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I have this set, represented by extension:

$\{1, 2, 4, 8, 16, 32, 64, 128, ...\}$

Now I must represent it by comprehension:

$$\{x \in \mathbb{N} : 2^{x-1} \text{ is power of } 2\}$$

Is it necessary to write "is power of 2" ?

diligar
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ESCM
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  • it is $${2^n|0\leq n\leq 7} $$ – Qurultay Mar 08 '18 at 20:49
  • $2^{x-1}$ is always a power of two (except perhaps for $x=0$ if negative powers don’t count). So ${x\in \mathbb N: \mbox{$2^{x-1}$ is a power of two} }$ is the set of all positive integers. – spaceisdarkgreen Mar 08 '18 at 21:10
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    My question is if i can write {$x \in \mathbb{N}$ : $2^{x-1}$} without "is pow of 2" – ESCM Mar 08 '18 at 21:24
  • What you have written there does not make any sense. Currently, your set is all $x$ in the natural numbers. The righthand side condition does not "restrict" your set the way you need it to. – rubikscube09 Mar 08 '18 at 21:46
  • Edited, is good now ? ${\mathbb{N}}$ start from 1,2, 3. – ESCM Mar 08 '18 at 23:15
  • $\mathbb{N}$ typically includes 0 – Q the Platypus Mar 08 '18 at 23:24
  • My teacher, said that the natural numbers were created to count, therefore, it does not include it – ESCM Mar 08 '18 at 23:53
  • Some people don’t include it but you should be aware that 0 is a natural number in most of mathematics. If someone asked you to count how many sheep are in an empty field you would say 0. – Q the Platypus Mar 09 '18 at 02:19
  • @QthePlatypus I disagree. It depends entirely on context, and (to some extent) the tastes of the mathematician who is working with $\mathbb{N}$. Personally, I don't like to include $0$, and write either $\mathbb{N}0$ or $\mathbb{Z}{+}$ for the nonnegative integers. In any event, I've witnessed near fisticuffs over the "proper" definition of $\mathbb{N}$. – Xander Henderson Mar 09 '18 at 02:51
  • Almost a duplicate of https://math.stackexchange.com/questions/2148354/set-builder-notation-can-i-switch-whats-written-on-the-left-and-right-of-the-v/ – Asaf Karagila Mar 09 '18 at 08:42

4 Answers4

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The set you must represent has the following form:

$$ \{2^n: \, 0 \le n \le 7\} $$

TheWanderer
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$$A=\{1, 2, 4, 8, 16, 32, 64, ...\}$$ $$A=\{2^{n-1}|n\in \mathbb {N} \}$$

$$A=\{x| x=2^{n-1},n\in \mathbb {N} \}$$

Mohammad Riazi-Kermani
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$\{1, 2, 4, 8, 16, 32, 64, 128... +\infty\}$
= { $2^n$ : n non-negative integer }
= { n : n = $2^k$ for some non-negative integer k }.

William Elliot
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Another way would be

$$\{ x \in \mathbb{N} \; | \; \exists n \in \mathbb{N}, \; x = 2^{n-1} \},$$

which would be read aloud as "the set of all $x$ in $\mathbb{N}$ such that there exists an $n$ in $\mathbb{N}$ where $x = 2^{n-1}$."

diligar
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