Converting "n is an odd negative integer" Into Formal Logic

Question

I am learning how to think about mathematical truth at the level of formal logic and I am tasked with converting the following basic statement into something formal: "n is a negative integer that is odd"

My first attempt at this was the following: Let $n \in \mathbb{Z}$ such that $2n + 1 < 0$

From my point of view this statement is saying that "n is an integer and it's values are restricted by the inequality $2n + 1 < 0$". But how this actually translated to a mathematician was "$n$ is an integer satisfying the inequality $2n + 1<0$" and after some thought, I concluded my initial statement was indeed wrong.

So I went back to the drawing board and came up with this:

$$\exists k \in \mathbb{N} \quad n = -2k + 1 $$

But I feel that this formulation is still missing something, and I just can't put my finger on it. Should something be said about $n$ as well? Another way I thought of writing this was:

$$ \exists k \in \mathbb{N} \quad (n = -2k + 1) \Rightarrow n \in \mathbb{Z} $$

$$\exists n \in \mathbb{Z} \quad \exists k \in \mathbb{N} \quad n = -2k + 1 $$ Any help thinking through this would be appreciated.

Answer 1

We can write "$n$ is a negative integer" as $$n\in\mathbb{Z}^-$$ And "$n$ is odd" as $$\exists k\in\mathbb{Z}.\ n=2k-1$$ Combine both: $$\exists k\in\mathbb{Z}.\ n=2k-1 \wedge n\in\mathbb{Z}^-$$ Or alternatively, $$\exists k\in\mathbb{Z}.\ \mathbb{Z}^-\ni n=2k-1$$

Hope this helps. :)

Answer 2

This answer follows the comment provided by Mauro ALLEGRANZA in the comments:

The statement "$n$ is an odd negative integer" can be translated to "$n$ is an integer AND $n$ is odd AND $n$ is negative". Converting this to formal logic we can write:

• "$n$ is an integer" == $n \in \mathbb{Z}$
• "$n$ is negative" == $n < 0$
• "$n$ is odd" == $\exists k \in \mathbb{Z} \quad (n = 2k +1) $ (edit : if one takes $k \in \mathbb{N}$, then $n > 0$)

and combining these statements about $n$:

• "n is an odd negative integer" == ($n \in \mathbb{Z}) \quad \land \quad (n<0) \quad \land \quad \exists k \in \mathbb{Z} \thinspace (n = 2k +1)$

edit : Since $\exists k \in \mathbb{Z} \thinspace (n = 2k +1)$ already implies that $n \in \mathbb{Z}$, one can omit it and state :

$$ (n<0) \quad \land \quad \exists k \in \mathbb{Z} \thinspace (n = 2k +1)$$