Thomas Andrews

Nobody writes an analytic continuation of the Harmonic series as a $\sum$ notation. Well, they might do so as informal motivation, not as definition. I have no idea what you are saying here, or what part of my comment you think you are contradicting. @KamalSaleh

No, $\sum_{n=f(a)}^{f(b)} a_n=0$ whenever $f(a)>f(b).$ You might find isolated cases where others use it differently, but communication is about clarity, and people choosing to allow $\sum$ to go backwards are using the notation in a non-standard way. They should be clear that they are going to do so, to avoid confusion. @KamalSaleh

I have never seen that notation, rather than $$\sum_{-\infty}^{\infty}c_n(z-a)^n.$$ @DermotCraddock

Certainly when we write $S_N=\sum_{n=1}^N a_n,$ we don't say $S_0=a_1+a_0.$ We say $S_0=0.$ @Mason

That notation is a bit of abuse of notation - it officially mean $0.$ Unline integrals, we don't accept sums that go backwards. What you mean is: $$\sum_{n=0}^\infty 2^{3-n}$$

The first order logic equivalent would be $\forall G:Group(G)\implies P_1(G),$ then $\forall G:(Group(G)\land P_1(G))\implies P_2(G).$ So it is a huge shorthand, because we don't have to keep restating all the stuff we have concluded or assumed before.

I don't think if means much, formally. It is similar to starting a sentence "if", but applying to a lot of sentences. "If $G$ is a group, ..." but the assumption applies across sentences is, and $G$ is the same group for the rest of the text or section of text or proof. That is, we,are giving a name for one supposed entity.

Ah, when you said "I removed," I thought it meant "accidentally." That happens a lot here - people forget conditions all the time.

Then edit the question. Make it not $p_{k_i}$ but $p_{k+i},$ and mention consecutive. @AdamRubinson That said, there is still always an infinite numberr of answers when $n=1,$ by Dirichlet.

But Dirichlet says you can find infinitely many examples for any $n,$ including $n=1.$

You can find $n-1$ primes $p\equiv 1\pmod a$ and one prime $q\equiv 2\pmod a.$

Doesn't this follow from Dirichlet on primes in arithmetic seauences?

I think we could say that the open sets in your ring of polynomials , $R,$ have a basis around $0$ of the form $Ux^n+x^{n+1}R,$ where $U$ is an open set of the $p$-adic number (either the $p$-adic integers or the rationals.) This is some join of the $p$-adic topology and the general $x$-adic topology on polynomial rings (which treats $x$ as an infinitesimal, roughly speaking, compared the elements of the ring.)

You don't define things after you've studied it. You need to try definitions then check if it is interesting. "But the behavior..." What behavior, before you have defined it? Why is this more interesting than adding $10^{-\infty}$ to the real numbers? You seeem to think it is interesting, because we 'd now have a "most significant digit," or something about infinite numbers, but you don't have either, not really.

In the end, the best algebraic stricture I can imagine is the subring or $p$-adic polynomials, $\f\in\mathbb Z_p[x],$ with the highest $x^{\deg f}$ having an coefficient. in $\mathbb Z.$ Here, $x$ would be a stand-in for $p^\infty,$ and this multiplication will be commutative and associative. Not sure how the topology would interact with this algebra, but there really isn't anything about $p^\infty$ that distinguishes it from any other indeterminate, algebraically.

Fundamentally, $p$-adic numbers are interesting because they have both an interesting topology and an interesting algebra, and because they help solve interesting problems. We lose the interesting algebra if you try to add digits at $\infty,$ even if you don't require a "last" non-zero digit. The same is true for real numbers of you add $10^{-\infty}.$ You might be able to give h a set a topology, though. It won't have an associated metric, but it is still a topology.

Thus a coefficient for $p^\infty$ could be the least significant digit, but almost never the most significant digit.

But the real problem with your formulation is still the term "most significant digit." First of all, in real numbers, it is technically the "most-significant digit in base $10.$" It doesn't have a lot of meaning, except in the base notation. In real numbers, you can write a number in any integer base $b>1$, but only bases $p^k$ can be used for $p$-adic numbers. You can't write $3$-adic numbers in base $10.$ Also, in $p$-adic numbers, $p$ is considered small. So it is like $10^{-1}$ in base $10.$

There is a problem with ordinals - they are not commutative under addition. So if $p^i\cdot p^j=p^{i+j}$ for ordinals, $i,j,$ then multiplication won't be commutative. So you still haven't defined the algebraic concept of these numbers in a way that looks remotely useful or interesting.

$p^\infty,$ for example, would not have an additive inverse. At least not one that also has a last non-zero $p$-adic digit.

Just ask yourself the same question for real. What would it mean to have a digit $d_\infty$ corresponding to $10^{-\infty}$ in the reals?

You say you know, but your whole concept of $p$-adics shows you don't.

For a $p$-adic integer, the digits $d_i$ are enumerated by $i\in\mathbb N.$

There are no digits after infinite digits. That is not how $p$-adic numbers work.

The $p$-adic integers that have a finite number of digits are just the integers, $\mathbb Z$ and the $p$-adic rationals with a finite number of digits are $\mathbb Z[1/p],$ the ring of rational numbers of the form $n/p^k$ for $n,k$ integers.

They are not infinite numbers. $\pi$ is not an infinite realnnumber. It is a misconception to thiink of $p$-adic numbers as infinite. $p$-adics only look infinite to you because you are so used to the real numbers. It does no good to think of them as infinite. Forr example, there are $p$-adic numbers with an infinite number of digits which, when multiplied together, yield $1.$

Left-most is a rather meaningless term. It depends on how you write $p$-adics, and we write $p$-adics in reverse, so there is always a left-most non-zero digit. The problem is our concept of "most significant digit."

