The language has three unary function symbols f,g and h. The domain is a square if the model is a model of this sentence and the domain is finite.
124023
(c) Let $\mathcal{L}^*\subseteq\mathcal{L}^{\text{FOPC}}$ be a first order language of predicate calculus consisting of three unary function symbols $f,g,h$.  (i) Write down an $\mathcal{L}^*$-sentence $\phi$ such if $\mathcal{A}$ is an $\mathcal{L}^*$-structure with $\mathcal{A}\vDash\phi$ then the restriction of $h_\mathcal{A}$ to the range of $f_\mathcal{A}$ gives a one-to-one function onto the range of $g_\mathcal{A}$.  (ii) Write down an $\mathcal{L}^*$-sentence $\psi$ such that if $\mathcal{A}$ is an $\mathcal{L}^*$-structure with $\mathcal{A}\vDash\psi$ and the domain $A$ of $\mathcal{A}$ is finite then the cardinality $|A|$ is a square.    Part (i) is easy enough, but I'm having trouble with part (ii). I assume that you can use part (i). My initial idea was to use $h$ to ensure $|Range(f_A)| =|Range(g_A)|$ but I'm not sure where to go from there. Thanks.
109014
In predicate logic, to define a language, we consider the following $3$ sets:  (1) $V$, the set of variables;  (2) $F$, the sets of function symbols;  (3) $P$, the set of predicate symbols with their arities.   Now I am concidering about determine the cardinality of a language over a set with arbitary cardinality, $κ$(Note that $κ$ can be either finite, countablely infinite or uncountablely infinite ).  May I please ask:  If the cardinality of $V$, $\#V = κ$, what conditions on $V$ , $F$ and $P$ will give us a language over $V$, $F$ and $P$ with cardinality $κ$?  And may I please ask if it is the fact that if $\kappa$ is finite, then to get a language with cardinality $\kappa$ we need $(V+F+P)^{&lt;\omega}=(\kappa+F+P)^{&lt;\omega}=\kappa$, so we can never have a language of cardinality $\kappa$. I think it is something weird here. Am I wrong with something?  I had googled it but I have get nothing from websites, now I am confused. Could someone please help? Thanks so much!  EDIT: Latter I am told that in my course, the cardinality of a language refers to the set of expression instead of this of the set of strings. So now I need to consider further about when a string makes sense.  EDIT':Now I am considering that if we can find out a general way to determine the cardinality of expression, which may related to combinatorics. See this question:  Any help would be appreciate.
2189
Since $X$ and $Y$ are countable, we have two bijections:   (1) $f: \mathbb{N} \rightarrow X$ ;   (2) $g: \mathbb{N} \rightarrow Y$.   So to prove that $X\cup Y$ is countable, I figure I need to define some function,   h: $\mathbb{N} \rightarrow X\cup Y$  Thus, I was wondering if I could claim something similar to the following:   since we have (1) &amp; (2) it follows that we also have the bijections   $\alpha : \{n\in \mathbb{N} : n = 2k + 1, k\in \mathbb{N}\} \rightarrow X$;   $\beta : \{n\in \mathbb{N} : n = 2k, k\in \mathbb{N}\} \rightarrow Y$;    because we have bijections from $\mathbb{N}$ to the evens and odds respectively.   Then define $h := \alpha$ for odd $n$, $h := \beta$ for even $n$.   Thus, since $\forall n\in \mathbb{N}$(either $n$ is even or $n$ is odd but not both) $h$ is a bijection from $\mathbb{N}$ to $X\cup Y$.   Thanks for reading, and for answering if you answer. I'm just really unsure if my logic holds here, or if I'm even approaching this right because I've been stuck on this problem for a little while today.   p.s sorry for asking so many questions recently, I'm trying to study on my own and apparently I get confused and stuck more easily than I thought I would without a teacher.  EDIT: (1) fixed the even and odd.  (2) I do mean countably infinite. My bad, going through some notes I found online they were using the notation of "countable" for what you call "countably infinite", and "at most countable" for what you called "countable" (3) Thanks for the good answers, I do see now that this makes breaks down when they are not disjoint, but at least I'm not too far off.
95937
This is not for homework. (I am a grader for a class.)  The case in which $G$ is finite is trivial. (That is, use a corollary to Lagrange's Theorem, and set $[G:H] = \dfrac{|G|}{|H|}$, and similarly for $[H:K]$.) How do you prove this for when $G$ is infinite?   The proof of this statement that I know well tries to show that $[G:K]$ is finite iff $[G:H]$ and $[H:K]$ are finite, but I am not a fan of that proof. For pedagogical purposes, I'm looking for a better proof. (The course uses right cosets by convention.)  The course has just started covering normal subgroups, so things like the Isomorphism Theorems are not known yet.
