If $f: \mathbb{Z} \to \mathbb{Z_{99}}$ is defined by $f(x) = [50][x]$ is not injective. Is it injective when $f: \mathbb{Z_{99}} \to \mathbb{Z_{99}}$?

Question

Prove that $f: \mathbb{Z} \to \mathbb{Z_{99}}$ defined by $f(x) = [50][x]$ is not injective.

Is it injective when $f: \mathbb{Z_{99}} \to \mathbb{Z_{99}}$?

For the first part, I used $f(45) = f(144) = [72]$ to show that $f : \mathbb{Z} \to \mathbb{Z_{99}}$ is not injective. I am now sure if this method is entirely correct as I don't know if I can simply multiply $[50]$ and $[x]$ to get $[50x]$ since $[50] \in \mathbb{Z_{99}}$ while $[x] \in \mathbb{Z}$.

For the second part, $\mathbb{Z_{99}} = \{{[0], [1], ... , [98]}\}$, so there are no repeat elements in $\mathbb{Z_{99}}$. And so there is only one possible combination of $[50]$ and $[x]$ to produce a particular $f([x])$. I am not entirely sure how to prove this is injective.

Your first part is fine. Note that $x \in \mathbb Z$, but $[x] \in \mathbb Z_{99}$. — Alexis Olson, Mar 06 '17 at 17:56
@AlexisOlson ah i see, thanks. Any thoughts on the second part. — u123435, Mar 06 '17 at 17:59
Basically, since $50$ is relatively prime to $99$ it is a generator for the cyclic group $\mathbb Z_{99}$. — Alexis Olson, Mar 06 '17 at 18:06

score -1 · Answer 1 · answered Mar 01 '23 at 23:27

-1

Your argument on the first part is correct!

For the $2$nd part, we first know that because $\gcd(50,99) = 1$, we know that $50^k$ for however large a $k$ we need will run through all possibilities of $\{1, 2, 3, 4, \dots, 99\}$ so indeed, our second function is injective.

Check out this page for a more in-depth explanation.

answered Mar 01 '23 at 23:27

Clyde Kertzer

2,904

Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Mar 02 '23 at 00:08
Bill I really wanna give you the benefit of the doubt but this is getting ridiculous – Clyde Kertzer Mar 02 '23 at 03:21
I don't understand what you doubt. It is clearly a dupe (of many posts). Almost all common elementary exercises (like this) are dupes by now. Please keep that in mind before answering. Digging up old unaswered questions is only useful if it adds new content to the site. – Bill Dubuque Mar 02 '23 at 03:47
In case you don't know, many users (including I) often sort questions using the "active" filter, so they will see (and review) all new activities (in their filtered tags), including all new answers to old quesions. This may cause old questions to receive new processing (they may now be dupes, or users may have missed them in the past, etc). That happens all the time. Not to mention that there was much less enforcement of dupes/psqs in the old days (before the mod team announced EOQS 2 years ago). – Bill Dubuque Mar 02 '23 at 04:27
Here's the deal. I answered the question BEFORE you marked it as a dupe. Then you marked it as a dupe after you saw it come up in active feed AFTER I answered it. You obviously have more knowledge of what question may be a dupe than I do, so you are able to quickly find a dupe of the question. I would just appreciate you being just a tad forgiving to a new user who is trying to answer old questions because he wants to help the site. It's just really demeaning when you comment the exact same copy paste on answers that you know I spent time finding and answering. – Clyde Kertzer Mar 02 '23 at 06:16
It seems you may misunderstand the primary goal of this site - to build a repository of answers to questions that can be efficiently searched by questioners. If the search results on common questions are cluttered with hundreds of duplicates then it is difficult if not impossible to locate the "best" answers - which means all the effort invested by the community over the past 13 years is wasted. To avoid that is is essential to nip in the bud rampant duplication. Otherwise the site will devolve into an ephemeral stream of (mostly) low-quality fgitw answers vs library of proofs from book – Bill Dubuque Mar 02 '23 at 06:45
^^(proofs from the book - hit char limit). Many old unanswered questions were not answered because they were known to be dupes, but users didn't have the time (or motivation) to search for dupes. That has changed a bit after EOQS since now there is strong moderator support for site organization, so more users are investing effort there. – Bill Dubuque Mar 02 '23 at 06:47
Yeah man I get it. All that I'm asking for is a bit of slack. I hope you can understand I am trying to do what I think is best for the site. – Clyde Kertzer Mar 02 '23 at 07:12

If $f: \mathbb{Z} \to \mathbb{Z_{99}}$ is defined by $f(x) = [50][x]$ is not injective. Is it injective when $f: \mathbb{Z_{99}} \to \mathbb{Z_{99}}$?

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