The "most significant digit" of the $p$-adic number $\sum a_ip^i$ is $a_n$ when $n$ is the smallest value where $a_n\neq 0.$ So there is a most significant digit, it just feels backwards compared to hoow we thing of most significant digits for real numbers.

What do you mean "we don't know what the most significant digit is?" A field doesn't have digits, so presumably you mean something about the digits of the elements. For non-zero elements, there is always a most significant digit. Whether we can tell what it is seems hard to say, but, for integers, the most significant digit different from "the most significant digit" when the same number is represented as a real number in base $p.$

@BenGrossmann I couldn't say that has much to do with chess, since it can also be said to be a mutilated checkerboard problem. But there are a lot of problems related to chess like rook and queen placements, knight's tours, etc, being all specific instances of graph theory questions.

Of course, the reason we use Pythagoras is precisely because our intuition is that the angles between the axes are all right angles. So, from that intuitive picture in our minds, we get the leng of vectors, and from Euclidean geometry we get the dot product, and from there we get a generalization to any inner product. A different inner product on $\mathbb R^n$ will give the space a different (but still Euclidean) geometry.

How are vectors dilations of the vector space? @Fnacool

The reason for the dot product is that we want intuitively that the length of the vector is defined by Pythagoras, so we want $|a|^2=\sum a_i^2.$ The law of cosines says that $$|a-b|^2=|a|^2+|b|^2 -2|a|\cdot |b|\cos\theta$$ where $\theta$ is the angle between $a,b.$ 'Tis is thus equivalent to $|a|\cdot |b|\cos(\theta)=a\cdot b.$ And then we realize that $a\cdot a=|a|^2,$ so the dot product makes much more sense as a kind of important geometric operation.

Addition and multiplication often come in pairs, but they aren't really dual. The big example is the distributive law. We have $(a+b)\times c=(a\times c)+(b\times c),$ but we definitely don't have $a+(b\times c)=(a+b)\times(a+c).$ In other math, multiplication is often not commutative, while addition almost always is.

Given two continuous functions on $f,g:[0,1]\to\mathbb R,$ we have an inner product of the form $$f\cdot g=\int_0^1 f(x)g(x)\,dx$$ which is very much like the dot product, and it turns out studying this dot product leads to the concept of Fourier series - the sine and cosine functions are like the vectors $(1,0,0,\dots),(0,1,0,0,\dots),\dots$ Inner products are everywhere. Your function is an example of thinking by analogy, which can sometimes yield useful ideas, but often not. Addition and multiplication, while seeming related, are not "dual" operations.

The key is that the dot product is actually something more general on realvector spaces, called an inner product. It gives a geometric structure to a vector space - a sense of length to vector spaces and a sense of what an angle is in a vector space. There are lots of inner products that are not of this form, at least intuitively, yet we get that notion of length and angle from an inner product.

You might find individual discussions of your function here and there, but, if they give it a name, it will be a one-off, or maybe the name will be picked up here and there, but not have reached critical mass.

What does $\log(a+b)$ even mean? @ydd We don't apply functions to rectors term by term, usually, and $\log$ is not a very well-defined function on negative numbers.

Right, but you don't have years of mathematical knowledge. "Interesting" here means "Of frequent usage by mathematicians as something with elucidates something useful." What you are interested in is not important for the discussion - we give names so we can discuss in a group of people, and it is not interesting to the vast majority of mathematicians because it has neither use nor geometric meaning.

No, it is not, because the product of two vectors $(a_ib_i)_i$ is not an interesting binary operation in itself - we do not use it as a fundamental operation on vector spaces.

There really isn't much use for the operator $$f(\mathbf v)=\prod_i \mathbf v_i.$$ It doesn't have any interesting geometric meaning, and, vitally, it is not invariant under rotations, like the dot product.

That would just be an operation on a single vector, really, since we already have a notion of $\mathbf a+\mathbf b.$

I guess you might take the remainder of the polynomial $x^n+1$ when divided by the polynomial $x^m+1.$ You can certainly show that if there is a counter-example, then there is a counter-example with $n<2m.$

There really isn't a way unless you know what the order of $a$ mod $d$ mean. This is a theorem about multiplicative orders.

Indeed, your theorem isn't true if $a=0,1.$

We don't even have a notion of polynomials with rational coefficients, so no. As I've pointed out, $2+2\mid 2^n+2^n$ for even and odd $n>0,$ so the fact that $x+y$ is not a polynomial factor of $x^2+y^2$ doesn't say anything about particular integer values of $x,y.$

If you know that polynomial $p(x,y)$ is not a factor of $q(x,y)$ it is still possible for specific integers $a,b$ for $p(a,b)\mid q(a,b).$ It is easier to see with one variable: $x^3+2x-2=(x^2+1)x+(x-2),$ so $x^2+1$ is not a factor of $x^3+2x+2,$ but when $x=2,$ $$5=2^2+1\mid 2^3+2\cdot 2-2=10$$

Your lemma proved no such thing. Your lemma showed $x+y\mid x^{2n+1}+y^{2n+1},$ it did not show it can't divide any other $x^q+y^q.$

No, the lemma proves that if $n$ is an odd integer, then $x+y$ is always a factor of $x^n+y^n.$ It doesn't prove the reverse, that if $x+y$ is a factor of $x^n+y^n$ for some $x,y$ integers, then $n$ is an odd integer, even if you know $n$ is an integer. You certainly can't say it when $n$ is not an integer.

What you've shown is a polynomial division, $x+y\mid x^n+y^n$ for $n$ an odd natural number. But, even if you proved the polynomial wasn't a factor when $n$ even, that wouldn't mean that particular integers $x=a,y=b$ can't make $a+b\mid a^n+b^n.$