I definitely feel like an idiot because of the fact that I have to ask you the following question, but anyway I just don't know any other option. If just proven two things which are the opposites of each other, so at least one of my proofs have to be wrong. I tried to solve the following problem:       Definition: $a \boxplus b = 2a+2b$. $\boxplus$ is commutative.   Is $\boxplus$ assosiativ?   So I found two ways to solve the problem: 1. per definition We want to show $(a \boxplus b) \boxplus c = a \boxplus (b \boxplus c)$. $(a \boxplus b) \boxplus c = (2a+2b)*2+2c = 4a+4b+2c.$ And on the other side $a \boxplus (b \boxplus c) = 2a+2(2b+2c)= 2a+4b+4c$ $4a+4b+2c \neq 2a+4b+4c$. So $\boxplus$ is not assosiative.    2. Show that commutativity implicates associativity We want to proof $ (a \circ b) \circ c = a \circ (b \circ c)$ [$\circ$ shall be commutative] $ (a \circ b) \circ c = a \circ b \circ c = b \circ a \circ c = b \circ c \circ a = (b \circ c) \circ a. $    I know that my mistake have to look stupid and obvious, but I just can't see where I'm wrong. Thank you all for helping me!
98608
Does commutativity imply associativity? I'm asking this because I was trying to think of structures that are commutative but non-associative but couldn't come up with any. Are there any such examples?  NOTE: I wasn't sure how to tag this so feel free to retag it.
119233
I have already tried with $S_3$, and indeed, the product is $(13)$, and $(13)^2=e$    But what about this: I define + in this way:$45=2$,$26=3$, $1$ is the identity. therefore, $123456=2326=233=3$, and as you can see from the picture, the inverse of $3$ is not itself, thus contradicts the statement.  My question: What is wrong with my argument?  And How do you prove the statement? it seems that my set of numbers perfectly forms a group under the operation I defined via the table above. first it's closed, second there is an identity, third every elements have an inverse. By the way it's commutitive. I tried (45)2=22=6=42=4(52), (45)6=26=3=43=4(56), so associativity seems to be satisfied.
123815
I have some difficulties with a task in algebra. I guess it's trivial and really easy but I can't figure out how to solve it.  I have a set $G$ and a binary operation on it, let it be $\circ$. I have that the operation is associative and that the equations $a\circ x = b$ and $x\circ a = b$ have unique solutions. I have to prove that $(G, \circ)$ is a group.  I already have that the operation is binary and associative, so I have to prove that there is unique identity element and unique inverse element and it will come from the equations, but how exactly?
53963
Just need some verification for this question or another way to approach it.  Given: a~b is defined on the rationals by a~b iff $b = ak^2$ for $some$ rational number $k$.  Is it reflexive? Symmetric? Transitive?(Show the Equivalence relation)  Reflexive: Must show that a~a   $a=ak^2$  $1=k^2$ which checks out. Its Reflexive  Symmetric: Must show that a~b and b~a  a~b: $b=ak^2$, $\frac ba$=$k^2$  b~a: $a=bk^2$, $\frac ab$=$k^2$  Not symmetric.  Transitive:   a~b and b~c than we must show that a~c  a~b: $b=ak^2$,  b~c: $c=bk^2$  a~c: $c=ak^2$  This is were I am sorta stuck not sure if I am allowed to do the following:  since $c=ak^2$ and $c=bk^2$ I can find that a=b?not sure any help would be appreciated.  Noticing that if zero is excluded then we would have an equivalence relation. However, when zero is included would symmetric fail and both reflexive and transitive still hold?
52033
Me and my friend were arguing over this &quot;fact&quot; that we all know and hold dear. However, I do know that $1+1=2$ is an axiom. That is why I beg to differ. Neither of us have the required mathematical knowledge to convince each other. And that is why, we decided to turn to Math Stackexchange for help. What would be stack's opinion?
32316
So basically this proof centers around proving that (S,*) is a group, as it's quite easy to see that it's abelian as both addition and multiplication are commutative. My issue is finding an identity element, other than 0. Because if 0 is the identity element, then this group won't have inverses.  The set explicitly excludes -1, which I found to be its identity element, which makes going about proving that this is a group mighty difficult.
