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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.

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    $\begingroup$ “n is an odd negative integer” is a predicate; thus, the formula must be of form $P(n)$ with $n$ free. $\endgroup$ – Mauro ALLEGRANZA Jan 19 at 14:41
  • $\begingroup$ @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". $\endgroup$ – GrayLiterature Jan 19 at 14:48
  • $\begingroup$ $\text {OddNegInt}(n) \leftrightarrow [n \in \mathbb Z \land (n < 0) \land \exists k \in \mathbb N (n=2k+1)]$ $\endgroup$ – Mauro ALLEGRANZA Jan 19 at 14:56
  • $\begingroup$ @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" $\endgroup$ – GrayLiterature Jan 19 at 15:02
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    $\begingroup$ @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? $\endgroup$ – GrayLiterature Jan 19 at 15:15
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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. :)

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  • $\begingroup$ No, $\mathbb{Z}$ also contains all positive integers. "$n$ is a negative integer" is writen $n \in \mathbb{Z} \wedge n < 0$. $\endgroup$ – Olivier Roche Jan 19 at 15:29
  • $\begingroup$ @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. $\endgroup$ – ultralegend5385 Jan 19 at 23:40
  • $\begingroup$ Oh sorry I didn't see the superscript! $\endgroup$ – Olivier Roche Jan 20 at 0:50
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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 :

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

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  • $\begingroup$ Thank you for the edit. $\endgroup$ – GrayLiterature Jan 19 at 16:01

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