Can Veeky Forums solve this simple task?

the part where you pulled those parentheses out of your ass. That you, OP?

>That you, OP?

Lel but this is a component of the definition of a algebraic ring, so it must simply be true by definition.

However, let me see if I can elucidate the problem a little:
(∀x, y) (∃z) [x(y + z) = x(y + (1-y)) = x(y + 1 - y) = x(1) = x = x(y) + x(1-y) = xy + xz]

However, this only shows that that that multiplication is distributive over addition when the sum of the terms being added is equal to one. How can we generalize this result?

Give me a minute. I just got out of the hospital (I'm fine, but a little groggy). I might be able to come up with something.

it is a self evident true

x(y+z) = (y+z) + (y+z) + (y+z) + ... (x number of additions, by definition of multiplication)

x(y+z) = (y+y+y+...) + (z+z+z+...) (by commutative property of addition)
x(y+z) = x(y) + x(z) (by definition of multiplication)

It's also easy enough to prove that this is the case when y and z are additive inverses, that is, when (y + z = 0), or to say the same, when z = -y.

To simplify things, I'll eliminate any complex logical operations and/or symbolism, leaving it to be inferred by the reader. Thus we have:
x(y + z) = x[y + (-y)] = x(y-y) = x(0) = 0 = xy - xy = xy + (-xy) = xy + x(-y) = xy + xz

I should have slightly modified the proof in my original my post so as to quantify of all z. I leave this up to the reader to see. Having thus modified the first proof, we have established that:
(∀x, y, z) [(z= -y) v (z = 1 - y)] ⇒ [x(y+z) = xy + xz]

BTW in writing this up, I've been assuming we're dealing with generalized algebraic ring. However, I just realized that assuming we're working with integers, we could just be really autistic and prove the distributive property on the basis of the Peano Axioms and the rules of first order logic. I'll probably do this for you guys in a later post.

The reason, btw, that I'm making such long posts with explanations and shit is because like I said, I'm just back from the hospital and it helps with my own reasoning, but also because I think it will helps any other anons understand the underlying reasoning and intuition.

Lol you must be stupids shit. Thats not how math works, nothing is "self evidently true" except logic. Everything else must be proven. Even "1 + 1 = 2" isn't self evidently true. Well, perhaps it is with respect to human psychology and the sort of, so to speak, well behaved or ordinary algebraic and topological systems we deal with on an everyday basis. However, the fact that something seems to be extremely plausible and intuitive, doesn't make it true. Hence in math we must prove any not basic, logical rules. In fact, there are mathematical systems in which we have
x(y + z) ≠ xy + xz

In fact you post is indicative of a rather undeveloped level of mathematical understanding. I'm pretty much positive that you've never taken any courses higher than a Calculus I-III sequence, if even that.

tl;dr you sound like a pleb. Nobody honestly cares what does or doesn't seem "self-evident" to you. That's simply not a scientific concept, and as it turns out, you're not the center of the universe.

>Thats not how math works, nothing is "self evidently true" except logic

Axioms are not proven - they are given as true. What you choose as your axioms does not matter as long as you are consistent. I could take the Pythagorean theorem as an axiom and then derive several conventionally chosen axioms from there.

OP's problem is not even answerable in any meaningful way because he never specified what x,y,z are elements of.

sorry for rustling your jimmies

...

would be right for vectors?

The dot product of two vectors is the product of their magnitudes and the cos of the angle between the vectors. Therefore, that solution is only correct for vectors that have an angle of 0 between them.

oops wrong picture. that pic isn't the dot product.

Pardon all the typos, my keyboard is mildly fucked up.

Anyway, to continue where I left off in we finally need to prove that x(y + z) = xy + xz for (z ≠ -y)^(z ≠ 1-y)

Beyond this point I'm not sure if such a proof can be constructed purely algebraically. Once again, in the context of abstract algebra and ring theory, this would simply be taken as a component of the definition of a ring, and so it's not something one would prove for all systems, but rather something that is, by definition, satisfied (potentially in a different manner, so to speak) by any system that qualifies as a ring. So at this point I'll begin relying on the Peano axioms. We are thus dealing with the extension of the non-cyclic Abelian group on one generator (namely the group generated by 〈1〉) to an integral domain.

