# Tag Info

0

You know that $1>0$ (because $1$ is a square), so $3=1+1+1 > 0+0+0=0$. On the other hand, you have $$3=4-1=(\sqrt{2}-1)(\sqrt{2}+1).$$ So you know that $$(\sqrt{2}-1)(\sqrt{2}+1)>0 \ \text{and} \ \sqrt{2}+1>1+0=1>0,$$ hence (by dividing) $$\sqrt{2}-1>0, \ \text{ie.} \ \sqrt{2}>1.$$

0

It's really not possible to give a satisfactory answer to this question without knowing exactly what the 15 axioms are to which the OP refers and which theorems (the OP's "whatnot") have already been proved from them and are at our disposal, but presumably one theorem says that if $0\lt a$ and $b\lt c$, then $ab\lt ac$. From that one can argue that if $0\lt ... 5 We know, by the usual ordering axioms, that$2 = 1+1>1$. Now, suppose that$\sqrt{2} \leq 1$. It would follow that $$\sqrt{2}\cdot \sqrt{2} \leq 1 \cdot 1$$ since given positive$a,b,c,d$, we have$a\leq c$and$b\leq d$implies$ab \leq cd$. However, this leads to the conclusion $$2 \leq 1$$ Which is a contradiction. 1$\sqrt2+2>0$since both$2$and$\sqrt2$are positive. If we prove that$2>\sqrt{2}$than we've proven that$\sqrt2>1$(because from$2>\sqrt{2}$we have$2\sqrt{2}>2$or$\sqrt{2}>1$). Let we suppose that$2<\sqrt{2}$, than$\sqrt2-2>0$. Now from$\sqrt2+2>0$and$\sqrt2-2>0$we have that$2-4>0$or$-2>0$or$2<0$... 0 If$0 \leq x \leq 1$, then you also have$0 \leq x^2 \leq 1$. But$\left(\sqrt2\right)^2 = 2 > 1$, so... 0 The problem here is definitional. Since you have infinitely many formulas, you need to define the percentage as some kind of limit. We can define it for finite axiomatic systems by the limit as n goes to infinity of the percentage on at-most length-n propositions. Note that sometimes people consider very large axiomatic systems (much more symbols or much ... 5 Posted as an answer, as requested by the OP. Writing $$(λx.\ x)\ M=M$$ is wrong (or yields falsehood), those two terms are not equal. I think what you want is the$β$-reduction, that is, $$(λx.\ x)\ M\ {\leadsto}_\beta\ M,$$ however, as the name suggests, reduction is not equality. In case of$\beta$-reduction, the definition clearly states why ... 1 This is a "simplified" proof [see (T9) below]. Note. I will call the four Axioms as (Ax1)-(Ax4) and the five following results as (R5)-(R9). I will introduce the definition :$A \rightarrow B$stands for$\lnot A \lor B$. [*1.01] We can have also :$A \land B$stands for$\lnot (\lnot A \lor \lnot B)$[*3.01], and$A \leftrightarrow B$stands for$(A ...

6

Hilbert's axioms predate the development (or at least the wide adoption) of fully symbolic logic, so they are expressed in partially informal language -- though Hilbert strove to make them as precise as he could. They include the Axiom of Archimedes, formulated in language that presupposes that the natural numbers are already known. As such, if we want to ...

-5

Is the above accurate […] ? Clearly not. Your interpretation of Gödel's completeness is very specious. This theorem only says : any first-order consistent theory admits a model.

1

The first part, the set is $\{ M + N\sqrt 3 |M,N \in \mathbb Z\}$. Think of it as a formal sum, like the complex numbers, if you have seen complex numbers. This is closed under both addition and multiplication, since $(M + N \sqrt 3 ) + (M' + N'\sqrt 3) = (M+M') + (N + N')\sqrt3$, and both $M+M'$ and $N+N'$ are integers, so we have an element of the set. ...

