We represent an element of the domain as an equivalence class $\bar x$, and use the notation $ \left[ x \right]$ for equivalence classes in the codomain. Show that this is not well defined.     $ f: \Bbb Z_{4}  \rightarrow \Bbb Z_{6} \; \text{ given by, } f(\bar x) = \left[ 2x+1\right]$   My question is how do does one determine/ show that this is not a function? How does one show that this not one to one?  $f(\bar 0) =  2(0) +1 = [1]$  $f(\bar 1) = [3]$  $f(\bar 2) = [5]$  $f(\bar 3) = [7]$  $f(\bar 4) = [9]$  $f(\bar 5) = [11]$  $f(\bar 6) = [13]$  I also guessing here that if a number is a multiple of another then that may play a role in determining it if is one-one.
90614
By naming an equivalence class in the domain that is assigned at least two different values prove that the following is not a well defined function.     $$f : \Bbb Z_{3} \to \Bbb Z_{6} \;\;\;\text{ given by } f(\overline x) = [x]  $$      In this case we represent an element of the domain as an $\bar x$ and use the notation $[x]$ for equivalence classes in the co-domain.   $f(\overline0) =  [0] \;,$  $ \Bbb Z_{3}  \quad (3x+0)\;\; \overline 0 = \{ ...-6,-3,0,3,6...  \}, \; \Bbb Z_{6}\;  (6x+0)\; \overline0 =\{ ...-12,-6,0,6,12...\}$   $f(\overline1) = [1], $ $\qquad   \; (3x+1) \; \;\;\;\overline   1 = \{ ...-5,-2,1,4,7 ... \},\; \;  (6x+1)\;\overline1 =\{...-11,-5,1,7,13.. \}$  $f(\overline2) = [2], $ $\qquad  \qquad \qquad \;\overline 2 = \{ ...-4,-3,2,5,8 ... \},\;\;\overline 2 = \{ ...-10,-4,2,8,14  ...\},\;$  $f(\overline3) = [3] ,$  $\qquad \qquad \qquad \qquad \qquad \qquad,\; \quad \quad \quad \; \; \; \;\overline 3 = \{ ...-9,-3,3,9,15 ... \},$  $f(\overline4) =  [4],\qquad \qquad \qquad\qquad \qquad\qquad \; \quad \quad \quad \quad \; \;\overline 4 = \{ ...-8,-2,4,10,16...  \}, $   $f(\overline5) = [5], \qquad \qquad \qquad\qquad \qquad\qquad \; \quad \quad \quad \quad \;\;\overline 5 = \{ ...-7,-1,5,11,17...  \},  $  $f(\overline6) = [6] ,$  So my main question for this problem is how to find out if this question is not a function. From the information I have gathered here I still cannot see why this is not a function any help on showing how this is not function would be much appreciated.     The set of equivalence classes for the relation $\cong_{m}$ is denoted $\Bbb Z_{m}$   The $ 3x+0 \text{ and } 6x+0$ are just showing how I got $\overline 0 $
23980
I'm reading .     The graph of $f: A \rightarrow B$ is the subset of $A × B$ consisting   of all ordered pairs $(a, b)$ with $f (a) = b$. If $A$ happens to be   $B^n$ for some $n ∈ N$, then we say that $f$ is a function on $B$ and   $n$ is the arity of $f$.   I don't understand the notation for the exponent of a set, I'm aware of what  is, it's the number of arguments or operands the function takes. I'm just not so sure if I can assume that a set with an exponent such as in $B^n$ means that a function has two operands.  I know the question may be too naive and perhapes the answer is the one I suggested and it's right in front of my face, I'm just searching for a confirmation or the negation of it. Thanks.
16246
I am asked to consider f(x) and g(x) which are both 1-1 functions on their respective domains. I have been asked to show f ∘ g is a 1-1 function and then a followup statement of if f ∘ g is 1-1 does that mean f(x) and g(x) are both 1-1 on their domains.  The problem I am having with this question is how do I show the result? I understand when a function is 1-1 over a certain domain and I have tried composing my own functions but am unsure how I prove the result is true because I do not believe providing an example of two functions f(x) and g(x) where the first statement holds is really what I am supposed to do.  Sorry for posting again so soon, especially for such a similar and straightforward question.  Thank you.
48020
From      Every linear function on a finite-dimensional space is continuous.   I was wondering what the domain and codomain of such linear function are?   Are they any two topological vector spaces (not necessarily the same), as along as the domain is finite-dimensional? Can the codomain be a different normed space (and may not be finite-dimensional)?  I asked this because I saw elsewhere the same statement except the domain is a finite-dimensional normed space, and am also not sure if the codomain can be a different normed space (and may not be finite-dimensional).  Thanks and regards!
23965
I have trouble understanding the following task:     Show that the set $X = \{f_1,\ldots ,f_6\}$ of functions $f_i : \Bbb Q\setminus \{0,1\} \to \Bbb Q\setminus \{0,1\}$ with   $x ↦ f_1(x) = x$,$x ↦ f_2(x) = 1/x$,$ x ↦ f_3(x) = 1-x,$ $x ↦ f_4(x) = 1/(1-x)$,$ x ↦ f_5(x) = (x-1)/x$, $x ↦ f_6(x) = x/(x-1)$   and the composition as the operation is a group. Is this group commutative?    I'm a little confused as I don't know what to do with all these functions. I can't apply the definitions of a group on this because of that. If the group is $(X,∘)$ how do I show that this is associative don't I need the information on how the single functions are connected or am I missing something?
