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From a high school (Algebra II) quiz:

If the domain of $f(x)=2^x$ is the positive rational numbers, what is the range?

Since "range" can be ambiguous at this level, let's assume it means the image, not the codomain.

The solution provided by the teacher was that the range was also the positive rational numbers.

No matter how you interpret range, that can't be right since there are obviously irrational numbers in the image (let $x=\frac12$).

If the answer provided was the positive real numbers I'd accept it as meaning codomain not image since $f$ maps the domain into but not onto the reals.

So my question is the following: the image is a countable subset of $\mathbb{R}$ and is described as

$\{y\in\mathbb{R}: y=2^x \space\text{for some}\space x\in\mathbb{Q}\}$

and actually a unique $x$ in this case [added: and $x>0$].

Is there another way to describe it or think about this problem at a high-school level, or it is just the teacher not fully considering what he/she asked and answered?

MathematicianByMistake
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AlgTop1854
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  • Hint: the given function is strictly monotone increasing. – Anton Vrdoljak Oct 21 '23 at 15:40
  • Yes and you can express with $log$ or $ln$ if you really want to, but I was thinking more conceptually, is there a better way to think about it and give a correct answer at the high school level or just assume the teacher meant something simpler and move on. – AlgTop1854 Oct 21 '23 at 15:46
  • The actual image of the function is as you described, but if it can be taken to be the same as Q because they're both countable, then both can be seen to be the same as N because N is by definition countable as well. So the function is a function from N to N? Probably best to just ask your teacher to clarify – Divide1918 Oct 21 '23 at 15:58
  • See $2^x$ can't be $0$ and -ve for any $x \in \mathbb{Q}$ so range is $(0,\infty)$ – Lucky Chouhan Oct 21 '23 at 16:01
  • Well there are bijections between N, Q and the image that make that work; I'm not sure the particular high school student I know here is going to appreciate that. I am pretty sure the objective was NOT illustrating countable sets and the (surprising at first) fact that N and Q have the same size! – AlgTop1854 Oct 21 '23 at 16:01
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    @LuckyChouhan clearly if you assume range means codomain, since this function maps into that interval not onto it. In high school range is often taught to mean image. – AlgTop1854 Oct 21 '23 at 16:04
  • @AlgTop1854 Oh yeah! our domain is not $\mathbb{R}$ it is $\mathbb{Q}$ – Lucky Chouhan Oct 21 '23 at 16:09
  • I don't think there is a "clean" answer here - it is what it is - but good discussion. – AlgTop1854 Oct 21 '23 at 16:13

1 Answers1

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You are right that there is no sense in which the range is the positive rational numbers. There are three issues of varying importance:

  • Most significantly, many outputs of $f$, such as $f(\frac12) = \sqrt2$, are not rational.
  • Also, many positive rational numbers are not values of $f$. (This is not a mistake, if we're talking about a codomain instead of a range, but it's still awkward.) For example, there is no rational $x$ such that $f(x) = 3$, because $\log_2 3$ is irrational.
  • Also, "positive" is not the best cutoff we can give: if $x$ is a positive rational number, then $f(x) > 1$.

The image definitely is $\{y \in \mathbb R: y = 2^x \text{ for some } x \in \mathbb Q \text{ with } x>0\}$, just by definition. We have to be careful about $x$ being a positive real number. Aside from that, we could also write $$\{y \in \mathbb R : y > 1 \text{ and } \log_2 y \in \mathbb Q\}.$$ This is arguably slightly simpler, but not a lot simpler.

If we wanted to define a nice codomain for the function, we could take the interval $(1,\infty)$, or $[1,\infty)$, or $(0,\infty)$ (the positive real numbers), or even all of $\mathbb R$. The image of $f$ is dense in the interval $(1,\infty)$, so there's no good "middle ground" between the image and the interval $(1,\infty)$.

Misha Lavrov
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