@PeterTamaroff Not so much - a bit of work here, a bit of work there. I've just been asked to teach a number theory course this summer, and I have to start thinking about that
I hated math prior - I was not interested in calculus and the like. I was under the (too common) impression that rote calculation was all that math was
@PeterTamaroff By this, I mean calculation without any thought. Many calculus and earlier classes are taught without any emphasis on understanding what you're doing, but instead the ability to perform certain calculations
this reminds me of another time, when I was trying to donate some math books to a public library in my hometown, and they rejected them because they didn't think anyone would use them
@mixedmath Well, from abstract algebra I know about group theory and ring theory.
Say groups, congruences, normal subgroups, the isomorphism theorems, sylow, cycles, cycle decomposition, sylow and other stuff. Basically, BA I's chapter on Group Theory, and I have just started out with Rings.
@PeterTamaroff That sounds pretty reasonable. I don't remember that book as much. I suppose if you like it, then there's not so much reason to leave it
I learned elementary number theory from a book by Kenneth Rosen (not Mike Rosen), and I loved it - but I felt a bit cheated when I revisited it after I learned some algebra
we spent all this time developing material to state Euler's theorem and what not, which is trivial after you know algebra
In fact, one of the things that I'm looking forward to about this summer number theory course is revisiting some of these old-fashioned elementary arguments
@anon That's funny. But to be fair, I understand why robjohn left a comment too. They've changed it so that the last mod comment on a deleted answer stay in the inbox even if the answer is deleted. So robjohn wrote that and then immediately deleted the answer, so that the answerer could read it whenever
it's nonetheless surprising that so many people left that comment
Wow I just found out that Albert Einstein said of Dirac "This balancing on the dizzying path between genius and madness is awful" referring to his autistic traits.
An anecdote recounted in a review of the 2009 biography tells of Werner Heisenberg and Dirac sailing on a cruise ship to a conference in Japan in August 1929. "Both still in their twenties, and unmarried, they made an odd couple. Heisenberg was a ladies' man who constantly flirted and danced,
while Dirac—'an Edwardian geek', as biographer Graham Farmelo puts it—suffered agonies if forced into any kind of socialising or small talk. 'Why do you dance?' Dirac asked his companion. 'When there are nice girls, it is a pleasure,' Heisenberg replied. Dirac pondered this notion, then blurted out: 'But, Heisenberg, how do you know beforehand that the girls are nice?'"
I love/hate it when I think days about something, I struggle and struggle and then it turns out the proof I come up with is a very simple proof and people are like "well yeah" :P
@mixedmath About why the following algorithm for finding the quadratic residues modulo a prime $p$ always works out, and prove there are $(p-1)/2$ non residues and $(p-1)/2$ residues.
@DanZimm Gotcha. After you prove that $A\setminus B$ is uncountable, it would be fun to prove that the union of countably many countable sets is countable.
@WilliamStagner so the first one I'm struggling on making rigorous, but given that each $A_n$ is countable then it's clear that you can form a sequence of sequences $a_{n_k}$ (the nth sequence) then you can form a sort of grid of the sequences iterating through each sequence's nth term similar to the rationals thing
Well, and $x^2\equiv n\mod p$ has at most two solutions in $1\leq x\leq p$, and by $(p-x)^2\equiv x^2$ it has at most one in $1\leq x\leq \frac{p-1}2$, yes?
Thus if it is solvable it has exactly one solution in $1\leq x\leq \frac{p-1}2$
@WilliamStagner so I would probably say... assume $A \backslash B$ is countable then it's clear that $A \backslash B \cup B$ should be countable as well (form a grid and go along diagonals... similar to the rationals proof)
but that contradicts that $A$ is uncountable so $A \backslash B$ must be uncountable as well
I can use that $\Bbb Z_p$ is a field, yes? If $x^2=y^2$ then $x-y=0$ or $x+y=0$. Since $x,y\leq \frac{p-1}2$ and $x\neq y$, we must have $x=p-y$ so $y\geq \frac{p+1}2$ and it is one of the others.
but this is more complicated than it has to be. If you know that $x^2 \equiv n \mod p$ has either 2 or zero incongruent solutions, then you're essentially there
@mixedmath Suppose it is solvable. By Lagrange, it has at most two solutions in $1\dots p$, and since $(p-x)^2=x^2$ it will certainly have the two. Thus it has one in $1\dots (p-1)/2$ and one in $(p+1)/2$; I see now, I guess.
@DanZimm If you're still interested in further cardinality arguments, the most important is probably Cantor's Theorem, i.e. $|P(A)| > |A|$ for any set $A$ where $P(A)$ is the power set of $A$.
Cardinality is just a rigorous way of studying the size of sets, particularly infinite ones.
@WilliamStagner so thats not how i was going to go about it at first, but isnt what i said exactly that there isnt an injection from $P(A)$ to $A$?
(just making sure I have my definitions right)
on a side note TIL that any riemann integral can be expressed as $\int_{\mathbb{R}} f(x) \cdot \chi_{(a,b)}(x) \, \mathrm{d} x$ where $\chi_{(a,b)}$ is the characterisic function of $(a,b)$ in $\mathbb{R}$ so then we simply learn how to integrate over all of $\mathbb{R}$ and we can integrate any interval!
@JayeshBadwaik I understand. That's my dilemma: I think best on an empty stomach...I feel sluggish after meals (especially after delicious meals), so I nibble...here and there, when I remember to! ;-)
