like it sounded as though if and only if coupled two statements and no other statements were related to them idk
hence why I said I'm just overanalyzing
Like $(A \iff B) \implies \nexists C$ s.t. $(C \implies A) \lor (C \implies B)$ and $\nexists C$ s.t. $A \implies C$ nor $B \implies C$ or something lol
what's the significance of a normal subgroup? The center of a ring/group is the set of elements that commute with the rest of the structure but I can't grasp the significance of equal left/right cosets
let $G$ be a group, $N$ be a normal subgroup, and $G/N$ be the quotient group. Is the significance that the cosets are reflexive&symmetric
nvm
this explanation was helpful
"If $N$ is a normal subgroup of a group $G$ and if $x^2 \in N$ for every $x \in G$, prove that every nonidentity element of the quotient group $G/N$ has order 2."
this problem doesn't explicitly mention if $|G|,|N|$ are finite or not
Let $N$ be a normal subgroup of a group $G$ and let $x^2 \in N$ for every $x \in G$.
I can immediately say that the quotient group $G/N$ exists right
and that the element are of the form $Nx$ for $x \in G$
I was going to argue the order of $N$ must divide $G$ and the order of the elements must divide the order of the group/subgroup but idk if that argument works for possibly infinite $|G|$
i guess also lurking in the bacgkround here is the general issue of when two cosets represented as Nx and Ny (for potentially different x, y in G) are equal to one another
for additive group $(\mathbb{Z}\times\mathbb{Z})/\langle(5,5)\rangle$ an element $(1,1)$ for example, has order $5$ right since $(1,1)+(1,1)+(1,1)+(1,1)+(1,1) = (0,0)$?
trying to find an element of infinite order
it's cool because $(a,b)$ have to be out of sync
like for no $k \in \mathbb{Z}$ do we have $a^k \equiv b^k \pmod 5$
obliv: someone nitpicky might not like your notation in that first equality (which is written an equality of ordered pairs, when you mean something like the elements of the quotient group represented by those ordered pairs) but yes
i don't have any opinion about what you should be using (there are any number of OK choices), most such abuses of notation are perfectly fine in contexts other than this exact one (i.e. exercises directed at understanding quotient groups). it is definitely important to be aware of when one is abusing notation. i don't know that it is as important to always use unambiguous, non-abused forms of notation
i don't think it's a coincidence that basically the only time you see \oplus and other funny operation symbols for the binary operation on a group is in those sections in a group theory book
hmm so in finding order of an element of a quotient group $G/N$ we have $a \in G$ has order $k$ if $Na^k = N$ so for the example above we need $(5,5)(a,b)^k=(5,5)$
"(1,0)^5" (maybe also a bad idea to mix additive and multiplicative notation?) is (5,0). this is not in the subgroup generated by (5,5) because [why? why would it be?]
reading over the above, you seem to observe (by implication) that if (x,y) is in the subgroup generated by (5,5), then x and y are congruent mod 5. this is certainly true, but the converse is not. this is not an if-and-only-if test for membership in the subgroup generated by (5,5)
it would be wise to care slightly more about how you are using notation, at least for now. try to make clear in notation whether a calculation refers to elements of the "non-quotiented" object ZxZ, or to elements of the quotient object (if a single calculation refers to both, make sure that it's clear which is which). it doesn't matter which one you pick, but it's important to pick one, and bad practice to do a half-and-half mix that leaves a reader to decide where it might be taking place
Thank you. I've spent four days in figuring out this problem: mathb.in/78273 (I have also proven the upper bound, too), and I tend to disbelieve even the basic laws of functions. Inequalities are evil sometimes.
On the other hand, I've learned a ton about AM-GM-HM inequality.
So yeah, a book of David A. Santos is a good exercise source
I'm like, 'hey, I've seen some ideas that made me doubt I understand the basics properly'. Of course I do accept them, I'm just asking myself 'maybe THAT is the exception'
Show that $(x_1+1)(x_2+1)...(x_n+1)=1+x_1 +x_2+...x_n+$ "other terms".
How can the following be expanded ?
$$(x_1+1)(x_2+1)...(x_n+1)$$
I have a problem where I have to expand and show that I'll have a form of $1+x_1 +x_2+...x_n+$all the other terms. I don't have a math background so I'm...
