# Using set notation, define the set of even natural numbers between 100 and 500.

Using set notation, define the set of even natural numbers between 100 and 500.

This is what I have so far:

$P$ is even numbers so the set of natural numbers between 100 and 500 would be

$$P = \{x:x \in\mathbb N, 100 < x < 500\}$$

Would this be correct?

• if $N$ is the set of natural numbers, then your set includes all natural numbers between 1 and 500. Commented Dec 5, 2014 at 5:29
• Your set includes $101$, which is odd. Also note that there are many ways of describing the set. Commented Dec 5, 2014 at 5:30
• So how would I write it? P = {x:x ∈ N, 100 ≤ x ≤ 500} Commented Dec 5, 2014 at 5:31
• @Harry: You say "I don't know how you would characterise an even interger", but you must have a concept of "even integer". What is this concept? For example, you might know that 14 is an even integer. How do you justify that believe? If you know how to express the argument that 6, 14 and 9002 are all even integers, then you know how to characterize it. The only thing left to do is replace that definition with set theoretical language. But go step by step. Commented Dec 5, 2014 at 10:50
• Welcome to Stack Exchange, @Harry. If one of the answers below answers your question, you should accept it (click the check mark next to the appropriate answer). That does two things. It lets everyone know your issue has been resolved, and it gives the person that helps you credit for the assist. See here for a full explanation. Commented Dec 5, 2014 at 15:43

Here is one way:

\begin{align} P= \{ &102, 104, 106, 108, 110, 112, 114, 116, 118, 120, 122, 124, 126, 128, 130, 132, 134, \\ &136, 138, 140, 142, 144, 146, 148, 150, 152, 154, 156, 158, 160, 162, 164, 166, 168, \\ &170, 172, 174, 176, 178, 180, 182, 184, 186, 188, 190, 192, 194, 196, 198, 200, 202, \\ &204, 206, 208, 210, 212, 214, 216, 218, 220, 222, 224, 226, 228, 230, 232, 234, 236, \\ &238, 240, 242, 244, 246, 248, 250, 252, 254, 256, 258, 260, 262, 264, 266, 268, 270, \\ &272, 274, 276, 278, 280, 282, 284, 286, 288, 290, 292, 294, 296, 298, 300, 302, 304, \\ &306, 308, 310, 312, 314, 316, 318, 320, 322, 324, 326, 328, 330, 332, 334, 336, 338, \\ &340, 342, 344, 346, 348, 350, 352, 354, 356, 358, 360, 362, 364, 366, 368, 370, 372, \\ &374, 376, 378, 380, 382, 384, 386, 388, 390, 392, 394, 396, 398, 400, 402, 404, 406, \\ &408, 410, 412, 414, 416, 418, 420, 422, 424, 426, 428, 430, 432, 434, 436, 438, 440, \\ &442, 444, 446, 448, 450, 452, 454, 456, 458, 460, 462, 464, 466, 468, 470, 472, 474, \\ &476, 478, 480, 482, 484, 486, 488, 490, 492, 494, 496, 498 \} \end{align}.

Another slightly shorter way:

$P= \{ n \in \mathbb{N} | 100 < n < 500 \text{ and } \sin ( n {\pi \over 2} ) = 0 \}$.

$P= (2 \mathbb{N}+\{100\}) \setminus (2 \mathbb{N}+\{498\})$.

[2*x | x <- [51..249] ]

• ... you're not wrong. Commented Dec 5, 2014 at 5:40
• I like this${}{}{}$ Commented Dec 5, 2014 at 6:06
• @rubik: It is just shorthand, if $\square$ is some binary operation, then $A \square B = \{ a \square b | a \in A, b \in B \}$. Then $\{a,b\}+ \{c,d\} = \{a+c,a+d,b+c,b+d\}$. Commented Dec 5, 2014 at 15:27
• @Meelo: If I do, I will be arrested by the MathJax police... Commented Dec 6, 2014 at 3:15
• @HenningMakholm: Why would I want to 'trick the OP...'??? You are seeing darkness where there is none. Commented Dec 6, 2014 at 15:07

For something very close to your proposal, you could say $$P=\{2x:x \in \Bbb N, 50 \lt x \lt 250\}$$ The $2x$ is one way to get rid of the odd numbers.

• Ahhh thank you, this is exactly the solution i was looking for :) Commented Dec 5, 2014 at 5:41
• @Harry: The protocol is that if you are happy with an answer that you accept it. There is no hurry, just letting you know ;-). Commented Dec 5, 2014 at 5:46
• thank you all for the help :) trying to learn this atm and the guides ive been given haven't been much help Commented Dec 5, 2014 at 5:48
• Good luck. Its a bit intimidating. My rule is generally ask forgiveness not permission... Commented Dec 5, 2014 at 6:00

Let $\mathbb N=\{1,2,3,...\}$. You want a set which includes the even members of $\mathbb N$ that lie between 100 and 500. Well, a member $n$ of $\mathbb N$ is even precisely when $n=2k$ for some $k\in\mathbb N$.