0

My first answer was based on my interpretation of the OP's axioms system being the one originally propesed in Principia Mathematica (1910) : five propositional axioms (then reduced to four by Barnays (1918)) plus modus ponens and substitution, without Deduction Theorem. As noted in some comments, in PM $p \rightarrow q$ is defined as $\lnot p \lor q$ ...

1

The argument is no circular. You can write the axiom of infinity in a much longer form, $$\exists S(\exists u(\forall z(z\notin u)\land u\in S)\land\forall x(x\in S\rightarrow\exists v(v\in S\land\forall w(w\in v\leftrightarrow v\in x\lor v=x))))$$ No reference to the empty set there. Of course from this assertion we can prove the existence of a set which ...

3

None of P1, P2, and P3 say anything about the number of things you have, nor do they say anything about time. P1 says that grouping doesn't matter in the order of addition. Rephrased in terms of apples, if I have a pile of $a$ green apples, a pile of $b$ red apples, and a pile of $c$ yellow apples all in a row, then I'll end up with the same number of ...

2

The fact that you can't do it for very small sets invariably means that you can't do it at all. First of all, it is important to note that the way we encode the pairs shouldn't depend on the set we are considering at the moment. So $\langle 0,0\rangle$ should be encoded to the same set regardless to use considering it as an ordered pair in ...

0

a) $a^{-1} \cdot a = 1 \Rightarrow \left(a^{-1}\right)^{-1} = a$. This follows from the definition of the multiplicative inverse. b) Let $a^{-1} = b$. Then $ab = 1$, so $(-a)(-b) = ab = 1$. This implies $(-a)^{-1} = -b = -(a^{-1})$.