Let there be a partition $P=\{x_1,\ldots,x_n\}$ of the domain $[a,b]$ of a Riemann integrable function $f$. Let $m_i(f)=\inf\limits_{I_i}f$ and $M_i(f)=\sup\limits_{I_i}f$, where $I_i=[x_{i-1},x_i]$. Define the oscillation of $f$ at $x$ to be $$\mathrm{osc}(f;x) = \lim_{h \to 0^+} \sup \{ |f(x')-f(x'')| : x', x'' \in (x-h,x+h) \cap [a,b] \}.$$
I'm reading proof of the Lebesgue criterion and I'm stuck on a basic claim. We want to prove that if $f$ is bounded on $[a,b]$ and Riemann integrable, then the set of discontinuities has measure zero. If a function is discontinuous at $x$, then $\mathrm{osc}(f;x)>0$ and so we construct the sets $$A_k=\left\{x\in[a,b]:\mathrm{osc}(f;x)\geq\frac{1}{k}\right\}.\tag1$$
It is then claimed that for a given $i$, if $x\in A_k$ and $x\in (x_{i-1},x_i)$, then $$(M_i(f)-m_i(f))\Delta x_i\geq\frac{1}{k}\Delta x_i \quad\text{by } (1).$$
I don't understand this inequality. The oscillation is not $M_i(f)-m_i(f)$. How does it follow?
@JohnZimmerman Just handing the answer to someone doesn't really help them to learn the material that they are trying to learn. "Teach a man to fish..."
I'm honestly trying to do these on my own and have done many of them alone. The ones I've been asking about are the ones I've struggled to figure out alone.
this boils down to identities that any real numbers satisfy like distributivity: $a(c+d) = ac+ad$
So now if you have a product of sums of terms $(a_1+...+a_n)(b_1+...+b_m)$ then you should try to think on how to multiply such an expression out and what would it lead you to
I could just tell you what $(a_1+...+a_n)(b_1+...+b_m)$ is and that would probably be enough but that wouldn't give an explanation on how I arrived at the result
right, and maybe this is not a completely clear, but from those distribution laws above we also have more general distribution laws $$a(b_1+b_2+...+b_m) = ab_1+...+ab_m,\ (a_1+a_2+...+a_n)b = a_1b+...+a_nb$$
think of above distrubtional law applied over and over again to arrive at above formulas
@Jakobian Actually he was struggling on whether -(a+b)=-a-b or a+b yesterday. So I assume it would be hard for him to answer what you wrote. I hope I didn't made you angry by directing him to a YouTube video
Sorry, the biggest thing is that while not a short one I am kind of on a time limit for studying. A week from tomorrow is the final, and the study guide is due a week from yesterday. And this may very well be the last time I tackle algebra, at least for a grade. If I go into it again I'm planning on doing it as a personal challenge.
okay but whats the most important in a sum? Its not the order in which we sum but what we sum and here we are summing all the terms $a_ib_j$ for $i = 1, ..., n$ and $j = 1, ..., m$
so the conclusion is that $(a_1+...+a_n)(b_1+...+b_m)$ is a sum over all those products $a_ib_j$
and this is actually what we mean with multiplication term by term
remember you don't have to introduce any $a_i$ or $b_j$, just remember to multiply everything in the first batch with everything in the second batch and add all of the terms together
@Jakobian Will do, and I'll get to answering that problem once I return from dogsitting for our neighbore. And lunch. I should be back close to 12, brb. And seriously man, the help is appreciated.
If a real analytic function $f$ is involutive i.e. $f(f(x))=x$ and its Mellin transform can be taken, does this imply that $\tilde f$ inherits a functional equation as well?
I'm reminded by Poisson summation involving the Jacbobi theta function and its functional equation corresponding to a new f...
@Obliv thanks a lot for the help. The answer is from the time when my teacher had solved it in class. Maybe both the answers are different representation of the same thing
@Obliv they are equivalent but as I am not great at handling inverse trigo ratios so it took me some time to derive it. And yeah, I'm not fond of my handwriting either
Show that $(x_1+1)(x_2+1)...(x_n+1)=1+x_1 +x_2+...x_n+$ "other terms".
How can the following be expanded ?
$$(x_1+1)(x_2+1)...(x_n+1)$$
I have a problem where I have to expand and show that I'll have a form of $1+x_1 +x_2+...x_n+$all the other terms. I don't have a math background so I'm...