92571
Let $A$ be a ring and let $A[x]$ be the ring of polynomials in an indeterminate $x,$ with coefficients in $A.$ Let $f=a_0 + a_1x+\cdots+a_nx^n \in A[x].$ $f$ is said to be primitive if $(a_0,a_1,\ldots,a_n)=(1).$     Prove that if $f,g\in A[x],$ then $f$ and $g$ are primitive $\Rightarrow$ $fg$ is primitive.   I know Gauss lemma holds true for UFDs, but in this case it isn't required $A$ to be UFD: is it possible to drop the assumption, or is some hypothesis missing here?
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.
39864
Let $$f(x)= \frac{x}{x-9} \text{ and } g(x) = \frac{-6}{x+5} $$   Find the domain of $f \circ g.$    Please show the steps and substitution/algebra to the problem. Thank you.
22403
Let $f:\mathbb{N}\cup\{0\}\to\mathbb{N}\cup\{0\}$ be a function which satisfies $f(x^2+y^2)=f(x)^2+f(y)^2$ for all $x,y \in\mathbb{N}\cup\{0\}$. It' easy to see that $f(0)=0$ and $f(1)=0$ or $f(1)=1$. Suppose let's assume that $f(1)=1$. Then it's very easy to see that $f(2)=2$ and since $f(2)=2$, we can see that $f(4)=4$ and $f(5)=5$ because $5=2^2+1^2$. To see $f(3)=3$, note that $$25=f(5^2)=f(3)^2+f(4)^2 \implies f(3)=3$$ One can even see that $f(6)=6$, $f(7)=7$ and so on. Is it generally true that $f(n)=n$ for all $n$?  Edit. If anyone wondering how $f(7)=7$ here is the way. Note that $f$ satisfies $f(n^2)=f(n)^2$. One can see that $$25^{2}=24^2+7^2 \implies 25^{2}=f(24)^{2}+f(7)^{2}$$ We have to show $f(24)=24$. For this note that $$26^2 = 24^2+10^2 \implies f(26)^{2} = f(24)^{2}+f(10)^{2}$$ But $f(26)=26$ because $26=5^2+1^2$ and $f(10)=10$ since $10=3^{2}+1^{2}$. In short if $n=x^2+y^2$ for some $x,y$ and $f(1)=1$, then we can see that $f(n)=n$.
82384
The answer seems to be yes, judging from the following exercise I found in the book Mathematical Logic by H.D. Ebbinghaus, J. Flum, and W. Thomas:     Let $S$ be a finite symbol set and let $\mathfrak{U}$ be a finite $S$-structure. Show that there is an $S$-sentence $\varphi _{\mathfrak{U}}$ the models of which are precisely the $S$-structures isomorphic to $\mathfrak{U}$.   I think I have an idea of how to solve this exercise, but I seem to be unable to materialize it. Thanks.
31727
This is from Fraleigh's First Course in Abstract Algebra (page 82, Theorem 8.16) and I keep having hard time understanding its proof. I understand only until they mention the map $\lambda_x (g) = xg$. Can someone explain this proof step by step? Thank you so much! Here is the proof:  Let $G$ be a group. We show that $G$ is isomorphic to a subgroup of $S_G$. Define a one-to-one function $\phi: G \to S_G$ such that $\phi(xy)=\phi(x) \phi(y)$ for all $x,y \in G$. For $x \in G$, let $\lambda_x: G \to G$ be defined by $\lambda_x (g) = xg$ for all $g \in G$. The equation $\lambda_x (x^{-1} c) = x(x^{-1} c) = c$ for all $c \in G$ shows that $\lambda_x$ maps $G$ onto $G$. If $\lambda_x (a) = \lambda_x (b)$, then $xa=xb$ so $a=b$ by cancellation. Thus $\lambda_x$ is also one to one, and is a permutation of $G$. We now define $\phi: G \to S_G$ by defining $\phi(x) = \lambda_x$ for all $x \in G$. To show that $\phi$ is one to one, suppose that $\phi(x) = \phi(y)$. Then $\lambda_x = \lambda_y$ as functions mapping $G$ into $G$. In particular $\lambda_x (e) = \lambda_y (e)$, so $xe=ye$ and $x=y$. Thus $\phi$ is one to one. It only remains to show that $\phi(xy) = \phi(x) \phi(y)$, that is, that $\phi_{xy} = \lambda_x \lambda_y$. Now for any $g \in G$, we have $\lambda_{xy} (g) = (xy)g$. Permutation multiplication is function composition, so $(\lambda_x \lambda_y)(g) = \lambda_x (\lambda_y (g)) = \lambda_x (yg) = x(yg)$. Thus by associativity, $\lambda_{xy} = \lambda_x \lambda_y$.
