Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In set $T = \{(0+2)^{i+5}|\space i \in \mathbb{N}\}$, what is the meaning of the $(0+2)^{i+5}$?

share|cite|improve this question
doesnt it take on the normal meaning? – Carry on Smiling Aug 31 '12 at 0:09
what is the difference between 0+2 and just 2? – Carry on Smiling Aug 31 '12 at 0:10
that is what I was wondering. but apparently 0,2 should be considered as characters that make up a string.. (why they wouldn't just use 0,1 idk) – James Aug 31 '12 at 0:11
where did you see that? – Carry on Smiling Aug 31 '12 at 0:14
It is from my problem set, not a text. To add some context, all the previous problems considered numbers as characters. That is, 10 is not ten but one-zero. – James Aug 31 '12 at 0:17
up vote 3 down vote accepted

A likely possibility, given your comment, is that $(0+2)$ is to be interpreted as a regular expression, namely "either 0 or 2" and the apparent exponentiation would be read as concatenation of the elements ({0, 2}) of that set. Under this reading, the set $T$ would be (assuming $\mathbb{N}$ doesn't include zero) $\{000000, 022222,0202020200002, , \dots\}$, namely all strings of $0$s and $2$s of length greater than or equal to 6.

share|cite|improve this answer

(For me $0 \in \mathbb{N}$: adjust to personal taste)

I would read $\{(0+2)^{i+5}|\space i \in \mathbb{N}\}$ as $\{(0+2)^{0+5}, (0+2)^{1+5}, (0+2)^{2+5}, (0+2)^{3+5}, \ldots\}$, i.e. as $\{32,64,128,256 \ldots\}$.

I might wonder why the writer did not just say $\{2^{i+5}|\space i \in \mathbb{N}\}$ but I would not worry too much.

share|cite|improve this answer
Probably because, as James commented, the problem is looking at digits, rather than their values as expressions as numbers. – Rick Decker Aug 31 '12 at 0:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.