65136
I am just looking for some clarification on this exercise:  Let $A = \{a,b,c,d\}$. Give an example of a relation $R$ on $A^2$ which is reflexive, symmetric, and not transitive.  I understand that if I was only dealing with $A$ and and not $A^2$ I could give an answer similar to:  $R = \{(a,a),(a,b),(a,d),(b,a),(b,b),(c,c),(d,a),(d,d)\}$  Here $R$ is reflexive, symmetric, and not transitive.  What is confusing me is the fact that we have $A^2$. I am not sure what I am missing here. I am looking through my book to see if I can figure this out, but if someone could help shed some light on this it would be appreciated.  Thanks,  Tony
116588
I am trying to solve an exercise from Vakil's lecture notes on algebraic geometry, namely, I want to show that $\require{AMScd}$ \begin{CD}     X_1\times_Y X_2 @&gt;&gt;&gt; X_1\times_Z X_2\\     @V V V @VV V\\     Y @&gt;&gt;&gt; Y \times_Z Y \end{CD}  is a cartesian diagram. I think I've figured out where the maps come from: The lower map is the diagonal induced from the universal property of $Y \times_Z Y$, the upper map is coming from the universal property of $X_1\times_Z X_2$ (this is one of the previous exercises in Vakil's notes), the map on the right also comes from the universal property of $Y \times_Z Y$ since we have two maps from $X_1\times_Z X_2$ to $Y$ whose composition with $Y\rightarrow Z$ coincides, and finally the map on the left side is either composition $X_1\times_Y X_2 \rightarrow X_i \rightarrow Y$. At this point, I don't even see why this diagram should be commutative.  But even discarding the issue of commutativity, given some $T$ instead of $X_1\times_Y X_2$ making the above diagram commute, I have no clue how to obtain a canonical map $T \rightarrow X_1\times_Y X_2$.  I could solve all the other problems in this section of his notes, but this one eludes me. Any help or solution (preferably just using the universal properties that we're given) would be greatly appreciated.
47706
From Dr. Pinter's "A Book of Abstract Algebra":     Given $a$ and $b$ are in $G$ and $ab=ba$, we say that $a$ and $b$ commute.         Prove $a^2$ commutes with $b^2$      I tried:  $$ ab=ba $$ $$ a^{2}b^{2}=b^{2}a^{2} \text{ // to prove that $a^2$ commutes with $b^2$} $$ $$ aabb=bbaa $$  Then, knowing that $$a=b^{-1}ab,$$ I substituted that '$a$' into '$aabb=bbaa$'.  But that simplified to '$aabb=bbaa$', i.e. I got nowhere with it.  How can I prove that $$aabb=bbaa$$ per this problem?
10591
If $A,B,C$ are matrices I am thinking how to show that $$ A(B + C) = AB + AC$$  Is possible to show without sums like $\sum_i a_i, ..., \sum_j b_j$? It seems if I do the proof with many indexes then is tedious and I don't learn much from it.
22190
So far I have that $a|b$ implies $b=ax$ for some x in the integers and $c|d$ implies that $d=cy$ for some y in the integers.  From here I can see that $gcd(a,c)|gcd(b,d)$ is logically equivalent to $gcd(a,c)|gcd(ax,cy)$ but I am not sure where to go from there.
35983
1) How do I prove the following:   Let A = {6a + 4b ∈ Z : a, b ∈ Z} and B = {2a ∈ Z : a ∈ Z}. Show that A = B.  Thank you all for the help!
93154
I'm having problems with this; the strategy is to show the left side is in the right side, and vice versa. Thus far I can only show that both sides of the equality are in A\B. Can I use "the empty set = A and not A" as a fact? Because then it works. Or just show me how to do it!
71823
Let $X$ de a vector space and $A\subset X$. If we define $A+B :=\{a+b|a\in A, b\in B\}$ I am trying to show that    $$ \begin{equation} 2A \subsetneq A+A, \end{equation} $$  i.e., it might happen that $2A \neq A+A$. I was able to show that $2A \subset A+A$, but I am not able to come up with a counterexample to show that the converse is not necessarily true. If I assume $A+A\subset 2A$ and let $x\in(A+A)$, then by its definition we have $x=a_1 + a_2$ for some $a_1,a_2\in A$ I guess I could naively say that $a_1+a_2 = 2(a'_1+a'_2)$ where $a_i=a_i/2$ for $i=1,2$ and I guess $(a'_1+a'_2)$ might not be an element of $A$ in some concrete case but I am not able to come up with anything and I've spent more time than I would like to admit. Any suggestions?  Thank you.
31698
I'm having problems trying to figure this out. It seems very straight forward, but I don't know what I should do to prove it. Let $X$ be a set with an associative binary operation $\circ$ with identity $e$. Suppose every element of $X$ satisﬁes $x \circ x = e$. Prove that $\circ$ is commutative.
123850
Through the use of Boolean algebra, show that the XOR operator ⊕ is both commutative and associative. I know I can show using a truth table. But using boolean algebra? How do I show? I totally have no clue. Any help please?
62124
For all $ a, b \text{ and } c \in \mathbb{R}$ and $a&gt;1$, Prove that $a^b\cdot a^c=a^{b+c}$  I have come across this question and its  bugging me. Its a basic property that we learn in HS  and I was hoping someone can enlighten me