morrowmh

Wait! Why do I have to use strong induction to prove the inequality holds?

Showing $m^2<p(pq2+2pq+p)$ is not the same as showing $m^2<p$. Easier is to use the fundamental theorem of arithmetic to show that if $p$ divides $m^2$ then $p$ divides $m$.

Okay, then maybe you can use the fundamental theorem of arithmetic instead, see math.stackexchange.com/questions/3061922/…

Why don't you want to use Euclid's lemma? It's a pretty elementary lemma with an elementary proof.

Well, you can also view integration as area under a curve. So the integral $\int_0^1 xdx$ represents the area of the triangle formed by the line $y=x$, the $x$-axis, and the line $x=1$. If you graph this, visually the area should be $\frac{1}{2}$. So, when calculating the integral, we get $\int_0^1 xdx=\left[\frac{x^2}{2}\right]_0^1=\frac{1^2}{2}-\frac{0^2}{2}=\frac{1}{2}$. So you see that if the factor of $2$ was not in the denominator, the area would not be correct. So here, the factor of $2$ comes because the area of the triangle is half the area of the square.

Since the derivative of $\frac{x^{n+1}}{n+1}$ is $x^n$, the integral of $x^n$ is $\frac{x^{n+1}}{n+1}$. You can think of the integral as an "antiderivative". This is why the factor of $n+1$ shows up in the denominator. It doesn't just "come out of the blue".

Do you know what an integral is?

Ok!

you may also already know this, but here is the relevant theorem which says the compact subsets need to be bounded: en.wikipedia.org/wiki/Heine%E2%80%93Borel_theorem

then since the open interval is not compact, the image need not be compact

Good question. I think you need to modify your definition to be $f:(0,1)\to\mathbb{R}^2$ such that $f(0,1)=\mathbb{R}^2$

I'll get the relevant references, one moment

Well then I think compactness is still an issue. If $\mathbb{R}^2\subseteq K([0,1])$ then $K([0,1])$ is not bounded, so it can't be compact.

Because the continuous image of a compact set must be compact. So, since $[0,1]$ is compact, whatever $K([0,1])$ is, it needs to be compact. Which, since $\mathbb{R}^3$ is a metric space, means it needs to be bounded, hence cannot fill space.

Ah ok. I think I misunderstood. So your knot doesn't have to fill $\mathbb{R}^3$, just $\mathbb{R}^2$?

I may be wrong, but I don't see how a continuous map $K:[0,1]\to\mathbb{R}^3$ could satisfy $K([0,1])=\mathbb{R}^3$. The interval $[0,1]$ is compact but $\mathbb{R}^3$ is not.

You can integrate $\int_{-\infty}^\infty e^{-x^2}dx$ using a nice trick of integrating the square of the integral with polar coordinates. See en.wikipedia.org/wiki/Gaussian_integral

Also check out "flammable maths" youtube channel, he has a lot of videos featuring weird and wacky integration techniques accessible to strong high school students or first year mathematics degree students.

One of the most prominent things in math is the idea of combining two things to make a new thing with similar properties. This gives a notion of a group. However, when this "combination" operation is commutative, i.e. $a$ combined with $b$ is the same as $b$ combined with $a$, we can talk about an $abelian$ group. This idea shows up in almost every field of math.

Have a nice day!

Ah ok, that makes sense now. And no worries, I appreciate you explaining it. Maybe you could add that last part in to your answer about the p^7 so others can see?

Ok, I am following

And the p^7 appears because Z_p^{9} has p^7 elements of order dividing p^8?

I understand p^{15}(p^8-p^7), but I'm not quite sure about the extra p^8. Namely, how is it instead p^{15}p^8(p^8-p^7)?

Okay, so my last question is, on each term, where does the factor of $p^8$ (on the first term) and $p^7$ (on the second term) come from?

Hi, so are we getting the sum because we can pick a generator from $\mathbb{Z}_9$ or $\mathbb{Z}_11$?

Okay, I see the factor of $p^{15}$ now, so where do we get the sum from?

I mean, what specifically do you mean by "taking products"? I understand the part about the totient function, but where are you coming up with the expression $p^{15}p^8(p^8-p^7)+p^{15}(p^8-p^7)p^7$? Does this have to do with generators for $\mathbb{Z}_{p^3}$ etc.?

How are you getting the $2p^{15}p^8$ part?