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4

As Andres points out in the comments, $\sf CH$ is not the actual player here. Consider the case that there is an inaccessible cardinal $\kappa$ which is itself a limit of inaccessible cardinals. By forcing we can arrange $\kappa$ to be $\omega_1$. This will certainly say that $\omega_1$ is large in the sense that it is the limit of inaccessible cardinals in ...


3

You do not need to know any calculus to do geometry. I think you are all set to read it. By the way, I am also reading Hartshorne!, but his other geometry book, also no calculus required there either.


4

Proved from what? Even logic is formalized in a mathematical fashion, so it has a mathematical backbone. We need to first assume that first-order logic does not include a contradiction. Then we need to decide on which system we are going to use to develop our mathematics. Arithmetical systems will suffice for syntactical manipulations of strings, and set ...


1

For a first order theory, basically a set of "axioms" formulated in a first order language, you have Gödel's completeness theorem. The theorem establishes that a first order theory is consistent(i.e. non-contradictory) is and only if you have a model for that theory(i.e. a "real" mathematical object satisfying the axioms of the theory). From this you can ...


-1

You can prove contradiction by constructing results which stem from them that contradict. Right? If A leads to Z and B leads to NOT Z then something is obviously fishy. Or equally showing that C leads to both Y and NOT Y simultaneously. The inability to show contradiction does not prove anything. Indeed even the ability to demonstrate consistency in any ...


0

$\forall x\exists y(z\in y\iff (z\in x\wedge \forall w(w\in z\iff w=z)))$


0

The separation schema takes a formula as a parameter, and says that for every $z$ there is a $w$ which is made of exactly the elements of $z$ satisfying the formula. Here the formula is given informally as $x=\{x\}$, and you are tasked with writing the relevant axiom in details.


-2

My intuitive model of sets includes only finite sets. My intuitive idea of "natural" numbers includes the idea that there is no maximal number. To handle the "set" of all these numbers I visualize a sequence of finite sets $\{0\},\{0,1\},\{0,1,2\},\dots$ that never ends and imagine a mathematical object that I call $\mathbb N$ and that is such that ...


3

Start with two inaccessible cardinals and $V=L$. Then add Cohen reals so the continuum equals to the larger of the two, and consider $L_\kappa$ where $\kappa$ is the smaller inaccessible. Of course you don't need inaccessible cardinals for that, a Worldly cardinal of uncountable cofinality will suffice, since it is the limit of smaller worldly cardinals. ...


1

The basics of what you're asking deal with operations. Addition is a binary operation (uses two terms), and is typically indicated with the symbol $+$ between the two terms. The same symbol can also be a unary operation. For example, $+1$ is not addition per se, even though it uses the same symbol as for addition. In this case, it's making explicit the ...


1

The short answer is that the symbols and syntax work the way they do because that's how we've defined them. This may not seem satisfying, but it makes a certain amount of sense, considering that mathematics is based on axioms and definitions. As for why they are defined that way in particular, there are a number of different reasons, depending on which ...


1

By definition, $a/b$ is the unique solution of the equation $x\cdot b=a$. That is, if there is no solution or if the solution is not unique, we do not have a quotient. That being said, how can anything not be divisible by itself? How can $x\cdot a=a$ not have $1$ as the unique solution? Firstly, $1$ could fail to be a solution if there is no $1$ in our ...


2

Well, what is a 'number'? Though among complex numbers indeed only $0$ is the only one that cannot be a divisor, if instead of an imaginary number $i$ whose square is $-1$, we extend $\Bbb R$ the same way by an 'imaginary number' $E$ whose square is $1$, then $(1+E)(1-E)=0$, hence if we had $(1-E)/(1-E)$, then $$1+E \ =\ \frac{(1+E)\cdot(1-E)}{1-E}\ = ...


0

No. All numbers, real or complex, are divisible by themselves... the sole exception being zero. Someone might suggest that infinity is another, but infinity is not considered a number... at least not a real number. There are of course types of mathematics that dont involve numbers at all, and operations on them can be defined rather trickily. Vectors, ...


1

There is another common model where the points of the projective plane consist of the points of the Euclidean plane and equivalence classes of lines for the equivalence relation "is parallel to". The mental image is that the latter types of points are to be thought of as the "point at infinity" that the class of parallel lines intersects at. (and the lines ...



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