Anyway, we have:
(∀m, n ∈ℕ) m(n+1) = mn + m
(∀m∈ℕ) (m = 0) v (∃n∈ℕ n + 1 = m)
(∀m, n ∈ℕ) (m + n)∈ℕ

∴ (∀x, y, z ∈ℕ) [(∃m∈ℕ) (x(y + z) = xm) ^ (∃n∈ℕ m = n + 1)]

∴ (∀x, y, z ∈ℕ) [(∃m∈ℕ) (x(y + z) = xm) ^ (xm = x(n+1) = xn + x)]

Finally because all values in the set are recursively defined by the addition operator on 〈1〉, we have:
(∀x, y, z ∈ℕ) x(y+z) = x(y + 1 + 1 + . . . + 1) = (((xy) + x) + x) + . . .) + 1 = xy + (x + x + . . . + x)
= xy + xz

∴ (∀x, y, z ∈ℕ) x(y + z) = xy + xz

I like you

In general the axioms of any formal/mathematical system are expressed in set-theoretic notation and can in fact be proved from the axioms of set theory and logic, provided a set of arbitrary definitions for the system under consideration. For example, the axioms of number theory and finite ordinal arithmetic can be proved from the axioms of set theory and formal logic provided we use the standard set theoretic definition of the natural numbers and the peano axioms. I.e.
0 = O
1 = {O}
2 = {O, {O}},
. . . etc.

and (∀x) (x + 1 = x ∪ {x})
etc. . .

What does the O symbol mean?

Oh that was supposed to be the null set (i.e. the empty set), but I guess Veeky Forums doesn't recognize that symbol - at least not the form of it that I was using. It normally looks like an O with a slash through it (kind of like "not equals"). It can also be written simply as {}

Another typo here btw. This:
(((xy) + x) + x) + . . .) + 1 = xy + (x + x + . . . + x)

Should instead be this:
(((xy) + x) + x) + . . .x) + x = xy + (x + x + . . . + x + x)

>Is probably taking Calculus II or differential equations this semester
>Studying for an engineering major
>Literally gay
>Makes a mathematical mistake on a Vietnamese tennis image board
>Feels self conscious
>Ayyy lmao goys I was just trolling xD, arent you pissed

You're only fooling yourself my friendo. In all seriousness though, its okay. We were all new to everything at some point. Most people can't be expected to have an in depth knowledge of abstract, axiomatic mathematics. Shit my knowledge also happens to be fairly superficial. I basically know what I need to know for linguistics type shit including generative grammar and Montague semantics. I'm only in my undergrad though, and the field is becoming increasingly mathematical, and in particular, algebraic. Thus I'm currently working on shit like category theory, information theory, and lambda calculus. Most people on Veeky Forums don't really know much about that shit, but there are also a lot of people know way more than me, and some of them are even working on Ph.D.s on homological algebra or homtopy theory (and perhaps even a few anons are conversant in Homotopy Meme Theory and Univalent Memedations).

>all values in the set are recursively defined by the addition operator on 〈1〉

It was all fine and dandy, untill that last part, what do you mean by that? The book I'm using starts off by defining adition and multiplication on ℕ and then proves the distributive law by induction. I'll show you what I mean:

(∀m, n ∈ℕ) m+n=successor of [m+(n-1)]
(∀m, n ∈ℕ) mn=m(n-1)+m

Then, it is necessary to prove that: k(m+n)=km+kn (∀ k, m, n ∈ℕ)

The base is irrelevant (n=0), I'll skip that.
For n=n+1 the expression becomes:

k[m+(n+1)]=km + k(n+1)

Working out the first member:

k{[m+(n+1)]-1} + k
=k(m+n)+k
=(km+kn)+k (HYPOTHESIS)
=km+(kn+k) (ASSOCIATIVE OF ADDITION, PROVED EARLIER)
=km+k(n+1) (HYPOTHESIS)

Q.E.D?

Is there any fallacy here? I mean, it's fine to assume the hypothesis as being true as long as you don't arrive in a contradiction, right?

>OP has already left the thread

What I mean is that the set of integers is what we call "generated by a single element", in this case the numeral one.

What I was doing was essentially a bastardized proof by induction. In fact, my proof was very informal, and some professors might not even accept it (most would in say a 200 or 300 level class, but they might not at the 400 level, and certainly not in a graduate level class). Yours on the other hand was more acceptable.

Basically though, we were employing the same idea because mathematical induction in fact relies on recursion. Basically since the integers are recursively defined ordinal numbers (or cardinals, but the the distinction doesn't matter in the case of finite arithmetic), if we show that f(1) is true (or in some cases we'll use f(0)), and then show that for every n, f(n) => f(n+1), we've thereby effectively proved that f(n) is true for ALL the integers, because
[f(n) => f(n+1)] [f(n) => f(S(n))]
And we know that S(n) produces the set of integers.

Another way of looking at it is to keep in mind that the set of integers is (A) what we call "closed with respect to addition" (B) that every integer besides 0 is the successor of some other integer, and (C) the successor, S(n), of any integer, n, is equal to n+1.

