# Tag Info

## New answers tagged axioms

-1

A1 is P(C(x, C(y, x))). A2 is P(C(C(x, C(y, z)), C(C(x, y), C(x, z)))). A3 is P(C(C(N(x), N(y)), C(C(N(x), y), x))). Here's a Prover9 proof. ============================== PROOF ================================= % -------- Comments from original proof -------- % Proof 1 at 38.91 (+ 4.42) seconds. % Length of proof is 57. % Level of proof is 18. % ...

4

You will get probably a lot of responses disagreeing with me on this site, because there are a lot of people here that focus a lot on these types of logical foundations. But in my experience, most mathematicians don't pay a lot of attention to this. Yes, logical consistency is important, but most mathematicians take the LEM and the AoC as given. Sets can ...

0

There is no mathematical defination of LEFT and RIGHT.There is neither a position called LEFT nor RIGHT in the real world. In a big space your own position tells that what is right and what left.It is with respect to you.So left and right positions are thought of with respect to something.Think that you are in the space,then symmetric law of the ...

1

You shouldn't take Euclid's axioms too seriously. It was the first known TRY to axiomatise geometry, but as it turned out Euclid in his proofs derived some properties from diagrams and tacitly assumed some facts. The first theory which is purely based on deduction from axioms was given by David Hilbert in Grundlagen der Geometrie in 1899. You can deduce all ...

2

It became clear a couple of hundred years ago that Euclid's axioms were flawed and incomplete. Even in his first theorem he implicitly assumes somthing that is not stated as an axiom (circles that "look like" they should intersect actually do have a point of intersection). The two books I've heard are good are "Geometry: Euclid and Beyond" by Robin ...

0

I think one can learn a ton by simply reading and working through $\textit{The Elements}$. Seeing how Euclid built upon everything, from the original definitions to the five axioms, is astonishing when you think about everything that can be shown. The algebraic principles you know can be derived and shown to be true geometrically through these definitions ...

0

Examine it again: $\langle\mathbf u, \mathbf v\rangle = 2\underline{u_1v_2} + u_2v_1 + \underline{u_1v_2} + 2u_2v_2$ The underlined terms are a clue that there may be a typo in the expression.   Why else are they separated? The symmetry requirement suggests that the first term was intended to be $u_1v_1$. $$\langle\mathbf u, \mathbf v\rangle = ... 0 You see, writing the inner product in a more illuminating manner will explain why your textbook either has a typo, as others point out, or is simply wrong, which i salute you for pointing out. So we know \langle u,v\rangle = 2u_1v_2 + u_2v_1 + u_1v_2 + 2u_2v_2. We rewrite this as:$$ \langle u,v\rangle = \begin{pmatrix} u_1 \ u_2 \end{pmatrix} ...

0

You're right that as stated it is not an inner product for exactly the reason you gave. However, most probably the first term is typed incorrectly and it should have been "$2 u_1 v_1$", which then makes sense as the four terms correspond to the four possible products of $u_1,u_2$ with $v_1,v_2$.

2

To complement the fine answers above, it is worth mentioning that Mathias does not merely criticize a neglect of logic and set theory, but also documents serious errors in Bourbaki's Theory of sets. There are both conceptual errors, when an outdated formalism involving Hilbert's $\varepsilon$ is used throughout, and also technical errors when theorems are ...

4

Hint: By replacement, $Y = \{ \operatorname{rank}_{\in}(x) \mid x \in X\}$ is a set and it's now easy to check that $\omega \subseteq Y$.

1

The requirement is usually stated as: the theory is able to represent all the computable functions. In reality, the theory only needs to be able to represent the particular computable functions that are used in the proof. Usually, the easiest way to show that a particular theory is able to represent these particular functions is to show that it represents ...

1

We cannot necessarily construct a choice function, but be can show that a choice function exists, i.e., that the set $C$ of choice functions is non-empty (in other words: that the direct product of finitely many non-empty sets is non-empty). Once we know this, we can pick a choice function from this non-empty set and argue with it, e.g., if we have ...

0

Sometimes it's possible to explicitly pick an element, sometimes it is not. It depends on what you know about your sets. For all $x$ in $S$, $\neq\emptyset$ by assumption. Therefore $\exists y\in x$. Then consider the pair $(x,y)$. The set of all such pairs is a choice function on $S$. If $y$ can be explicitly chosen, then $C=\{(x,y)\}$ can be. If not, ...

1

Prove smaller props along the way. If x > 0 and y > 0 then xy > 0 is an ordered field axiom. 0x = 0 is a proposition to be proven: 0x +0x = (0+0)x=0x (distributive) so 0 = 0x - 0x = 0x +0x - 0x = 0x. If x > 0 then -x < 0 and vice versa can be proven: x>0 then by ordered field axiom 0=x-x>0-x =x.(same for <) If x > 0 and y < 0 then xy < 0. ...

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