1

We can reconstruct, with the help of Wiki Principia Mathematica propositional logic , the original proof of PM's *2.77 --- $(p \rightarrow (q \rightarrow r)) \rightarrow ((p \rightarrow q) \rightarrow (p \rightarrow r))$. We establish first some useful theorems : *(2.05) --- $\vdash (q \rightarrow r) \rightarrow [(p \rightarrow q) \rightarrow (p ... 4 In an axiomatic system you cannot derive it at all (you need to add it as an axiom) , but if you have a natural deduction system you can derive it from nothing at all (ex nihilo) 1) It is easely proved in a natural deduction system: 1 | |______ P -> (Q -> R)) Assumption 2 | | |____ P -> Q ... 2 The term "Archimedean axiom" was introduced by Otto Stolz around 1883. Later, Johan Heiberg noticed that this is closely related to an axiom appearing in Euclid V.4. The term has become commonly accepted since then. It should be noted that none of the European mathematicians from Simon Stevin and until the end of the 19th century used the term in this ... 2 Here is the recipe: Take any complete first-order theory defining the rational numbers, in which every standard rational number is definable. Add a new constant symbol "$\omega$" For each rational number$q$, add an axiom "$\omega > q$" The resulting theory is consistent, and thus has a model. A simple recipe for the first point is: Put whatever ... 1 Any continuous function from a connected space to a totally disconnected space must be constant, and thus any continuous function from$[-1,1]$to a totally disconnected vector space$V$(such as a finite-dimensional vector space over$\mathbb F_p$or a number field, with the topology induced by the base field) is constant. So the set of continuous functions ... 1 "Extension" refers to a set being defined by its content, as opposed to "intension" which is a term to say it is defined by some form of specification. Let A = {x in R such that -1 <= x <= +1} Let B = {y in R such that y = sin(x) for some x in R}. The sets have different intensions (specifications), but contain the same elements and therefore have ... 1 Suppose I want to focus on G. Among the things that are true about G is Every set is regular extensional so, in the course of focusing on G, I will invoke this "dictatorial" fact which is true about the objects I'm studying. 1 Something nice I learned from Wikipedia is that, even in logical systems that don't have an equality predicate built in, it's possible to express the axiom of extensionality as $$\forall a,b: (\forall x: x \in a \iff x \in b) \iff (\forall y: a \in y \iff b \in y), \tag1\label{ik-eqn:1}$$ or, in words: "if$a$and$b$have all the same members, then they ... 2 I think that can be useful to start from the debate about extensionality in modern mathematics and logic, that dates back to Cantor and Frege and their researches about classes and extensions. We can start with the (very simplified) picture of a concept or “universal”, we can symbolize it as$\phi(x)$, and its extension, defined as the set of objects such ... 1 Because it says a set is determined by its extension. 2 To add to the other answers: (1) does say that every set the quantifier ranges over is extensional, full stop. There's no room in any model of extensionality for non-extensional objects. But at the same time there are other theories of which ZF is a subtheory, or in which ZF is interpretable, including some with very different properties (some strong ... 1 The axiom of extensionality can be written as follows. $$(∀a,b) \;\;a \sim b \iff ∀x(x∈b \leftrightarrow x∈a)$$ Notice that, syntactically, this is shaped exactly like a definition. Thus, extensionality can be seen as defining what it means for sets to be equivalent. It then readily follows that$\sim$is an equivalence relation. However, ... 5 1) is correct. There are things which are not extensional, but in a theory which accepts the axiom of extensionality, those things are not called "sets". This is not "a statement of a dictator", whatever you mean by that, this is how language works: every term in every language has a definition, which describes what the term means and what it does not mean. ... 2 You should interpret it the first way. Axioms of formal systems are to be satisfied without explanation. A model which instantiates a formal system must satisfy the axioms of the formal system. Thus any model of ZF does not have a set that is not extensional, they are not merely 'not considered'. I mean that they cannot be neglected because non-extensional ... 4 The logical predicates ∀ and ∃ are defined using the concept of a Domain of Discourse, which itself is defined as a set. This much is true: to fix the content of e.g.$\forall xFx$, we need to know which objects the quantifier is ranging over -- i.e. which objects are such that each of them supposedly satisfies the predicate$F$. But to understand ... 4 We need not think of$\forall$as defined in terms of a domain of quantification. Rather, we can take it to be a${\it primitive}$of our language. Definition has to stop somewhere, and the quantifiers seem like a pretty basic place to stop! Of course, a quantifier may${\it have}$a domain of quantification even if it isn't defined in terms of it. "is red", ... 0 this is a very interesting question! When it comes to$\mathbb R$, there is basically two options: 1) Define it by an axiom:$\mathbb R$is the unique (up to isomorphism) of Archimedian totally ordered field satisfying the least upper bound property. With this axiom, the property you want to prove is part of the definition of totally ordered set. 2) Define ... 0 The basic fact is that product of two positive numbers is positive and$x>y$is shorthand to$x-y>0$. Hence $$(x>y), ~~\text{and}~~b>0 \Rightarrow b (x-y) > 0 \Rightarrow bx -by >0 \Rightarrow bx > by$$ 4 From$x>y$you know that$x-y>0$, since$b>0$you will have$b(x-y)>0$(the product of two positive numbers is positive). So$bx>by$. I'm not sure if it is what you want. 1 This is generally not true. Recall that given a set$X$, its transitive closure is the smallest set$A$such that$X\subseteq A$and$A$is transitive (i.e.$a\in A\implies a\subseteq A$). We say that a set is hereditarily countable, if its transitive closure is countable. The collection of all sets which are hereditarily countable, denoted by ... 1 For elementary geometry without the Pasch Axiom, every model is isomorphic to a Cartesian plane over a formally real Pythagorean semi-ordered field$\mathcal{F}$. More can be said if our geometry has the full second-order continuity axiom. For then$\mathcal{F}$is (as a field) isomorphic to the reals. Now let$f:\mathbb{R}\to \mathbb{R}$be a non-linear ... 0 The proof you suggest is wrong. The axiom of regularity doesn't imply that every set is well-ordered by inclusion. This is false in every possible aspect. The axiom of regularity says that$\in$is well-founded. This means that every non-empty set$x$has$z\in x$such that$z\cap x=\varnothing$. The proof should be by$\in\$-induction, which we can preform ...

Top 50 recent answers are included