# Is there a such function? (Math question) **URL:** \ **Category:** Factual Questions\ **Created:** [March 15, 2007, 12:38am UTC](https://boards.straightdope.com/t/is-there-a-such-function-math-question/395971 "2007-03-15T00:38:19Z")\ **Posts on this page:** 8\ **Page:** 1
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [March 15, 2007, 12:38am UTC](https://boards.straightdope.com/t/is-there-a-such-function-math-question/395971/1 "2007-03-15T00:38:19Z")
Let F be a function such that: F is a one to one corresondence from the positive real numbers onto the positive real numbers, and If x = 1.5y, then F(x) = 2F(y) (If x equals one and a half times y, then F of x equals two times F of y.) Intuitively, I thought there could be such a function. But when I tried to work out a set of actual correspondences for such a function, it started to feel like the kind of construction that ends up being impossible. Is there a proof that such a function can not exist? Alternatively, is there some formula that someone here can provide which, for some F, and for all positive real numbers r, takes r and yields the number which F maps them onto? There is probably a very simple answer to this question,\* and if that is the case, then my apologies in advance. -FrL- \*I _think maybe_ it has to do with logarithms, but its been over ten years since I have had to think about those, so my fuzzy intuition that they may be relevant is not anything I take very seriously. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [March 15, 2007, 1:06am UTC](https://boards.straightdope.com/t/is-there-a-such-function-math-question/395971/2 "2007-03-15T01:06:59Z")
Try f(x) = 4x/3. ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [March 15, 2007, 1:14am UTC](https://boards.straightdope.com/t/is-there-a-such-function-math-question/395971/3 "2007-03-15T01:14:48Z")
[QUOTE=ultrafilter] Try f(x) = 4x/3. [/QUOTE] f(12) = 16 f(18) = 24 18 = 1.5 \* 12 but 24 /= 2 \* 16. (That was the first one that popped in my head, too, though.) -FrL- ---
**Author:** ![Omphaloskeptic](https://avatars.discourse-cdn.com/v4/letter/o/bcef8e/32.png) [@Omphaloskeptic](https://boards.straightdope.com/u/Omphaloskeptic)\ **Post date:** [March 15, 2007, 1:15am UTC](https://boards.straightdope.com/t/is-there-a-such-function-math-question/395971/4 "2007-03-15T01:15:40Z")
Yes, of course such a function exists. In fact there are a lot of them. The simplest is just F(x) = 2[sup](log x)/(log 1.5)[/sup] ; the exponent here can alternately be written as (log x)/(log 1.5) = log[sub]1.5[/sub]x , so you see that multiplying x by 1.5 adds one to the value of the exponent and hence multiplies the result by 2. Now you can multiply this function by any (positive) constant k; k F(x) is obviously a solution whenever F(x) is. And since you haven’t specified continuity, you can choose different values of k (not independently, if you want to make F bijective) over all real values in, say, [1,1.5) to get as ugly a function as you like. But this probably doesn’t prove anything interesting about octaves or fifths. ---
**Author:** ![MaxTheVool](https://avatars.discourse-cdn.com/v4/letter/m/9fc29f/32.png) [@MaxTheVool](https://boards.straightdope.com/u/MaxTheVool)\ **Post date:** [March 15, 2007, 1:21am UTC](https://boards.straightdope.com/t/is-there-a-such-function-math-question/395971/5 "2007-03-15T01:21:26Z")
I think you want to first take x, then take its log base 1.5, then raise 2 to that power. So you’d have 2^(log-base-1.5(x)) Log base 1.5 of x is the natural log of x divided by the natural log of 1.5. So we want 2^(ln(x) / ln(1.5)) Let’s try some sample values: f(1) = 1 f(1.5) = 2 f(2.25) = 4 Bingo! I think there’s probably a way to clean that up further, but it’s also been a while since I’ve done logarithms… Let’s see, some playing around gets us: x^(ln(2) / ln(1.5)), so x^1.709… ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [March 15, 2007, 1:22am UTC](https://boards.straightdope.com/t/is-there-a-such-function-math-question/395971/6 "2007-03-15T01:22:38Z")
[QUOTE=Omphaloskeptic] Yes, of course such a function exists. In fact there are a lot of them. The simplest is just F(x) = 2[sup](log x)/(log 1.5)[/sup] ; the exponent here can alternately be written as (log x)/(log 1.5) = log[sub]1.5[/sub]x , so you see that multiplying x by 1.5 adds one to the value of the exponent and hence multiplies the result by 2. Now you can multiply this function by any (positive) constant k; k F(x) is obviously a solution whenever F(x) is. And since you haven’t specified continuity, you can choose different values of k (not independently, if you want to make F bijective) over all real values in, say, [1,1.5) to get as ugly a function as you like. [/quote] Thanks! I purposefully didn’t specify continuity, though at the same time, I was hoping there would be continuous functions of this type. If I am reading you correctly, that first example you give is continuous, correct? Also, please remind me, does “is bijective” mean “is a one-to-one correspondence?” If not, what does it mean? (Well, I guess I could look it up…) > [@](#): > > But this probably doesn’t prove anything interesting about octaves or fifths. Heheh, don’t draw attention. 🙂 No, seriously, I’m not working on a proof of anything, I just started trying to really work out the ‘mechanics’ of one of the things I proposed in that thread. -FrL- ---
**Author:** ![Rysto](https://avatars.discourse-cdn.com/v4/letter/r/ecccb3/32.png) [@Rysto](https://boards.straightdope.com/u/Rysto)\ **Post date:** [March 15, 2007, 1:28am UTC](https://boards.straightdope.com/t/is-there-a-such-function-math-question/395971/7 "2007-03-15T01:28:35Z")
[QUOTE=Frylock] Also, please remind me, does “is bijective” mean “is a one-to-one correspondence?” If not, what does it mean? [/QUOTE] It means that it is both **one-to-one** and **onto**. ---
**Author:** ![Omphaloskeptic](https://avatars.discourse-cdn.com/v4/letter/o/bcef8e/32.png) [@Omphaloskeptic](https://boards.straightdope.com/u/Omphaloskeptic)\ **Post date:** [March 15, 2007, 1:35am UTC](https://boards.straightdope.com/t/is-there-a-such-function-math-question/395971/8 "2007-03-15T01:35:00Z")
[QUOTE=Frylock] I purposefully didn’t specify continuity, though at the same time, I was hoping there would be continuous functions of this type. If I am reading you correctly, that first example you give is continuous, correct? Also, please remind me, does “is bijective” mean “is a one-to-one correspondence?” If not, what does it mean? [/QUOTE] Yes, the first example is continuous; it (and its constant multiples) are the smoothest solutions. But you can make other continuous solutions by choosing an arbitrary continuous bijective function f on [1,1.5] with f(1)=1, f(1.5)=2 as the generator, and reducing all values to this range. (A bijection is one-to-one and onto. For a function with F(1.5x)=2F(x) over the positive reals, continuity already implies onto.)