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

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.

• “n is an odd negative integer” is a predicate; thus, the formula must be of form $P(n)$ with $n$ free. Jan 19 at 14:41
• @MauroALLEGRANZA Could you perhaps provide an example demonstrating what you mean? It doesn't need to be related precisely to my question, but just to give a bit more insight. I understand that $P(n)$ is a statement about $n$, but I am missing what you mean by "$n$ free". Jan 19 at 14:48
• $\text {OddNegInt}(n) \leftrightarrow [n \in \mathbb Z \land (n < 0) \land \exists k \in \mathbb N (n=2k+1)]$ Jan 19 at 14:56
• @MauroALLEGRANZA I see now. "N is an odd negative integer" can be translated as "n is an integer AND n is odd AND n is negative" Jan 19 at 15:02
• @MauroALLEGRANZA Although, I think that the last part in your formulation for $k$ would require that $k \in \mathbb{Z}$. Since if $n = 2k + 1$ then n is strictly positive for any $k \in \mathbb{N}$. Would you agree with this observation? Jan 19 at 15:15

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. :)

• No, $\mathbb{Z}$ also contains all positive integers. "$n$ is a negative integer" is writen $n \in \mathbb{Z} \wedge n < 0$. Jan 19 at 15:29
• @OlivierRoche: As far as the popular convention says, $\mathbb {Z}^+$ is the set of positive integers, and $\mathbb {Z}^-$ is the set of negative integers. Jan 19 at 23:40
• Oh sorry I didn't see the superscript! Jan 20 at 0:50

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 :

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

• Thank you for the edit. Jan 19 at 16:01