# Tag Info

0

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 ...

0

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 ...

0

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 ...

1

You don't need the axiom of the empty set for the axiom of infinity. $\exists S(\exists x(x\in S\land\forall y(y\notin x)\land\forall z(z\in S\rightarrow\exists u(u\in S\land\forall w(w\in u\leftrightarrow u\in z\lor u=z))))$ The axiom states that there exists $S$ such that there is an element of $S$ which has no members, and $S$ is closed under ...

0

The existence of the empty set can be proved from predicate calculus, the Rule of Generalization, and the Axiom of Separation; you don't need the Axiom of Infinity. You can find a formalized proof here: http://us.metamath.org/mpegif/axnul.html Of course in many treatments it is simply taken as an axiom.

0

The Axiom of Existence states that the empty set exists. If you don't accept the Axiom of Existence as axiomatic, the Axiom of Infinity implies the existence of $\varnothing$, though you need another axiom to "extract" it from $S$.

3

One cannot prove this from $Z$ + Foundation. Indeed, one cannot prove that $V = \bigcup V_\alpha$ even in $Z$ + Foundation + $\forall \alpha (V_\alpha$ exists). For details, see this mathoverflow answer.

1

Your example for $n>3$ fails axiom 1.

1

If I take Peano arithmetic and interpret $1$ to mean "the sky", $2$ to mean "color", every other number to mean "blue", and $x+y$ to mean "the $y$ of $x$", then we see that our interpretation of "$1+2=3$" holds: the color of the sky is indeed blue. Of course, most arithmetic statements would be nonsensical under this interpretation....

2

Hint: $∠ABC=180^o=∠ABD+∠CBD$ Since $∠ABD$ is obtuse, $∠ABD>90^o$ Case I: $∠CBD>90^o$ $∠ABC=∠ABD+∠CBD>90^o+90^o=180^o$ which is a contradictiona as $∠ABC$ is a straight line angle. Case II: $∠CBD=90^o$ $∠ABC=∠ABD+∠CBD=∠ABD+90^o>90^o+90^o=180^o$ which is a contradictiona as $∠ABC$ is a straight line angle. You could alse see that in the ...

0

The set of integers satisfy the least upper bound property and is thus complete, as Andres Caicedo pointed out already. The empty set is explicitly excluded in the upper bound property.

1

As pointed out in the comments, it is definitely a theorem. Here is the shortest proof I could think of. If the parallelogram is a rectangle, then the equality holds. Therefore we can assume the parallelogram has an acute angle. Label the parallelogram as in the figure, where the lines of length $h$ are perpendicular to the top and bottom. Also, notice that ...

Top 50 recent answers are included