So $\{n\in\mathbb N:(\exists k\in \mathbb N)(n=2k)\text{ and } 100<n<500\}$ works.

(use the weak inequality $\leq$ if you want to include 100 and 500 in the set).

• What does the k stand for? Commented Dec 5, 2014 at 5:39
• What I am saying in the parentheses is "$n=2k$ for some $k\in\mathbb N$." Commented Dec 5, 2014 at 5:41
• Ahhh ok thank you for the help :) Commented Dec 5, 2014 at 5:43
• This notation is more useful than the other expressions since it offers many advantages in proving something. Commented Nov 28, 2020 at 8:59

I would tend to use one of;

$P = \{ n \in \mathbb{N}: 100 < n < 500, 2\mathop{|}n \}$

or

$P = \{n \in \mathbb{N}: \text{$n$even}, 100 < n < 500 \}$

The latter I suppose is slightly less formal, but it would be perfectly normal to see it used in a lecture or talk, say, and it is probably the clearest possible when speaking or reading.

Note $m \mathop{|} n$ means that $m$ divides $n$, i.e. $n$ is an integer multiple of $m$.

• I think it would be better to use $\&$, $\wedge$, or "and" instead of the commas. Commented Dec 6, 2014 at 0:02
• Another similar alternative would be to use $n \equiv 0 \pmod 2$ as the predicate. Commented Dec 7, 2014 at 5:32
• I have always seen | used as an equivalent for such that, so at first this confused me. Commented Aug 18, 2019 at 1:40

The most succinct way I know is $$\{x|100\leq x \leq 500, x\in2\mathbb{Z}\}$$

Inspired by copper.hat and NicolajK, inter alia, here are some further valid answers: $$\left\{2x+100\Bigg|\prod_{i=0}^{200}(x-i)=0\right\}$$ $$\{x|x\text{ is an even integer between 100 and 500}\}$$ $$\left\{x\Bigg|x=\frac{p^{(2n)}(0)}{p^{(2n-1)}(0)}, n\in\mathbb{N}\right\},\quad p(x)=\sum_{i=100}^{500}x^i$$ $$\{x|x=\log_2[G:H],\,H\subseteq G,\,|H|\leq 2^{100},\,|G|=2^{500}\}, \quad G\text{ is a group}$$ \begin{aligned}&\{((q\text{ incr})0)|\exists p\in\mathbb{H}, \\&q= \\&\lambda pfx.((\lambda MNfx.N(Mf)x)(\lambda MNfx.(NM)fx)f(f(f(f(fx))))f(f(x)))f((pf)pfx) \\&\text{and} \\&\lambda M.\lambda N.(\lambda n.n(\lambda x.(\lambda ab.\lambda b)) (\lambda ab.a))(\lambda mn.n(\lambda nfx.n(\lambda gh.h(fg))(\lambda u.x)(\lambda u.u)\\& m) M N)q(\lambda MNfx.(NM)fx)f(f(f(f(fx))))f(f(f(x)))f((pf)pfx)\}\end{aligned} where $\mathbb{H}=\{\lambda fx.x,\lambda fx.fx,\lambda fx.f(fx),\lambda fx.f(f(fx)),\dots\}$.

• I like this. It has inspired me. Commented Dec 5, 2014 at 6:05
• Me too, it looks like the first solution i had in my mind $\{x|100\leq x \leq 500, \frac{x}{2}\in\mathbb{N}\}$
– kai
Commented Dec 5, 2014 at 8:36
• Nikolaj, with k. @kai: If you write the arithmetic operation $\tfrac{x}{2}$, you should maybe specify a sensible domain for $x$ to range over. $\mathbb Q$ will be the shortest answer. Commented Dec 5, 2014 at 22:21
• I would +1 again if I could. Quidquid Latine dictum sit viditur altum. With apologies to Fr. Roderick, my latin teacher. Commented Dec 6, 2014 at 2:51
• :-) ${}{}{}{}{}$ Commented Dec 7, 2014 at 17:05

Some more versions:

1. $\{x\in 2\Bbb N|100<x<500\}$ or $\{x\in 2\Bbb Z|100<x<500\}$

2. $2\Bbb N \cap (100;500)$ or $2\Bbb Z\cap (100;500)$

Computer scientists know that any binary number that is greater than zero is even, if the least significant bit is zero $$\{x:x \in \Bbb N, 100 \lt x \lt 500, x=(b_8\dots b_10)_2\}$$ Cryptographers might do it like this $$\{x:x \in \Bbb N, 100 \lt x \lt 500, x\equiv 0\ (\mathrm{mod}\ 2)\}$$

