Is it always true that    $(A_1 \cap A_2) \times (B_1 \cap B_2) = (A_1\times B_1) \cap (A_2 \times B_2)$ and $(A_1 \cup A_2) \times (B_1 \cup B_2) = (A_1 \times B_1) \cup (A_2 \times B_2)$?   **I've already showed the first part, just trying to prove true/false for parts a and b, thanks!
123016
Is it always true that  $(A_1 \cup A_2) \times (B_1 \cup B_2)=(A_1\times B_1) \cup (A_2 \times B_2)$?  I don't believe this is true. I have tried to draw pictures to help me get on the right path, but I think that the union makes this untrue. for example, if $a \in A_1$ and $b \in B_2$, then $(a,b)$ would not be in $(A_1\times B_1) \cup (A_2 \times B_2)$. Is this a correct assumption?
75101
I tried to prove this question but without a success: Let $K_1$ and $K_2$ be definable sets, prove that $K_1\cup K_2$ is definable.  What I tried to do is to assume: $K_1=Ass(X)=\{ v|v \vDash X \}$ and $K_2=Ass(Y)=\{ v|v \vDash Y \}$ and to say that $K_1\cup K_2=Ass(X \cap Y)$. but I couldn't prove that: $K_1\cup K_2 \supseteq Ass(X \cap Y)$ .   I would like to get help with this proof
72883
Is it true that if $ U \oplus W_1 = U \oplus W_2 $, then $ W_1 = W_2 $? I think that if $ U \oplus W_1 = U \oplus W_2 $, then u+w1=u+w2, so W1=W2. But did I make any mistakes?
57225
Can $(X_1,X_2) \cap (X_3,X_4)$ be generated with two elements in the ring $R=k[X_1,X_2,X_3,X_4]$? Can it be generated with three elements? (Here $k$ is a field.)    Thanks for any help.
9602
If $A$ and $B$ are nonempty sets, prove that $A \times B = B \times A$ if and only if $A = B$.   Proving the first direction of this is easy. That is, if $A = B$ then it is obvious that $A^2 = A^2$. I am wondering how to prove the other direction now. Maybe proving it by contradiction will be easier than directly.
71386
I am fairly sure, given examples $\Bbb{R}\times \Bbb{R},\Bbb{R}\times \Bbb{Q},\Bbb{Q}\times \Bbb{Q} $, that this is correct, but do not know how to prove it.   In my cited examples the proof has always used some specific property of $A,B$, leaving me clueless as to the general method of solving this. I am especially interested in the case were $|\Bbb{R}|&gt;A&gt;|\Bbb{Q}|,|B|=|\Bbb{Q}|$, though ideally the answer would be more general.  To be explicit, I'm asking for a proof of the statement in the question title, where $A,B$ are sets with infinite cardinalities.
82823
Prove that for any sets $A$ or $B$, if $\mathcal P(A) \cup \mathcal P(B)= \mathcal P(A\cup B)$ then either $A \subseteq B$ or $B \subseteq A$. ($\mathcal P$ is the power set.)  I'm having trouble making any progress with this proof at all. I've assumed that $\mathcal P(A) \cup \mathcal P(B)= \mathcal P(A\cup B)$, and am trying to figure out some cases I can use to help me prove that either $A \subseteq B$ or $B \subseteq A$.   The statement that $\mathcal P(A) \cup \mathcal P(B)= \mathcal P(A\cup B)$ seems to be somewhat useless though. I can't seem to make any inferences with it that yield any new information about any of the sets it applies to or elements therein. The only "progress" I seem to be able to make is that I can conclude that $A \subseteq A \cup B$, or that $B \subseteq A \cup B$, but I don't think this gives me anything I don't already know. I've tried going down the contradiction path as well but I haven't been able to find anything there either.  I feel like I am missing something obvious here though...
10524
I am given this equation:  $f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)$  I want to prove it: what i did is  I take any $a \in f^{-1}(B_1 \cap B_2)$, then there is $b \in (B_1 \cap B_2)$ so that $f(a)=b$. Because of $b \in (B_1 \cap B_2)$, it is true that $b \in B_1$ and $b \in B_2$, so $a \in f^{-1}(B_1)$ and $a \in f^{-1}(B_2)$.    this means $f^{-1}(B_1 \cap B_2) \subseteq f^{-1}(B_1) \cap f^{-1}(B_2)$.   is it ok?
69606
$A$ and $B$ are $n\times n$ matrices and $A+B=AB$. I have an interesting proof that this implies $A$ commutes with $B$, but the proof only works when $||B|| \lt 1$. I'm looking for a way to salvage the proof so that it works for all $B$.  $$A=AB-B$$ $$A=(A-I)B$$ by repeatedly substituting $A$ with $(A-I)B$ on the RHS, $$A= (A-I)B^N - \sum_{i=1} ^{N-1} B^i$$  Since the space of $n\times n$ matrices is Banach (hence complete) under the standard operator norm, the limit $\sum_{i=1}^{\infty}B^i$ is well defined given $||B|| \lt 1$. Hence, taking the limit as $N \rightarrow \infty$, $$A =  -\sum_{i=1}^{\infty}B^i$$  Now the LHS commutes with B, hence so does the RHS.    Is there some trick to extend the method to work with any $B$ where $||B|| \geq 1$? I know that the result is true due to another simple algebraic proof, but I'd like to see if this method can be salvaged.
