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How would you describe the set $\{1, 5, 9, 13, 17, 21,\dots\}$ in the style of $x:P(x)=$? I know that the sequence is "the last number + 4" or $4n-3$.

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$\{n \in \mathbb N\; :\; n \equiv 1 \mod 4\}$

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  • $\begingroup$ How do you distinguish between (n≡1) mod 4 and n≡(1 mod 4) ? $\endgroup$ – Joshua Aug 31 '15 at 20:38
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    $\begingroup$ If you're going to put in parentheses at all, it should be $n \equiv 1 \ (\text{mod}\; 4)$. See en.wikipedia.org/wiki/Modular_arithmetic#Congruence_relation $\endgroup$ – Robert Israel Aug 31 '15 at 21:01
  • $\begingroup$ (1 mod 4) is valid. (7 mod 4) + 3 = 6. $\endgroup$ – Joshua Aug 31 '15 at 21:23
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    $\begingroup$ @Joshua You're confusing modular congruence with the binary modulus operation. The usage above is designating modular congruence. $\endgroup$ – apnorton Sep 1 '15 at 0:19
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    $\begingroup$ @MarioCarneiro "Correct" here is a very vague and difficult notion. Once you state clearly at the beginnin, that you work in $\mathbb{Z}/n\mathbb{Z}$ a lot, you can safely use $k\equiv 4\operatorname{mod} n$ and everybody knows what you mean. $\endgroup$ – yo' Sep 1 '15 at 8:17
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$\{4k-3 \mid k \in \mathbb{N} \}$ or $\{4k+1 \mid k \in \mathbb{N} \}$, depending on whether you consider $0$ a natural number.

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    $\begingroup$ I think of the options presented so far, this is the nicest in that: (i) it doesn't require an understanding of modular arithmetic; (ii) it avoids confusion with quantifiers; and (iii) it most clearly captures the sequence via the mapping $\{1\to 1, 2\to 5, 3\to 9,\dotsc\}$. $\endgroup$ – Joshua Taylor Sep 1 '15 at 2:29
  • $\begingroup$ Thanks! This is how I write it in MAGMA, I feel it is very intuitive. $\endgroup$ – Morgan Rodgers Sep 1 '15 at 5:23
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    $\begingroup$ The only problem I have with this is the idea that $0$ is not a natural number. Well, maybe I have to yield this time. Have your upvote but please consider writing $k\in \mathbb{Z}_+$ instead! $\endgroup$ – JiK Sep 1 '15 at 8:51
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    $\begingroup$ @MorganRodgers Yes. And I have a problem with your idea that $0$ is not a natural number. $\endgroup$ – JiK Sep 1 '15 at 9:09
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    $\begingroup$ @JiK I see, I misunderstood. Edited to encompass all viewpoints, since I like to think of this set as being congruent to $1 \bmod 4$ and not to $-3$. $\endgroup$ – Morgan Rodgers Sep 1 '15 at 9:35
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$$\{n \; \mid \; \exists k \in \mathbb N: n =4k-3\}$$

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    $\begingroup$ No. Try to find such $k$ for $n=2$, please. Also, if I write $\forall k \in \mathbb N : n=4k-3$, and n is in this set, then we have n=4*1-3=4*2-3=4*3-3=4*4-3=.... which is clearly not true for any $n$. $\endgroup$ – wythagoras Aug 31 '15 at 18:07
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    $\begingroup$ Secondly, if I write $\forall k \in \mathbb N : n=4k-3$, then I am searching for $n$ that statistify $n=4k-3$ for ALL natural $k$. So we have $n=4\cdot1-3=1$ because it must hold for $k=1$. We also have $n=4\cdot2-3=5$ because it must hold for $k=2$. So there can't be such n. $\endgroup$ – wythagoras Aug 31 '15 at 18:42
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    $\begingroup$ @El'endiaStarman In that route, it'd be even easier to say $\{4n-3 \mid n \in \mathbb{N}\}$ $\endgroup$ – Joshua Taylor Sep 1 '15 at 2:27
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    $\begingroup$ @JoshuaTaylor: That's the best. $\endgroup$ – El'endia Starman Sep 1 '15 at 2:29
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    $\begingroup$ @El'endiaStarman I didn't see it before my comment, but Morgan Rodgers posted it as an answer, too. $\endgroup$ – Joshua Taylor Sep 1 '15 at 2:31
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An arithmetic progression with first term 1 and common difference 4

(No I haven't described it in the form of $P(x)$, but this is how I would describe this number sequence...)

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$\{n\in\mathbb N:\frac{n+3}4\in\mathbb N\}$

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It's not exactly the form you're looking for, but close: $$ \{n\in\mathbb N \text{ [or }\mathbb R \text{ or whatever relevant]}: (n-1)/4 \in \mathbb N\} $$ If you really need an equality, then you might write $$ \{n\in\mathbb N: \mathbf1 _{\mathbb{N}} (n-1)/4 = 1\} $$ where $\mathbf1 _{\mathbb{N}}$ is the indicator function of the set $\mathbb N$.

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You can be as simple as this, given you don't insist on the $\{:\}$ set declaration:

$$4\mathbb{N}-3 = \{1,5,9,13,17,\dotsc\}$$

if $0\notin\mathbb{N}$ by your convention; and

$$4\mathbb{N}+1 = \{1,5,9,13,17,\dotsc\}$$

if $0\in\mathbb{N}$ by your convention.

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You can represent your set as the Range set of a Function from $\Bbb N$ to $\Bbb N$ itself defined as:

$$f(x)=4x+1$$

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  • $\begingroup$ The codomain of a function $\Bbb N\to\Bbb N$ is always $\Bbb N$. The notion you mean is called image or (sometimes) range. Please don't use the term "codomain" which is specifically meant to be something different, namely the set $Y$ for a function that was specified as being $X\to Y$, regardless of the actual values the function takes. $\endgroup$ – Marc van Leeuwen Sep 2 '15 at 11:49
  • $\begingroup$ Yeah sry for the ambiguity, now it should be clearer @MarcvanLeeuwen $\endgroup$ – Renato Faraone Sep 2 '15 at 11:51

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