I.e.
(A) (∀m, n ∈ℕ) (m+n) ∈ℕ
(B) (∀m ∈ℕ)(∃n ∈ℕ) n + 1 = m
(C) (∀m ∈ℕ) S(n) = n + 1

Thus we have [B, C ⊢ (∀m ∈ℕ)(∃n ∈ℕ) m = S(n)]
I.e. every integer is the immediate successor of some other integer (lets call this proposition D).

(continued)
Thus combining A and D, that is because of the fact that (D) every integer is the immediate successor of some other integer, and (A) the sum of any two integers is always also an integer, we find that for any value that is the sum of two integers, this value is the immediate successor of some other value (e.g. S(n + m - 1) = n + m). Finally because any integer is finite (a step that was left implicit in my proof), and integer can be represented by a finite sequence of additions of the number 1, or equivalently, by a finite sequence of recursively applied successor functions.

Part of the confusion here is that I'm sort of presupposing a certain knowledge of set theory and ordering principles (and very basic topology) that one probably wouldn't know in the first place if they were trying to prove the distribution of multiplication over addition.

*Any integer is finite and can be represented. . .

All my writing here is kinda sloppy. If we were working in person this shit would be a lot easier. I tend to leave out a lot of details and overlook the little stuff. So to further elaborate on what I was saying because I didn't really fully explain myself, once we show that

[B, C ⊢ (∀m ∈ℕ)(∃n ∈ℕ) (m≠0) ⇒ (m = S(n))]
and
[A, D ⊢ (∀m, n ∈ℕ) (m≠0 v n≠0) ⇒ (∃p ∈ℕ) m + n = S(p)]

all we need are the Peano axioms for multiplication to prove the distribution principle. Namely, we need the axiom that says
m(n+1) = mn + m
and the axiom that says
m(0) = 0
The first axiom is, of course, semantically equivalent to m(S(n)) = mn + m. Lets call this proposition E.
Thus we have
[A, B, C, D, E ⊢ (∀m, n, p ∈ℕ) (n + p) ≠ 0 ⇒ (∃r ∈ℕ) m(n + p) = m(S(r)) = mr + m)]

For any two integers, an integer multiple of there sum is equal to an integer multiple of some integer, namely their sum, since there sum itself must be an integer.

Finally, since we know that every integer is equal to a finite sequence of successor functions, and that [(∀m, n ∈ℕ) m(S(n)) = mn + m], we therefore know that multiplication can be thought of as iterated addition. I.e. suppose (∀x ∈ℕ) f(x) = m. Thus we have
(∀m, n ∈ℕ) mn = Σ f(x) on the interval (1≤x≤n)

In other words mn is equal to n additions on m.

Anyway this has been super confusing and probably more trouble than help.

I think I got it. I thought it had something to do with induction since that last expression of yours (x(y + 1 + 1 + . . . + 1) seemed to be expressing exactly that, but wasn't sure about it given the unfamiliar nature of that last statement. This idea of recursion has never occurred to me, and although I get the idea, I'm still not sure why you phrased things that way(addition operator on ?). I guess that has to do with topology or ordering principles? Hopefully, I'll get to that someday. Thanks for the insight!

Pretty sure everyone ITT is wrong for very pedantic reasons and the only correct way to do this is using three nested inductions.

I'm taking calc III, am bi, and doing EE.
pretty good guess
anyway, I know that there are algebraic structure that don't have distribution, or are nonabelian or whatever, I was just fun posting I swer!

nice b8

Yeah topology/abstract algebra. The "" thing is just the standard shorthand notation for the group generated by 1.

You haven't even told me where x,y,z come from or what dot and plus are meant to represent

GJ user. I hate your notation but at least you proved that somebody here at least knows basic formal number theory.

Honestly I didn't even intend for it to sound that baity, but its actually pretty awesome as far a bait goes; I sort of touched all the bases, so to speak. . . really covered a lot of ground. To be honest though it was like 80% sincere.

You should have had the doctor check you for autism while you were in the hospital

PROOF:
The statement is a thing-in-itself. Simply apply Hegel's phenomenological dialectic.

I thought the same thing. So equation is just 1 = 1 right?

Lel why do you say that?

Lel is this bait, or are you just retarded? It doesn't say that 1=1, nor does it say anything about any particular number. As long as you substitute the same integer values for each occurrence of any given variable, the expression will be true.

(z+y)+ (z+y)...x times=z+z+...x times +y+y+....x times.

-1/12

Do you really think 2 is somehow independent of 1?

lol

R is a field.
QED.