53929
Let $g$ be a function from $(0, 1)$ to $\mathbb{R}$ and $\int_0^1 g^2(x)dx$ finite. Does this imply that $\int_0^1 g(x)dx$ is finite?   So I know that $| \int_0^1 g(x)dx | \leq \int_0^1 |g(x)|dx = \int_0^1 \sqrt{g^2(x)}dx$. Not sure how to continue.
106353
I am very confused about how to interpret the following.  If we write $\{0,1\}^{+}$=$\cup_{n \in \mathbb{N}} \{0,1\}^{n}$  then I am told that elements  of $\{0,1\}^{+}$ are called words and subsists of it are called languages,  Then I was asked,   if we consider $$g: P( \{0,1\}^{+}) \to P( \{0,1\}^{+})$$ defined as  $g(X)=X \cup \{w01:w \in X\} \cup \{\tau\}$  where $\tau$ represents the word of zero length, then find the least fixed point of g ordered by inclusion.  I have been trying to make sense of it but cant make progress. I know that a fixed point is an element $X$ such that $g(X)=X$  I know that since we want the least, the fixed point has to be less then any other fixed points ordered by inclusion to.  Can anyone help to make sense of this? Thank you
75449
Just would like to see if I'm right.  Proof: Let $G$ be generated by $x$ (Let $G=\langle x \rangle$ for some $x$ in $G$.) We wish to show that there exists an $x'$ in $G'$ such that $G'=\langle x' \rangle$. First, we will show that for all $x^n$ in $G$, $f(x^n)=f^n(x)$. We proceed by induction.   Base case: $n=1$. $f(x*e_G)=f(x)f(e_G)=f(x)$. Now, assume that $f(x^k)=f^k(x)$ for some arbitrary positive integer k. We show that $f(x^{k+1})=f^{k+1}(x)$. Observe that $f(x^{k+1})=f(x^kx)=f(x^k)f(x)=f^k(x)f(x)=f^{k+1}(x).$   Next we will show that for all $x^{-n}$ in $G$, $f(x^{-n})=f^{-n}(x)$. We proceed by induction. Base case: $n=-1$. $f(x^{-1}*e_G)=f(x^{-1})f(e_G)=f(x)$. Now, assume that $f(x^{-k})=f^{-k}(x)$ for some arbitrary positive integer k. We show that $f(x^{-k-1})=f^{-k-1}(x)$. Observe that $f(x^{-k-1})=f(x^{-k}x)=f(x^{-k})f(x)=f^{-k}(x)f(x)=f^{-k-1}(x).$ Because $f$ is onto, $Im(f)=G'$ and therefore $Im(f)=(...,f^{-2}(x),f^{-1}(x),e_G,f(x),f^2(x),...)=G'$.  Thus, there exists an $x'$ in $G'$ such that $G'=\langle x' \rangle$ and the proof is complete by the principle of mathematical induction. Q.E.D.  I have a feeling the proof shouldn't be this long. Any suggestions for a shorter, cleaner proof?
38065
Let $G = \left\{e, a, b \right\}$  Since $G$ has $3$ elements, $a \not= e , \,b \not= e, \,a \not= b$  $G$ is closed so $ab \in G$  If $ab = a$, this means that $b = e$ and if $ab = b$, $ a = e $  So we are left with $ab = e$  Define a function  $\varphi (a) = [1]$  $\varphi (b) = [2]$  To show that this function is an isomorphism,   $\varphi (e) = \varphi (ab) = \varphi (a) + \varphi (b) = [1] + [2] = [0]$  So $\varphi (e) = [0]$  So this function takes the identity to the identity, so it's an isomorphism.
69862
I'm trying to understand the UMP of the product functor as right adjoint to the diagonal functor but I'm having difficulties understanding the notation in Steve Awodey's . On page 210 he uses $(f_1, f_2)$, $\langle f_1, f_2 \rangle$ and $(f_1 \times f_2)$ all within one paragraph. What are those? How are these defined and how do they differ?
21524
Let $L$ be a regular language. Prove that $\sqrt{L}:=\left\{ w : ww\in L\right\}$ is also a regular language.    I suppose I need to modify state machine for $L$ to accept $\sqrt{L}$, but I've been thinking how to do that for a few hours and still don't know. I would be very grateful for help.
11452
I just started learning model theory on my own, and I was wondering if there are any interesting examples of two structures of a language L which are not isomorphic, but are elementarily equivalent (this means that any L-sentence satisfied by one of them is satisfied by the second).  I am using the notaion of David Marker's book "Model theory: an introduction".