# 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. – Forever Mozart Dec 5 '14 at 5:29
• Your set includes $101$, which is odd. Also note that there are many ways of describing the set. – copper.hat Dec 5 '14 at 5:30
• So how would I write it? P = {x:x ∈ N, 100 ≤ x ≤ 500} – Harry Dec 5 '14 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. – Nikolaj-K Dec 5 '14 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. – charlotte Dec 5 '14 at 15:43

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 :) – Harry Dec 5 '14 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 ;-). – copper.hat Dec 5 '14 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 – Harry Dec 5 '14 at 5:48
• Good luck. Its a bit intimidating. My rule is generally ask forgiveness not permission... – copper.hat Dec 5 '14 at 6:00

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:

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

Inspired by Charlotte's answer:

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

Et iterum (Haskell's take on Ross' answer):

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

• ... you're not wrong. – Mike Pierce Dec 5 '14 at 5:40
• I like this${}{}{}$ – charlotte Dec 5 '14 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 $\{a,b\}+ \{c,d\} = \{a+c,a+d,b+c,b+d\}$. – copper.hat Dec 5 '14 at 15:27
• @Meelo: If I do, I will be arrested by the MathJax police... – copper.hat Dec 6 '14 at 3:15
• @HenningMakholm: Why would I want to 'trick the OP...'??? You are seeing darkness where there is none. – copper.hat Dec 6 '14 at 15:07

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? – Harry Dec 5 '14 at 5:39
• What I am saying in the parentheses is "$n=2k$ for some $k\in\mathbb N$." – Forever Mozart Dec 5 '14 at 5:41
• Ahhh ok thank you for the help :) – Harry Dec 5 '14 at 5:43
• This notation is more useful than the other expressions since it offers many advantages in proving something. – Hermis14 yesterday

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. – Trevor Wilson Dec 6 '14 at 0:02
• Another similar alternative would be to use $n \equiv 0 \pmod 2$ as the predicate. – wchargin Dec 7 '14 at 5:32
• I have always seen | used as an equivalent for such that, so at first this confused me. – Pro Q Aug 18 '19 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. – copper.hat Dec 5 '14 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 Dec 5 '14 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. – Nikolaj-K Dec 5 '14 at 22:21
• I would +1 again if I could. Quidquid Latine dictum sit viditur altum. With apologies to Fr. Roderick, my latin teacher. – copper.hat Dec 6 '14 at 2:51
• :-) ${}{}{}{}{}$ – copper.hat Dec 7 '14 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? – charlotte Dec 5 '14 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. – Nikolaj-K Dec 5 '14 at 19:30
• Sure, but I won't be able to really get into it until my exams are done. – charlotte Dec 5 '14 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 $0$. – Nikolaj-K Dec 5 '14 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). – Nikolaj-K Dec 5 '14 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? – hmakholm left over Monica Dec 6 '14 at 11:15
• Oh, I see your point, I'm being redundant. Thanks, I'll edit my answer. – Felipe Gavilan Dec 6 '14 at 20:36