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Re: MATH2601 Linear Algebra/Group Theory Questions $\noindent Let $a,b \in G$. We need to show that $ab = ba$. We know from the question's assumption that $x = x^{-1}$ for all $x\in G$. Thus$ $$\begin{align*} ab &= \left(ab\right)^{-1} \\ &= b^{-1}a^{-1} \\ &= ba.\end{align*}$ $\noindent...

Re: MATH2601 Linear Algebra/Group Theory Questions $\noindent What's the context? Assuming it's in a group say, note that by definition, $aa^{-1} = a^{-1}a = e$, where $e$ is the identity element. Hence $a$ is an inverse of $a^{-1}$ (ask yourself what it really means for something to be an...

Actually, it does always work here, and will also allow us to find an optimal permutation for any finite set (in fact, multiset) of real numbers (here the set was {1, 2, 3, 4, 5}). The interested student may wish to try proving this as an exercise.

We can also contstruct an optimal permutation by basically using a greedy algorithm or ideas from the rearrangement inequality.

Yep.

One shortcut you can employ for the brute force method is to note that only the cylic order of a to e matters for the value of the given expression (rather than the actual permutation). So it suffices to check 4! = 24 cases.

Re: Several Variable Calculus $\noindent That example is showing that a multivariable limit can exist even though the corresponding iterated limits may not exist. If we hold $x$ fixed and non-zero and limit $y$ to $0$, the limit won't exist because $f(x, y) = x \sin \frac1x \sin \frac1y +...

Re: Statistics Greater than or equal to. Events can have 0 probability.

If you know a bit of abstract algebra, you can read the top answer here about why (-x)(-y) = xy in any ring (though in that case you may have already thought about this and had it answered before): http://math.stackexchange.com/questions/878836/product-of-two-negative-numbers-is-positive .

No. Hint: it is northwest.

Correct, a negative divided by a negative is a positive. (Also a negative times a negative is a positive.)

The -125/(-8) is actually a positive quantity and can immediately be simplified to 125/8 (negative signs on top and bottom "cancel out"). So the original equation is equivalent to r-3 = 125/8.

$\noindent The first step isn't correct. The correct step is: $r^{-3} = \frac{1}{r^{3}}$ (shouldn't be $\frac{-1}{r^{-3}}$). So we have $\frac{1}{r^{3}} = \frac{-125}{-8}$. This becomes $\frac{1}{r^3} =\frac{125}{8}$ (in the fraction $\frac{-125}{-8}$, we can cancel negative signs: this fraction...

Re: Announcement from BOSTES - significant change to calculus courses They weren't referred to by the syllabus in that bullet point.

Re: Several Variable Calculus Here's some hints. $\noindent Assume the given equivalence, i.e. $\exists c_1,c_2 > 0:\\ c_1\rho (\textbf{x},\textbf{y}) \le \delta (\textbf{x},\textbf{y}) \le c_2 \rho(\textbf{x},\textbf{y})$ for all $\mathbf{x},\mathbf{y}$.$ $\noindent Consider an...

Re: Announcement from BOSTES - significant change to calculus courses I took a brief look just now and it says "with P(A) = 0 if A is an impossibility and P(A) = 1 if A is a certainty (ACMMM053)" (at least on Page 41 of this document...

Re: Announcement from BOSTES - significant change to calculus courses I think also a lot of high school teachers think an event having 0 probability is equivalent to that event being "impossible".

Hypothetically speaking, they could try and get a computer to help do these via Machine Learning algorithms etc. (so they don't need to spend personal time doing it by hand etc.) if they use scanned copies of essays. But I'm guessing they don't.

If technology is good enough / if they are willing to invest in it, they could use things like image recognition / machine learning to try and compare to past exam responses on a database or other resources, but I'm guessing they don't do this.

I think what it's saying is that since t is a function T of X, the probability Pr(X = x, t = y) is equal to Pr(X = x) if y = T(x) and 0 otherwise. This is because if y = T(x), then the event {X = x, t = y} (i.e. {X = x, t = T(x)}) is equivalent to the event {X = x} (because X equals x and t...