# Tag Info

## New answers tagged axioms

1

Other answers give equations that define sine and cosine without specifying the units. If you want radians, you'll want: \begin{align} \sin x<x<\tan x&&&x\in\left(0,\frac\pi2\right) \end{align} There are other equivalent ways to get radians, such as: \begin{align} \sin x&\le x^2+x&x&\in\mathbb R\\ \cos x&<\frac{\sin ...

15

To quote a previous answer of mine: Robison, "A new approach to circular functions, π, and lim sin(x)/x", Math. Mag. 41.2 (March 1968), 66–70 [jstor]. In this paper it is shown that the addition law for cosine (and a couple other simple assumptions) uniquely determines cosine and sine. (This is the paper I cite most often on StackExchange.) If ...

2

A definition not involving power series, integrals, differential equations, or geometric intuition is:$$\cos x+\mathrm i\sin x=\lim_{n\rightarrow\infty}\left(1+\frac{\mathrm ix}{n}\right)^{\!n}\quad(x\in\Bbb R).$$

11

From what you do not want that pretty much leaves functional equations. There seems to be a system of two functional equations for sine and cosine: (link) $$\Theta(x+y)=\Theta(x)\Theta(y)-\Omega(x)\Omega(y) \\ \Omega(x+y)=\Theta(x)\Omega(y)+\Omega(x)\Theta(y)$$

1

You can get any finite set of sets from the pairing axiom and unions. In the language of set theory, it is difficult to write the general axiom formally, but we can prove individual cases as needed just from pairing and unions. Basically, there is no need for it, and it adds a complicated "axiom scheme" to our theory. ...

1

It depends on what you mean by number of sets. If it is a finite natural number, then you can get it by the axioms of pairing and induction. If you want possibly infinitely many sets, then you either already have some set containing all of those sets as elements, or you have some sequence of sets, namely a function on an index set, in which case what you ...

0

How would you write it? An axiom is not just a philosophical assertion - it is a precise, formalizable statement. "Any number of sets can be combined into one" isn't really precise. The problem is the word "number." What do you mean, "any number of sets?" One way to approach this is to say, "Whenever I have a set $X$, and $X$ 'counts' some other family of ...

0

You are correct. Russell's Paradox exposed the inconsistency of a proposed formal set theory of his day. This so-called naive set theory allowed you to derive theorems of the form: $\exists S: \forall a:[a \in S \iff P(a)]$ for any unary predicate $P$. This was called the axiom schema of unrestricted comprehension. For $P(a)\equiv a\notin a$, ...

1

See : Joseph Shoenfield, Mathematical logic (1967), page 27, for the : Tautology Theorem : If $B$ is a tautological consequence of $A_1,\ldots,A_n$, and $\vdash A_1, \ldots, \vdash A_n$, then $\vdash B$. Thus, due to the fact that $A∨(B∨C)$ is a tautological consequence of $(A∨B)∨C$, by the above theorem we conclude with the "derived rule" : if ...

3

The axioms of the list you cite are not independent, but $1x=x$ is not the problem. Commutativity of addition follows from the other axioms; let $x,y\in V$ and set $$z=(1+1)(x+y)$$ Then $$z=1(x+y)+1(x+y)=x+y+x+y$$ (parentheses can be omitted because of associativity). On the other hand $$z=(1+1)x+(1+1)y=1x+1x+1y+1y=x+x+y+y$$ Therefore, being $V,+$ a ...

20

A mistake: You only showed that $\lambda=0$ or $x=0$ $\implies$ $\lambda x=0$. You did not show the reverse implication: $\lambda x=0 \implies \lambda=0$ or $x=0$. A proof of that implication uses the axiom $1x=x$. A standard counterexample of a structure that satisfies all the other axioms save $1x=x$ is the following: $V=\Bbb{F}^2$ ...

14

In order to prove $$\lambda x=0\implies \lambda=0\vee x=0\tag1$$ you need to use the axiom $1x=x$. Here is how you prove (1): if $\lambda\neq0$, then $\lambda x=0$ implies $$\lambda^{-1}(\lambda x)=\lambda^{-1}(0)$$ $$(\lambda^{-1}\lambda)x= 0$$ $$1x =0$$ $$x=0$$ so $\lambda\neq0$ implies $x=0$, or equivalently, $\lambda=0\vee x=0$. Thus, your ...

