if and only if $n\in\Bbb N\setminus\{1\}$.

  • 0
    This answer assumes that $0\not\in\Bbb N$. – 2017-01-20
  • 0
    Cheers, much appreciated. – 2017-01-20