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Apr
26
comment Up to isomorphism
@yons: Yes. Basically it says the labels are effectively just that, names for the group elements; to use a non-group example, up to relabelling isomorphism, the sets $\{1,2,3,\ldots\}$ $\{\text{one}, \text{two}, \text{three}, \ldots\}$ and $\{|,||,|||,\ldots\}$ are the same if the arithmetic structure is defined appropriately.
Apr
26
comment $b^{\frac{m}{n}}=(b^{\frac{1}{n}})^m=(b^m)^{\frac{1}{n}}$ except $b$ is not negative when $n$ is Even.
OK, I've posted it as answer (with a slight extension).
Apr
26
answered $b^{\frac{m}{n}}=(b^{\frac{1}{n}})^m=(b^m)^{\frac{1}{n}}$ except $b$ is not negative when $n$ is Even.
Apr
26
comment $b^{\frac{m}{n}}=(b^{\frac{1}{n}})^m=(b^m)^{\frac{1}{n}}$ except $b$ is not negative when $n$ is Even.
The quoted sentence says "if the conditions are fulfilled, then the claim holds." That's known as a sufficient condition. It does not say "if the conditions are not fulfilled, then the claim doesn't hold". That would be a necessary condition. Since only the sufficient condition was claimed, the only way to disprove the claim would be to find a case where the conditions are fulfilled but the formula doesn't hold. None of your examples qualify, since they all use $b=-1\notin\{0,1,2,3,\ldots\}$.
Apr
26
comment $b^{\frac{m}{n}}=(b^{\frac{1}{n}})^m=(b^m)^{\frac{1}{n}}$ except $b$ is not negative when $n$ is Even.
I can't find the word "only" in the quoted claim, so it's only given as sufficient condition, not as necessary condition. Thus you haven't disproved the condition in the book; what you have disproved is the (not claimed) assumption that it is a necessary condition.
Apr
25
answered Up to isomorphism
Apr
25
answered $f:[a,b] \to R$ is continuous and $\int_a^b{f(x)g(x)dx}=0$ for every continuous function $g:[a,b]\to R$
Apr
25
reviewed Approve $\sum_{k = 1}^{n} \frac{1}{k^{2}} < 2 - \frac{1}{n}$
Apr
25
comment Proving that $(u+v)×w=u×w+v×w$
What do you know about $\vec a\times\vec b$? I'm sure it's more than "there is an operation $\times$ that takes two vectors and gives a vector" (which wouldn't be sufficient to prove that equation anyway). On the other hand, that equation seems not to be among what you were told about it (or otherwise asking to prove it would be pointless).
Apr
25
comment Convergence: infinite series
So what is your question? How to prove it?
Apr
25
revised Why are we defining the norms on certain vector spaces the way they are?
added 218 characters in body
Apr
25
answered Why are we defining the norms on certain vector spaces the way they are?
Apr
25
revised Prove a sequentially compact metric space is bounded.
fixed math formatting
Apr
25
awarded  Promoter
Apr
25
comment Prove $ (A \cup B) \cap C$ = $(A \cap C) \cup (B \cap C) $
That link gives a 404 not found.
Apr
24
answered Murder at Hilbert's Hotel!
Apr
23
awarded  Tumbleweed
Apr
22
comment Zero to the zero power - Is $0^0=1$?
@user21820: Thank you.
Apr
21
answered How to solve the derivative of $b^x$ using the defintion
Apr
21
comment How to solve the derivative of $b^x$ using the defintion
Thinking about it, for this task probably the best way to define it is as $\lim_{h\to 0}\frac{b^h-1}{h}$ :-)