23

The axiom system you quote does not have $$\lambda x=0\implies \lambda=0 \lor x=0$$ as an axiom. If we drop the axiom $1v=v$, the following becomes an example of a "vector space" over $\mathbb R$: $V=(\mathbb R,+)$, $F=(\mathbb R,+,\cdot)$ for $\lambda \in F$ and $v\in V$ let $\lambda v=0$. We do not want this to happen.

0

The upper n dimension axiom can be found page 182 of http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.27.9012&rep=rep1&type=pdf The Upper n-Dimensional Axiom for n = 2, 3, . . . asserts that any three points a, b, c which are equidistant from each of n distinct points p1, p2, . . . , pn must be collinear. The axiom does not directly ...

1

These axioms won't adequately describe the natural numbers. First, you should start with $s:\mathbb{N}\to \mathbb{N}$ or some equivalent. Your axioms will not rule out a structure consisting of two main sequences -- one that looks like the regular set of natural numbers going off to infinity in one direction starting from $0$, the other like the integers ...

4

When you give a recursive definition of a sequence, then you define a sequence uniquely. You're not making arbitrary choices, since $a_{n+1}$ is picked uniquely once you know the values of $a_0,\ldots,a_n$. So giving a recursive formula and a starting condition makes no appeals to the axiom of choice. This is different from the fake proof you presented, ...

1

If you left out the axiom of induction from Peano's Axioms you could not eliminate the possibility that $x+1=x$ for some $x\in\mathbb{N}$, and other oddities that each of the other axioms would allow. There would also be no guarantee that if you started at $0$, you would eventually be able to reach any other element of $\mathbb{N}$ by repeatedly adding $1$. ...

1

The answer is no. This follows from a result of Skolem, which implies that any sentence of monadic second order logic is equivalent to a (finite) propositional combination of sentences $L_n$, $n = 1, 2, \ldots$, where $L_n$ means the universe has at least $n$ elements. See my answer to Is there any formula of monadic second-order logic that is only satisfied ...

3

First Question: You say "in your opinion". Try to prove it, then, and we will show the flaw. Second Question: You proved every $I_n:=\{1,...,n \}$ is finite. "By induction", you just repeat it: you proved that for every $n \in \mathbb{N}$, the property holds, meaning: every $I_n:=\{1,...,n \}$ is finite. $\mathbb{N}$ is none of those $I_n:=\{1,...,n \}$, so ...

0

The notion of equivalence is relative to the theory. It is important to note that some of the equivalences cited above are not correct if you do not assume Archimedes axiom. For example, in Dehn's semi-euclidean plane, the sum of angles of any triangle is $\pi$ but the axiom of parallels fails: https://en.wikipedia.org/wiki/Dehn_plane Also, Hartshorne ...

1

If your ring has a unit or if your semigroup is a monoid, then $ac = 0$ implies $a1c= 0$ whence $a1 = 0$ or $1c = 0$, that is $a = 0$ or $c = 0$ and you are back to the traditional definition. So I am not sure this is really the definition you wish.

7

Break it up into smaller pieces: $\lnot \exists z (z \in y)$ says, "$y$ is the empty set". So $\forall y (\lnot \exists z (z \in y) \to y \in x)$ says, "if $y$ is the empty set, then $y$ is a member of $x$". $\forall v (v \in u \leftrightarrow (v = w \lor v \in w))$ says, "the members of $u$ are $w$ and the members of $w$", or more simply, $u = w \cup \{ w ... 1 In words, the statement is saying: there exists$x$such that$\emptyset \in x$and for all$w$, if$w \in x$, then the successor of$w$is an element of$x$. You should be able to take it from here. 1 This statement is equivalent to the following $$\exists x : ((\emptyset \in x) \wedge (\forall w \in x, \ \{w\}\cup w \in x))$$ i.e. the Axiom of Infinity. 2 Well, there's a first-order theory where every model has exactly one element, and a first-order theory where every model has exactly two elements, and so on. There are infinitely many of these first-order theories, none of them equivalent to each other. These first-order theories can be constructed explicitly by an algorithm. Here's an example of a ... 3 See : Alfred Tarski, Introduction to Logic and to the Methodology of the Deductive Sciences (4th ed Oxford - 1994; original edition : 1936), page 202 (Dover reprint : page 214). A preliminary comment : the axioms are second-order logic; thus the interpretation of the set$R\$ is not (necessarily) the domain of the model. The first six axioms can be ...

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