Timeline for Is this an injection from $\mathbb{R}_+$ to $(0,1)$? [closed]
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
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| Oct 3, 2023 at 16:16 | history | closed |
Anne Bauval lulu Daniele Tampieri José Carlos Santos amWhy |
Needs details or clarity | |
| Sep 24, 2023 at 2:47 | vote | accept | Princess Mia | ||
| Sep 24, 2023 at 1:18 | answer | added | Noah Schweber | timeline score: 3 | |
| Sep 24, 2023 at 1:15 | comment | added | K. Jiang | @Shmuel That is precisely the problem - there is no "greatest" natural number to speak of. | |
| Sep 24, 2023 at 1:11 | comment | added | Princess Mia | @M W I meant, rather, the amount of zeroes to be an arbitrarily large finite number, rather than infinite. It is simply how many digits are in the "greatest" natural number. | |
| Sep 24, 2023 at 1:07 | comment | added | M W | @Shmuel you cannot have a decimal with an infinite number of $0$'s preceding other digits. A decimal expansion is a sequence of digits, meaning each digit is a finite number of places to the left or right of the decimal. If you allowed expansions such as you propose it would violate the archimedean property of the real numbers. | |
| Sep 24, 2023 at 1:06 | comment | added | aschepler | "An infinite number of zeros followed by an infinite number of threes" is not a sequence of digits, if it means anything at all. Remember that decimal notation actually means sums like $3.14 = 3 \cdot 10^0 + 1 \cdot 10^{-1} + 4 \cdot 10^{-2}$. You can't have the first term in this sum be $3 \cdot 10^{-\infty}$ - if that's a defined real number at all, it equals zero. | |
| Sep 24, 2023 at 1:06 | review | Close votes | |||
| Oct 3, 2023 at 16:16 | |||||
| Sep 24, 2023 at 1:05 | comment | added | K. Jiang | Then indeed, $f \left(\frac{1}{3} \right)$ is not well-defined, so $f: (0, \infty) \to (0, 1)$ is not well-defined. | |
| Sep 24, 2023 at 1:03 | comment | added | Princess Mia | @NoahSchweber it would be of the form $0.0000000000 \ldots 33333333 \ldots$ where there amount of zeroes preceding the first $3$ is some infinite number- I myself cannot give a precise number, which is why I am asking my question. The amount of zeroes is simply how much the decimal is to the right in a specific real number, which real number has the decimal point moved the furthest point to the right. I had this idea because if I just moved the decimal point to the left of every real number to make the real number within $(0,1)$ then I would get repeats, as $3.14$ and $31.4$ would be the same | |
| Sep 24, 2023 at 1:02 | comment | added | aschepler | Or, the answer is "no, this is not an injective function from $\mathbb{R}_+$ to $(0,1)$ because it is not a function with domain $\mathbb{R}_+$ at all". | |
| Sep 24, 2023 at 1:01 | comment | added | lulu | The fact that you can't simply tell us what $f\left(\frac 13\right)$ is suggests that you don't understand your definition either. Frankly, your description is extremely hard to follow. | |
| Sep 24, 2023 at 0:59 | comment | added | Ben123 | I think you should edit your question @Shmuel. As others have pointed out, it is unclear exactly how you have defined $f$, which makes it hard to answer your question. | |
| Sep 24, 2023 at 0:58 | comment | added | K. Jiang | @NoahSchweber I agree; the description makes very little sense. | |
| Sep 24, 2023 at 0:57 | comment | added | Noah Schweber | @Shmuel OK, so then what is $f({1\over 3})$? It's possible I'm misunderstanding your idea; I'll be honest, your description " moves the decimal point to the left of each number an equal amount of times as how far the decimal point is from the right of the first digit of the real number which has the decimal point featured to the furthest point right from its first digit" is not very clear to me. | |
| Sep 24, 2023 at 0:53 | history | edited | Princess Mia | CC BY-SA 4.0 |
added 24 characters in body
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| Sep 24, 2023 at 0:51 | comment | added | Princess Mia | @NoahSchweber yes, is there is a reason why terminating decimals are necessary to define $f$? the greatest number of spots a real number has the decimal to the right of its first digit seems to depend only on the things to the left of the decimal, not the things to the right of it, so I am not sure how a non-terminating decimal affects $f$ | |
| Sep 24, 2023 at 0:49 | comment | added | Noah Schweber | @Shmuel Remember that ${1\over 3}=0.333333...$ | |
| Sep 24, 2023 at 0:43 | comment | added | lulu | I don't understand the definition of $f$. It seems to only be defined for terminating decimals. I have no guess how to define $f\left(\frac 13\right)$, for instance. If you think it's clear, what is the value? | |
| Sep 24, 2023 at 0:40 | comment | added | Princess Mia | @NoahSchweber would this not just convert $1 \over 3$ to its decimal expansion, and then move the decimal point to the left the appropriate amount of times? | |
| Sep 24, 2023 at 0:38 | comment | added | Noah Schweber | What's $f({1\over 3})$? | |
| Sep 24, 2023 at 0:36 | history | asked | Princess Mia | CC BY-SA 4.0 |