90939
In $\mathbb{Z}\left [ x \right ]$, let $I = \left \{ f\left ( x \right ) \in \mathbb{Z}\left [ x \right ]: f\left ( 0 \right ) \text{ is an even integer} \right \}$   In fact, $I=\left \langle x,2 \right \rangle$  How do I show that the order of $\mathbb{Z}\left [ x \right ]/I$ is 2?
81503
I know that  $A_{[a} B_{b]} = \frac{1}{2!}(A_{a}B_{b} - A_{b}B_{a})$  But how can write $E_{[a} F_{bc]}$ like the above?  Can you provide a reference where this notational matter is discussed?
120464
I've come to this: $$f: \mathbb{N} \to \{\ldots, -6,-4,-2,0,2,4,6, \ldots\},\qquad f(n) = \begin{cases} 2n &amp; \text{ if } n \text{ is odd} \\  -n &amp; \text{ if } n  \text{ is even} \end{cases}$$  I don't know what to do with this though. I never know how to format a proof correctly.
113124
What I mean is that $\mathbb{CRZ}$ etc. are used for different classes of numbers, allowing me to do stuff like this:   $$f:\mathbb{R}\to\mathbb{R}$$ $$f:x\mapsto 3x$$  But say I have an expression like this:  $$g:x\mapsto x&gt;4$$  What symbol would you use to represent a boolean result? a $\mathbb B$?
112798
How do I prove this function is bijective? $$ v(s,p)=2^{p-1}(2s-1). $$ The domain is natural numbers and the codomain is also the natural numbers.  And this one: $$ f(s)=2s-1. $$ The domain is the natural numbers, and the codomain is the odd numbers in the natural numbers.  With this one I would do this to show it's injective: $$ \begin{align} v(s)&amp;=v(s_1)\\ \implies 2s-1&amp;=2s_1-1\\ \implies (2s_1)/2&amp;=(2s_2)/2\\ \implies s=s_1 \end{align} $$ So it's injective since if $v(s)=v(s_1)$ then $s=s_1$.  And to show it's surjective $f(s)=y$: $$ \begin{align} y&amp;=2s-1\\ \implies s&amp;=(y+1)/2 \end{align} $$ Then the function must be surjective since every $y$ is the same as the codomain for $f$.  Am I correct?
69087
Give:  $fn(S)=\prod_{x\in S}x$  what is:  $fn(\emptyset)$  I can see reason that it would be defined as 1 or 0.    Note: I thought about restricting the domain of $S$ but that would make the problem less general. If there is no general answer then answers for restricted domains would be valid.
93658
Suppose $a$ and $z$ are in $\mathbb{C}$ with $|z|=1$ and $|a|&lt;1$. I want to show $$\left|\frac{z-a}{1-\bar{a}z}\right|=1$$ but I'm stuck on dealing with $a$. Is this just another complex number of the form $a+bi$?   How do I show this is true? Thanks in advance.
68159
Let $f$: $Z_m \rightarrow G$, $f(n)$ = $a^n$. Show that f is a function and an homomorphism.  Doing a review on the whole semester, and a little refresh on this topic would be great.   By definition I know $G$ has the same order of $Z_m$, but not sure how to attack this problem with ease.   I also know that cyclic groups of the same order are isomorphic.   (I cannot use the fact that cyclic groups of the same order are isomorphic, that's why I ask for another approach, because I can't use theorems or lemmas that have not been discussed in class.)
2529
I am having trouble with the following problem.      Let $R$ be an integral domain, and let $a \in R$ be a non-zero element. Let $D = \{1, a, a^2, ...\}$. I need to show that $R_D \cong R[x]/(ax-1)$.    I just want a hint.  Basically, I've been looking for a surjective homomorphism from $R[x]$ to $R_D$, but everything I've tried has failed. I think the fact that $f(a)$ is a unit, where $f$ is our mapping, is relevant, but I'm not sure. Thanks
93443
$$(\mathbb{Z}/(n))[x] \simeq \mathbb{Z}[x] / (n)$$  I've shown this by showing that the map that sends $\overline{1} \mapsto [1+(n)]$ (where the bar denotes the congruence class mod $n$) and $x \mapsto [x+(n)]$ is a homomorphism that is injective and bijective, which is a bit cumbersome. Is there an easier way to show this? I can't tell if this is trivial or not.
17342
Let $R$ be an integral domain and $f,g\in R[X_1,...,X_n]$ where $n&gt;1$. What is an example of a pair $f,g$ such that $\deg(fg)&lt;\deg(f)+\deg(g)$?   Moreover, I have proven that the units of $R[X_1,...,X_n]$ are just the units in $R$. (Proof is done inductively) Is it true?  EDIT    Even though the each term in the underlined summation is nonzero, isn't it possible that the summation is 0? Why is this impossible ?
21756
I know $$g(x) = \arctan(x)+\arctan(y) = \arctan\left(\frac{x+y}{1-xy}\right)$$  which follows from the formula for $\tan(x+y)$. But my question is that my book defines it to be domain specific, by which I mean, has different definitions for different domains:  $$g(x) = \begin{cases}\arctan\left(\dfrac{x+y}{1-xy}\right), &amp;xy &lt; 1 \\[1.5ex]  \pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &amp;x&gt;0,\; y&gt;0,\; xy&gt;1 \\[1.5ex]  -\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &amp;x&lt;0,\; y&lt;0,\; xy &gt; 1\end{cases}$$  Furthermore, When I plot the function $2\arctan(x)$, it turns out that the book definition is correct. I don't understand how such peculier definition emerges. Thank you.