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Given the function: $f: \Bbb Z_{44100} \rightarrow \Bbb Z_{150}\times\Bbb Z_{294}$ defined as follows $[x]_{44100} \rightarrow ([x]_{150},[x]_{294})$

  1. Calculate $f(12345)$ - Answered
  2. A preimage of (106, 250) - Answered
  3. Is f surjective or injective? - Answered
  4. Prove that $f$ is well defined

For (1) I answered with $(45,291)$ which are of course $12345\pmod{150}$ and $12345\pmod{294}$. For (2) I don't know how to exactly proceed. Is it sufficient to multiplicate the two numbers? For (3) I suppose $f$ is surjective but I don't know how to exactly prove the surjectivity.

2 Answers2

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For the last part, the kernel is nontrivial because the $\operatorname{lcm}(150,294)=2\cdot3\cdot 5^2\cdot 7^2=7350\not\cong0\pmod{44100}$.

For part $1$, go back and divide correctly and get the residues.

For $2$, you can use the Chinese remainder theorem. You need $x$ such that $\begin{align}x\cong0\pmod 2\\x\cong1\pmod3\\x\cong6\pmod {5^2}\end{align}$ and $x\cong5\pmod{7^2}$.

I get $-44\cong44056\pmod{44100}$.

  • Hello, thanks for your answer. I keep trying to answer to (3) and (4). What is the exact relationship between the kernel you mentioned and the injectivity/surjectivity? –  Jan 20 '19 at 11:15
  • Since the groups have the same order, surjectivity is equivalent to trivial kernel. Trivial kernel, on the other hand, is equivalent to injectivity. We can conclude that the map is neither injective nor surjective. –  Jan 20 '19 at 12:30
  • I'm relying on the first isomorphism theorem. –  Jan 20 '19 at 12:37
  • Oh, yeah, that's right. Thank you for reminding me. It completely slipped my mind. However, why do you also exclude surjectivity? (I have a hypothesis but I'm not sure and I would like to ask your opinion). –  Jan 20 '19 at 18:47
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    They have the same order, so, if the kernel is bigger than ${e}$, the image is too small. That's because the image is isomorphic to $\Bbb Z_{44100}$ mod the kernel. –  Jan 20 '19 at 18:53
  • Actually, for a map between finite sets of the same order, injective corresponds to bijective. No need for the theorem. –  Jan 20 '19 at 19:34
  • I'm sorry for my insistence, I'm new to these topics. I have understood what you mean but I still cannot understand how this can exclude the surjectivity of the function. –  Jan 21 '19 at 11:33
  • If the map isn't injective, there aren't enough elements of the domain left to map onto the range. This is a typical problem in Discrete Mathematics. –  Jan 21 '19 at 16:36
  • Once two points go to the same point, the image has order $\le n-1$, where $n$ is the order of the two sets. –  Jan 21 '19 at 16:40
  • Thanks to your suggestion I have deepened this topic and now it is clearer. Looking for more information on the web about point (4) I noticed that one of the ways to verify if the function is well defined is to check if it is an isomorphism. In such a case since the function is not bijective then it is not defined. Is it correct or is there no relation with verifying if the function is well defined? Thanks again. –  Jan 22 '19 at 17:50
  • It's certainly well-defined. Again there is a basic fact that a map between finite sets of the same order is surjective precisely when injective. Sure thing. –  Jan 22 '19 at 17:54
  • I'm a bit confused. So the function is well-defined despite it isn't injective nor surjective or...? –  Jan 22 '19 at 18:42
  • Sure. Those are different notions. –  Jan 22 '19 at 18:57
  • So, how can I check if it is well defined? –  Jan 22 '19 at 18:59
  • It's well defined. You've got a function there. There's nothing to stop you from moding out. –  Jan 22 '19 at 19:02
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Hint for (2), note $\ 106\equiv \color{#c00}{-44}\pmod{150}\ $ and $\ 250\equiv \color{#c00}{-44}\pmod{294}$ so it is equivalent to find a preimage of $(-44,-44),\ $ which is obvious (this is essentially CCRT = Constant-case of CRT = Chinese Remainder Theorem).

Remark $ $ Generally if $\,x\equiv a_i\pmod {m_i}$ and $\,\color{#c00}{a_i - m_i = c}\,$ is constant (independent of $i),\,$ the system reduces to a simple constant case of CRT $\,\ x\equiv \color{#c00}{a_i \equiv m_i+c} \equiv c\pmod {m_i}.\,$ Therefore $$ x\equiv c\!\!\!\pmod {m_i}\iff m_i\mid x\!-\!c\iff {\rm lcm}\{m_i\}\mid x\!-\!c\iff x\equiv c\!\!\pmod{{\rm lcm}\{m_i\}}$$

Above is the special case $\, 106-150 = \color{#c00}{-44} = 250-294$

Of course we can also try adding/subtracting other small multiples of the modulus to search for the solution before diving head-first into the general CRT algorithm.

Bill Dubuque
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