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Given a generic triangle $T$, draw the altitude $h_c$ orthogonal to the longest side $c$. It splits $T$ into two right triangles $T_A$, $T_B$ with legs $c_A$ and $h_c$, resp., $c_B$ and $h_c$, whereby $c_A+c_B=c$. The triangle $T_A$ is half of a rectangle with side lengths $c_A$ and $h_c$, and similarly $T_B$ is half of a rectangle with side lengths $c_B$ ...


0

For any finite valued logic, where we have some connective "C" where detachment is a rule of inference for that connective, and we have truth tables, we can basically turn what the truth tables tell us into an axiomization of the logic using functioral variables (we'll also want |=Cpp to hold, otherwise we'll run into a problem with the way the ...


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Imagine we have a sequence of numbers: $$\{3,3.1,3.14,3.141,3.1415\dots\}$$ (In fact, I could use any increasing sequence that approaches $\pi$.) $\pi$ could be defined as the smallest number $x$ that is greater than all of those numbers. That is, I could define $\pi$ to be the smallest number that is greater than $3$, $3.1$, and $3.14$, and $3.1415$, etc. ...


2

If you accept the relationship between algebra and geometry, you might argue like this: Every line $f(x) = a x$ with $a \ne 0$ has exactly one intersection with the line $g(x) = 1$.


4

The real numbers $\mathbb R$ are an example of a mathematical field. In other words, they satisfy a number of axioms (essentially, the basic laws of middle school mathematics (distributive property, commutative property of multiplication, etc...)). One of these field axioms is the following: For each nonzero $x\in\mathbb R$, there exists a number ...


0

distributive law says $(a+1)\mathbf{u}=a\mathbf{u}+1\cdot \mathbf{u}$. Identity law says that $1\cdot \mathbf{u}=\mathbf{u}$. As such you see that $$(a+1)\mathbf{u}+a\mathbf{v}=(a\mathbf{u}+\mathbf{u})+a\mathbf{v}$$ $$=a\mathbf{u}+(\mathbf{u}+a\mathbf{v}) = a\mathbf{u}+(a\mathbf{v}+\mathbf{u})$$ $$=(a\mathbf{u}+a\mathbf{v})+\mathbf{u} ...


3

From what is given, we have $y-x>0$. The sum of positives is also positive, hence and $y+x>0$. The product of positives is positive, hence $y^2-x^2=(y-x)(y+x)>0$.


0

First multiply by $x$ and then by $y$: $$0\leq x<y\implies0\leq x^2<xy$$ $$0\leq x<y\implies 0\leq xy<y^2$$ so $$0\leq x^2<y^2$$


1

Well, you are given $0 \leq x < y $. Hence we have $$ x^2 < yx $$ Similarly, we obtain $$ xy < y^2 $$ Since $xy = yx \; \; \;(Proof?) $, then by transitivity $$ x^2 < yx = xy < y^2 \implies x^2 < y^2 $$ In particular, $f(x) = x^2$ is increasing in the first quadrant of the $xy-plane$


2

The first sentence of your main question is ungrammatical (since removed by an edit) - I can't make any sense of it. But I think I know what you're trying to ask: In what way can we formalize the fact that the usual definition of topological space in terms of open sets is equivalent to the definition in terms of the Kuratowski closure axioms, despite the ...


3

Whatever axioms that you might be working with, someone will ask themselves what happens if we remove them, and a new branch will be formed. If we confine ourselves to mainstream mathematics, then I suppose that induction axioms, modus ponens, and existential instantiation (along with the Leibniz laws about equality) would make the fundamental axioms. But ...


3

As David Holden said in the comments, if $h=\frac{1-a}a=\frac1a-1$, then $\frac1a=1+h$, and for $n\in\Bbb Z^+$ you know that $\frac1{a^n}=\left(\frac1a\right)^n=(1+h)^n\ge 1+nh$ and therefore $a^n\le\frac1{1+nh}$. What happens to $\frac1{1+nh}$ as $n\to\infty$? $9$(b) asks you to show that if $x\notin\Bbb Z$, then there is exactly one $n\in\Bbb Z$ such that ...


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Hint: Prove that, if $a>0$ and $b<0$ then $ab<0$. Prove that $1>0$. Now suppose that $a>0$ and $\frac{1}{a} <0$


1

So I think the issue is a matter of perspective. The infinite recursion that you see is actually the contradiction. The Axiom of Regularity says that your construction is against the rules. Axioms are rule of set theory that dictates how one must define a set. Therefore your idea of 'infinite recursive' is simply a violation of the rules, or axioms, of set ...


1

Yes, it's true. Apply the axiom of regularity to the set $\{A, B\}$. You ask whether you can prove $B \in B$. Of course, this is not a meaningful question with the axiom of regularity, since your premise is false. However, without the axiom, I believe the relations $B \in A \in B$ do not imply that $B \in B$.


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If you put these into words you can read them as follows: "ϕ→ϕ" "If ϕ, then ϕ". "ϕ→(ψ→ϕ)" "If ϕ, then if ψ, then ϕ." "(ϕ→(ψ→ξ))→((ϕ→ψ)→(ϕ→ξ))" "If, if ϕ, then if ψ, then ξ, then if, if ϕ, then ψ, then if ϕ, then ξ." "(¬ϕ→¬ψ)→(ψ→ϕ)" "If, if not ϕ, then not ψ, then if ψ, then ϕ".


1

Well, if you really want to work with Whitehead's axioms, here are the axioms as he put them in The axioms of projective geometry page 7: As you can see, axiom $II$ asserts there is a point, and axiom $III$ asserts there is a point other than the first point you picked. Then axioms $IV$ and $VII$ give you that the point you picked lies on a line. As ...


0

Axioms are used to show that under some circumstances, some sets exist. Russell's paradox shows that not every definable collection can be a set. So we have to restrict what sort of definable collections we allow. The separation axiom schema essentially say that if $A$ is already a set, then every definable subcollection of $A$ is also a set. If we repeat ...


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I don't think the axiom scheme of separation "resolves" Russell's paradox at all, but restricts the way of using predicates to determine sets. The paradox is nothing but a proof that there is no one-to-one correspondence between predicates and classes: there are predicates that not defines a class. When writing sets as $\{x|p(x)\}$ that one-to-one ...


2

The guarantee that such a set can't exist is already given by the argument of Russell's paradox: its existence leads to a contradiction therefore it can't exist. The problem with unrestricted comprehension was that it guaranteed the set does exist, which causes a problem because of the conflicting guarantees.


0

It's clear that $G3$ always holds. If $A,B,C,D$ are four distinct simple submodules of $M$ such that $(A+B)\cap (C+D)\neq\{0\}$, then $A+B+C+D$ has length $3$. (You can see this by noting that $(A+B+C+D)/(A+B)\cong (C+D)/((A+B)\cap(C+D))$ and that the length of the right hand side is $1$ and then it must be that $len(A+B+C+D)-2=1$. Then it is impossible for ...



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