79825
I'm taking my first course in Probability and one of my homework problems is to prove that for any two sets:  $$P(A \cup B) = P(A) + P(B) - P(A \cap B) $$  Note that the function $P$ is the probability of an event happening.  From the third axiom, we are given that for any two disjoint sets, that is $ A \cap B = \emptyset$:  $$P(A \cup B) = P(A) + P(B)$$    My thought process follows from this third axiom, but I run into a problem near the end:  Since $A$ and $B$ are disjoint sets, $P(A)=P(A-B)$ and $P(B)=P(B-A)$, so  $$P(A \cup B) = P(A) + P(B)=P(A-B)+P(B-A)$$  From this, it appears that we are essentially taking the symmetric difference of A and B.  This, to me, seems to be equivalent to:  $$P(A-B)+P(B-A)=P(A)+P(B)-P(A \cap B)$$  Hence, by my logic:  $$P(A \cup B) = P(A)+P(B)-P(A \cap B)$$  Which, in essence, doesn't care whether two sets are disjoint or not as we are subtracting the intersection of them.  If the sets are disjoint, then the intersection is merely null.  If they intersect, then we are removing the "extra pieces".  I feel like this is inadequate proof despite my ability to convince myself that it is true..  Where, if anywhere, did I go wrong?  Thanks!
8479
Given a field $F$ and a subfield $K$ of $F$. Let $A$, $B$ be $n\times n$ matrices such that all the entries of $A$ and $B$ are in $K$. Is it true that if $A$ is similar to $B$ in $F^{n\times n}$ then they are similar in $K^{n\times n}$?   Any help ... thanks!
9242
I have a proof and need some feedback. It seems really obvious that the statement is true but it is always the obvious ones that are a little trickier to prove. So I would appreciate any feedback. Thank you!  Here is what I am asked to prove:     If $n$ is composite then $(n-1)! \equiv 0 \pmod n$.   Proof:  $n$ is composite $\implies n=ab$ where $a,b \in \mathbb{Z}$ and $0&lt;a,b&lt;n$.  Case 1: If $a=b$ then $n=a^{2}$. Now $n \mid (n-1)! \implies a \mid (n-1)!$, so $$\begin{aligned} (n-1)! &amp;\equiv 1\times 2\times \dotsb \times a \times\dotsb\times (n-a)\times\dotsb\times (n-1) \\ &amp;\equiv 1\times 2\times \dotsb\times a \times\dotsb\times -a\times\dotsb\times -1 \\ &amp;\equiv 0 \pmod n \end{aligned}$$  Case 2: $0&lt;a&lt;b&lt;n$.  Then, since $a \mid n$, $b \mid n$ and $n \mid (n-1)!$ we have that $a \mid (n-1)!$ and $b \mid (n-1)!$.  So this implies $(n-1)! \equiv 1\times 2\times \dotsb\times a \times\dotsb\times b\times\dotsb\times (n-1) \equiv 0 \pmod n$, Q.E.D.
2845
Let $A,B,C$ be sets, and $B \cap C=\emptyset$. Show $|A^{B \cup C}|=|A^B \times A^C|$ by defining a bijection $f:A^{B \cup C} \rightarrow A^B \times A^C$.  Any hints on this one?  Thank you!
107994
Let $A \times \emptyset = \{(x,y)| x\in A, y \in \emptyset \}$. We know there is no element in $\emptyset$. But how does it follow that $A \times \emptyset = \emptyset $?
8158
How to prove $( A \times B ) \cap ( C \times D ) = ( A \cap C ) \times ( B \cap D )$ ?
21499
Statement: Let $A$, $B$, and $C$ be sets. Then $(A \setminus B) \cap (A \setminus C) = A \setminus (B \cup C)$.   Proof: Let $(x,y) \in ( A \setminus B) \cap (A \setminus C)$, where $ x \in A$ and $y \in (B \cap C$ by definition of intersection. So, $(x,y) \in (A \setminus B \cap C)$ which show that $(A \setminus C) \cap (A \setminus C) \subseteq A \setminus (A \cup C)$. \ Now let $(x,y) \in A \setminus(B \cup C)$, where $ x \in A $. So, $ (x,y) \in (A \setminus B) \cap (A \setminus C)$ by definition of intersection, which shows that $A \setminus (B \cup C) \subseteq (A \setminus B) \cap (A \setminus C)$. Therefore, by the Axiom Set of Equality we know that $(A \setminus B) \cap (A \setminus C) = A \setminus (B \cup C)$ .   Am I on the right track to proving that this is a true statement or is there a better way to show this?
36141
Show that $(\mathbb{Z}[x]/(x^{n+1}))^{\times}\cong \mathbb{Z}/2\mathbb{Z}\times\Pi_{i=1}^n\mathbb{Z}$.   Anyway how and what method is used to prove this. I still have no idea now. really thanks for your help