I tried expanding $n$ parentheses such as $a b^{-1} a$ and then tried to replace the second $a$ with $a^2 a^{-1}$ but didn't know how to proceed further. Also, is it possible to use commutativity here somehow?
120826
If $a,b \in G$, show that $ab^{n}a^{-1}=(aba^{-1})^{n}$.    I have not found any question similar to this, and I am wondering how to prove this... My work so far seems very far from what needs to be shown.  I thought maybe this would work:  $ab^{n}a^{-1} = (a^{-1}a)^{n}ab^{n}a^{-1}(a^{-1}a)^{n}$  And this would somehow simplify? Can anyone show me the way?
59317
I was asked to show that $\forall n\in \mathbb N$ there exist a $p\in \mathbb N^\ast$ such that $$(1+\sqrt{2})^n = \sqrt{p} + \sqrt{p-1}$$   I used induction but it wasn't fruitful, so I tried to use the binomial expansion of $(1+\sqrt{2})^n$ but it seems I lack some insight to go further.   Any hint is welcomed.
72619
Suppose $A$ is invertible $n\times n$ matrix. Show that $A^{-1}$ can be written as a polynomial of degree at most $n-1$, i.e.  $$A^{-1} = c_{n-1}A^{n-1} + \dots + \ c_{1}A + c_{0}I$$  Are there any tips on how may this proof be approached?
45673
What's wrong with this demonstration?:  $$A \iff 1 = 1^1$$ $$A \implies 1 = 1^\frac{2}{2}$$ $$A \implies 1 = (1^2)^\frac{1}{2}$$ $$A \implies 1 = ((-1)^2)^\frac{1}{2}$$ $$A \implies 1 = (-1)^\frac{2}{2}$$ $$A \implies 1 = (-1)^1 = -1$$  So finaly we get  $$1 = -1$$
93595
I have this exercise which I found very difficult:     $(A,+,.)$ is a finite pseudo-ring (a structure satisfying all the axioms of a ring except for the existence of a multiplicative identity), and $(a,b)\in A^2$ such that $ab^2=b$, prove that $bab=b$.   I used every possible combination of powers of $a$ and $b$ but didn't find anything, the idea I was using is the construction of a sequence $a^nb^m$ for example and there for there exist $n,m,n',m'$ such that $a^nb^m=a^{n'}b^{m'}$ and I use the given equality to reduce it, but it' s very difficult.  Can someone help me find the right sequence or the right method to solve this , thanks you
6993
I didn't find any good explanation how to perform multiplication on permutation group written in cyclic notation. For example, if $$   a=(1\,3\,5\,2),\quad b=(2\,5\,6),\quad c=(1\,6\,3\,4), $$  then why does $ab=(1\,3\,5\,6)$ and $ac=(1\,6\,5\,2)(3\,4)$?
114303
From Rosen's Discrete Mathematics and Its Applications, 3ed, chapter 8.1 p. 506-507:    ...there are Ck ways to insert parentheses in the product x0 · x1 · · · · · xk  I would like to ask where this conclusion comes from.  I am at a loss here.  It feels that this proof does not give me the whole story.  Thank you.
118281
[Update: Based on the hints provided by @zcn and @whacka, I believe I have found a solution. See my answer below.]  Below, $R$ is a commutative ring with $1$.  In John J. Watkins' Topics in Commutative Ring Theory, the author observes that if $a$ and $b$ are associates in $R$, then they generate the same ideal: $(a) = (b)$. This is clearly true, because $a = ub$ and $b = va$ for some units $u$ and $v$, and so $a \in (b)$ and $b \in (a)$, whence $(a) \subset (b)$ and $(b) \subset (a)$. Indeed, we didn't even require $u$ and $v$ to be units to reach this conclusion.  The author then claims that if $R$ is not a domain, then the converse may not be true: we may have $(a) = (b)$ even if $a$ and $b$ are not associates. As a "counterexample" he gives $R = \mathbb{Z}/(6)$, with $a = 2 + (6)$ and $b = 4 + (6)$. Clearly these elements generate the same ideal: $(a) = (b) = \{0, 2 + (6), 4 + (6)\}$. The author asserts that $a$ and $b$ are not associates, but this is false: $5 + (6)$ is a unit, and $(2 + (6))(5 + (6)) = 4 + (6)$, so $a$ and $b$ actually are associates.  I figured that the author botched the example but that surely the basic fact must be true: if $R = \mathbb{Z}/(n)$ then there may be nonassociates $a,b \in R$ with $(a) = (b)$. However, I was not able to find a counterexample and eventually resorted to writing a computer program and found that there are no counterexamples in $\mathbb{Z}/(n)$ for $n \leq 100$.  So now I'm starting to think that there is no counterexample in $\mathbb{Z}/(n)$ at all, but I have not yet been able to prove this.  My questions:   If I am correct that $(a) = (b)$ in $\mathbb{Z}/(n)$ implies that $a$ and $b$ are associates, I would appreciate a nudge in the right direction toward a proof. (But if the proof is elementary, please give only a hint, not a solution :-) I'll post my solution once I have one. In either case, could you please point me to a genuine counterexample? If there is one in $\mathbb{Z}/(n)$ for some $n$, that would be ideal (no pun intended), otherwise a counterexample in any commutative ring with $1$ would be great.