Using remainder classes, you can express it as $$P = \lbrace 100 + [n]_{400} \mid n \in \mathbb{N} \rbrace$$ Which is equal to $$P = 100+\mathbb{Z}_{400}$$

If we really want short, we could do this:

$$2\mathbb Z\cap[100,500]$$

If you need the set brackets:

$$\{x:x\in 2\mathbb Z\cap[100,500]\}$$

where $[a,b]=\{x:x\in \mathbb R \land a \le x \le b\}$, the normal sense of $[a,b]$

A few more ways to express it:

$P=\{x\in\mathbb{N}:100<x<500\land\lfloor\frac{x}{2}\rfloor=\frac{x}{2}\}$

$P=\{x\in\mathbb{N}:100<x<500\land x\equiv0\mod{2}\}$

$P=\{x\in\mathbb{N}:100<x<500\land\frac{x}{2}\in\mathbb{N}\}$

$\{|n| \mid n\in E, 100<|n|<500\}$

where

$E\equiv\{\emptyset\}\cup\{x\mid\exists (y\in E).\,x=\left(y\cup\{y\}\cup\{y\cup\{y\}\}\right)\}$

• This is beautiful. So if there's a set $y$ of order $n-2$ in $E$, then $x$ has order $(n-2)+1+1 = n$. Can you prove, though, that there are no odd-ordered elements in the set? Commented Dec 5, 2014 at 15:49
• @charlotte: No, you're right I was sloppy. To guarantee there are no extra sets in it, I need to add the condition that the $y$'s are also of the form $y'∪{y'}∪{y'∪{y'}}$ for some $y'$ in $E$. I mirrored the von Neumann construction of the countable set of ordinals - I can give you some more references on these constructions if you like. Commented Dec 5, 2014 at 19:30
• Sure, but I won't be able to really get into it until my exams are done. Commented Dec 5, 2014 at 20:06
• @charlotte: The solution is skipping every second number in the von Neumann model of N. Given any finite set $X=\{a,b,\dots,s,t\}$ with cardinality $n$, the set $\mathrm{succ}(X):=X\cup\{X\}=\{a,b,\dots,s,t,X\}$ has cardinality $n+1$ (as no set in standard set theory contains itself, see this axiom). So $\mathrm{succ}(\mathrm{succ}(Y))$, which is $Y\cup\{Y\}\cup\{Y\cup\{Y\}\}$, adds the cardinality by $2$. The set $E$ is generated by $\emptyset$ of size $0$. Commented Dec 5, 2014 at 22:19
• @charlotte: I've discussed these concepts in my online notebook, see successor set, followed by first infinite von Neumann ordinal, followed by natural number. I try to elaborate on the axioms in the entry set theory and I say something on models here (skip right to the Motivation section). Commented Dec 5, 2014 at 22:20

Here is another way that has not yet been suggested here:

$$P=\{n\in\mathbb{N}|[100<n<500]\wedge[(-1)^n=1]\}$$

$$\left\{ x \mid x \in \mathbb{N},\ 2 \mid x,\ 100 \leq x \leq 500 \right\}$$

Since you haven't specified, I'm assuming between 100 and 500 is inclusive.

$2 \mid x$ means $2$ is a divisor of $x$, or alternatively, $x$ is evenly divisible by $2$.

I think this is a common way of listing a set of even numbers:

$P=\{2x + 100:x\in\mathbb{N}, 0\leq x\leq 200\}$

This just means: Let P be the set of numbers yielded from "2x + 100" where x is a natural number and x is between 0 and 200 (inclusive).

If you start evaluating 2x + 100 with the numbers from 0 to 200, you get:

• 2(0) + 100 = 100
• 2(1) + 100 = 102
• ...
• 2(200) + 100 = 500
• I don't think "$x\in[0\le x \le 200]$" is standard notation. What is "$x\in$" supposed to achieve here? Commented Dec 6, 2014 at 11:15
• Oh, I see your point, I'm being redundant. Thanks, I'll edit my answer. Commented Dec 6, 2014 at 20:36

This is a more longwinded approach to help clarify what's going on. But be explicit in how you define 'between'. In this sense, between could be interpreted as $$100 \leq x \leq 500$$ or $$100 < x < 500$$. As long as you are explicit about your assumption, you're fine.

The sentence "Define the set of even natural numbers between 100 and 500" can be broken into several components:

1. P = {even natural numbers between 100 and 500}
2. P = {elements of x in natural numbers such that (x is even) AND (x is greater than 100) AND (x is less than 500) }
3. P = $$\{ x \in \mathbb{N} : (\exists k \in \mathbb{N} \quad x = 2k) AND (x > 100) AND (x < 500) \}$$
4. P = $$\{ x \in \mathbb{N} : (\exists k \in \mathbb{N} \quad x = 2k) \land [(x > 100) \land (x < 500)] \}$$

This way makes the inequality $$100 < x < 500$$ explicit using the logical and operator.