Tag Info

New answers tagged

0

Here is another equivalent of the axiom of choice which I believe has not yet been discussed. Given a surjective function $f: A \rightarrow B$, there exists a function $g: B \rightarrow A$ such that $f \circ g = Id_{B}$. The idea is the following: (1) Let $A,B \neq \emptyset$. (2) Then given that $\forall x \in B \ \exists y \in A \ (f(y) = x)$, by the ...


1

To elaborate on Thomas Andrews' comment: Addition is well-defined: For all $a,b,c,d\in\mathbb{R}$, if $a=b$ and $c=d$, then $a+c=b+d$. The well-defined nature of addition in $\mathbb{R}$ is taken as an axiom. Hence, if $a=b$ and $c=d$, we have $$ a+c=b+d\Longleftrightarrow a+c=b+c, $$ but I think there is something more interesting you could show, namely ...


1

You start with $$ a \oplus c = x \quad (*) $$ where $x$ is the resulting value, then note that $a = b$, so you replace $a$ with $b$ in equation $(*)$. This gives $$ b \oplus c = x $$ Of course this means $$ a \oplus c = x = b \oplus c $$ and thus $$ a = b \Rightarrow a \oplus c = b \oplus c $$ Let us redo this for variables $a, b, c \in \mathbb{R}$ and the ...


1

There is nothing wrong with them. The problem is that until the 19th century they were thought to be the only ones possible, giving rise to a single possible geometry (the one called today "Euclidean"). During the 19th century advances in mathematics (Gauss, Bolyai, Lobachevski, Beltrami, Klein but most importantly Riemann and the Italian school of ...


3

There's nothing wrong with Euclid's postulates per se; the main problem is that they're not sufficient to prove all of the theorems that he claims to prove. (A lesser problem is that they aren't stated quite precisely enough for modern tastes, but that's easily remedied.) In every modern axiom system (e.g., Hilbert's, Birkhoff's, and SMSG), each of Euclid's ...


1

Jyotirmoy writes that: [To prove the Banach-Tarski theorem] requires the Axiom of Choice. This statement needs some qualification. A bit of background: the standard collection of set-theoretic axioms is called ZFC, where "C" stands for the axiom of choice. Deleting the axiom of choice yields a smaller collection of axioms called ZF. Here's the key ...


6

$\bf 1.$ The Axiom of Choice Given a set $S$, to say that $S$ is not empty is to say that $\exists x(x\in S)$ (in English: there exists some $x$ such that $x$ is an element of $S$). First-order logic has an inference rule which allows us to move from $\exists x(x\in S)$ to some [new] constant symbol $s$ such that $s\in S$. This process, called existential ...


0

The (¬ϕ→¬ψ)→(ψ→ϕ) [CCNpNqCqp in Polish notation] axiom can get interpreted as meaning that if a negation of a variable is equivalent to the negation of another variable, then the variables will be equivalent. (¬ϕ→¬ψ) says that if the negation of the first proposition is true, then the negation of the next proposition is also true. Given the rule of ...


0

In a proof (sequence of statements), the axioms can be moved around the sequence arbitrarily since they don't depend on any previous statements. The "conclusion" that is to be proved can be taken to be the last statement that depends on the statements preceding it.


0

It would be a proof that the axiom is true, given that the axiom is true. So it's a tautology. It would not be an interesting proof, although it is a correct one; most correct proofs do not prove anything interesting.


2

Yes. Axioms can have proofs. And they can be as short as a sequence of length $1$. This is only counterintuitive since in high school you learn that "axioms are things which do not have a proof" or something silly like that. No. Axioms are just statements which we assume to be correct in the first place. It does not mean they have no proof. Not to ...


3

You're on the right track. We can "talk about" a recursive transfinite ordinal $\alpha$ by defining a recursive relation $R$ on $\mathbb{N}$ such that $\{\mathbb{N},R\}$ is a well-ordered set with order type $\alpha$. We can then define $\text{TI}(R)$ as the schema $\forall \alpha (\forall \beta (\beta \ R \ \alpha \implies \varphi(\beta)) \implies ...


1

$x\cdot 0 = x\cdot (0+0) = x\cdot 0 + x\cdot 0$, now add $-(x\cdot 0)$ on both sides If $x>0$ then $x\cdot x>x\cdot 0=0$. If $x<0$, then $0<-x$, hence $x\cdot x=(-x)\cdot(-x)>(-x)\cdot 0=0$ (I assume you already know $(-x)(-y)=xy$). Only if $x=0$, we do not have $x\cdot x>0$. So you need to show $\frac1a\ne 0$. You can simplify your ...


1

Basically, a set of axioms defines a mathematical structure. For example, the Peano axioms define what a natural number is, and the group axioms define what a group is. Propositions are true statements about the mathematical structure that can be derived from the axioms. Now it may happen that different sets of axioms define the same mathematical structure, ...


1

Hint: Write $0^3 + 1^3 + ... + k^3 + (k+1)^3 = \left(\frac{k(k+1)}{2}\right)^2 + (k+1)^3$ using your induction hypothesis. Then check if $\left(\frac{(k+1)(k+2)}{2}\right)^2=\left(\frac{k(k+1)}{2}\right)^2 + (k+1)^3$.


1

Finite structures can always be described completely. You haven't specified what the successor of $Max$ is, so I'm giving it the name $M$; $M$ has to be an element of your domain. For a number $k$, let $\langle k \rangle$ be $suc^{(k)}(0)$, that is the result of applying $suc$ to zero $k$ times. The axioms for $(0,Max,suc)$ are: $\forall x (x = \langle 0 ...


1

I played around with following axioms for finite arithmetic a while ago. I couldn't do much with them. Maybe you will be able to do something with them: Axioms $0\in N'$ $Max\in N'$ $\forall x\in N':[x\ne Max\implies S(a)\in N'] $ (a partial function) $\forall x\in N': S(x)\ne 0$ $S(Max)\notin N'$ $\forall x,y:[S(x)=S(y)\implies x=y]$ $\forall P\subset ...


1

This answer is really just an elaboration of Mauro's answer. To address your question as to why the Generalization axiom is necessary, we need it (or some mechanism like it) in order to prove lots of theorems that contain universal quantifiers. Your example $(x = 2) \vdash \forall z (x = 2)$ is, as you describe, a correct but vacuous application of the ...


1

This axiom is "useful" in proving the Generalization Theorem : If $\Gamma \vdash \varphi$ and $x$ does not occur free in any formula in $\Gamma$, then $\Gamma \vdash ∀x \varphi$. See : Herbert Enderton, A Mathematical Introduction to Logic (2nd ed - 2001), page 117. There are other axiomatizations of first-order logic that avoid this "unnatural" ...


1

You usually assume in the Renyi formulation of conditional probability that $$1=P(Z|Z).$$ Then $P(Z^C|Z)=0$ and therefore $P(X\cap Z^C|Z)=0$. So $$P(X|Z)=P((X\cap Z)\cup (X\cap Z^C)|Z)=P(X\cap Z|Z)+P(X\cap Z^C|Z)=P(X\cap Z|Z).$$


1

Here's one straightforward formulation of the parallel postulate: "Given a point in the space, and a co-line not incident to that point, there is a unique co-line incident to the point which does not meet the given co-line." By 'co-line' here I mean a $(n-1)$-dimensional subspace; in the case where the 'space' is the normal 2d plane this is just the ...



Top 50 recent answers are included