78948
I have shown that if two $n\times n$ matrices $A$ and $B$ are invertible, then $AB$ is also invertible. I am having problems though with showing that $(AB)^{-1}=B^{-1} A^{-1}$. I thought of using their left-multiplication transformations but it seems that this is only associative and not commutative so I can reverse $L_A L_B$
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
5557
The mapping is supposed to be from $\mathbb{N}$ to $\mathbb{N}$.  I'm still trying to understand if this is possible, I mean if it was from $\mathbb{R}$ to $\mathbb{N}$, I guess $x^2$ would work.
4128
On Dummit's Abstract Algebra on p. 128, it says:     "It is easy to see that $(1 2)(3 4)$ ... does not commute with any non-identity element of odd order in $A_5$."   But I don't find it easy.  Any help, please?
123189
I need to prove that if $A$, $B$ and $(A + B)$ are invertible then $(A^{-1} + B^{-1})^{-1} = A(A + B)^{-1}B$  I'm a bit lost with this one, I can't find a way to make any assumptions about $(A^{-1} + B^{-1})$, Neither by using $A^{-1}$, $B^{-1}$ and $(A + B)^{-1}$.  If someone could clue me in I'll be grateful.    UPDATE:  Thanks for all of your input - it really helped, I got a solution, I'd like to know if my way of proving it is valid:  $(A^{-1} + B^{-1})^{-1} = A(A + B)^{-1}B$      multiplying both sides by $A^{-1}$ and $B^{-1}$    $A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1} = (A+B)^{-1}$    now multiplying both sides by $(A + B)$, I get to  $(A+B)(A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1}) = I$  Lets call $(A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1})$ -> $C$    Now I determine that $C$ is $(A+B)^{-1}$, because $(A+B)C$ equals the Identity Matrix.    So lastly to verify that I place $C$ in the original equation:  $(A^{-1} + B^{-1})^{-1} = A(A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1})B$  and from this I get:  $(A^{-1} + B^{-1})^{-1} = (A^{-1} + B^{-1})^{-1}$
124091
No. of ways to choose $n$ things from $n$ alike "a" things $n$ alike "b" things and $n$ different things.  Answer is $(n+2)2^{n-1}$. But how to prove it..?
13698
I am given that $R$ is a commutative ring, $A$ is an ideal of $R$, and $N(A)=\{x\in R\,|\,x^n\in A$ for some $n\}$.  I am studying with a group for our comprehensive exam and this problem has us stuck for two reasons.  FIRST - We decided to assume $n\in\mathbb{Z}^+$ even though this restriction was not given. We decided $n\ne 0$ because then $x^0=1$ and we are not guaranteed unity. We also decided $n\notin\mathbb{Z}^-$ because $x^{-1}$ has no meaning if there are no multiplicative inverses. Is this a valid argument?  SECOND - We want to assume $x,\,y\in N(A)$ which means $x^m,\,y^n\in A$ and use the binomial theorem to expand $(x-y)^n$ which we have already proved is valid in a commutative ring and show that each term is in A so $x-y$ is in $N(A)$. The biggest problem is how to approach the $-y$ if we are not guaranteed unity. Does anyone have any suggestions?  Thank you in advance for any insight.
1896
This seems to be a common result. I've been trying to follow the bijective proof of it, which can be found easily online, but the explanations go over my head. It would be wonderful if you could give me an understandable explanation of the proof and let me know how I'd go about finding such a bijection.
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.