Can someone provide a formal definition of addition for the natural numbers directly from the axioms of ZFC? If this can't be done, why not?
121427
Surprisingly, the Wikipedia article on addition doesn't contain the answer. I looked  elsewhere online for it, but didn't find it.   Intuitively, the cardinal of the union of two sets seemed appealing. But that approach doesn't work, given that the union of 2 and 2 is 2. I wondered whether ordered sets might be essential to the answer.   So, how does ZFC describe addition?
27617
Show (within ZF Set Theory without choice) that for any sets $A,B,C:$   If $|A|^{|B|} = |B|$, then $|A| = 1 = |B|.$   That is, if there is a bijection from $|A|^{|B|}$ to $|B|$, then  $|A| = 1 = |B|.$  Please help me..
67383
Is multiplying by a constant m (integer) on group of set of all integers on addition an automorphism?   If so why does the 2nd example in  says that the unique non trivial automorphism is negation?
74273
A similar question was asked in .  A paper by Roberto Di Cosmo and Thomas Dufour () asserts in footnote 2 (p.3) that it has been shown "that there are no nontrivial equations for $⟨\mathbb{N}, \mathrm{Ack}(n, \_, \_)⟩$ if $n &gt; 2.$" References make it clear that $\mathrm{Ack}(n,x,y)=H_{n+1}(x,y)$ in the , so the assertion is as follows:     $\text{There are no nontrivial equations for the binary function }(\_\uparrow^y\_)\text{ if }y&gt;1\\ \text{ (i.e., for hyperoperations above exponentiation)}\tag{*}.$   (For this, the above paper cites Charles F. Martin, "Axiomatic bases for equational theories of natural numbers", Notices of the Am. Math. Soc., 19(7):778, 1972, which I have not found online.)  Q.1: What, precisely, is meant by a "nontrivial equation" here?  (The defining recursion $$x\uparrow^{y+1}(z+1)=x\uparrow^y (x\uparrow^{y+1}z),$$ which certainly holds for $y&gt;1$, is evidently considered trivial.)  Q.2: Is the following a valid proof based on assertion $(*)$? ...  Claim:     If $x,y,z$ are integers $\gt 1$, then the value of $\ x\uparrow^y z\ $ uniquely determines $x,y,z$; i.e.,  $$x\uparrow^y z = x'\uparrow^{y'} z' \quad\implies\quad x=x',\ y=y',\ z=z'.\quad \tag{1}$$  (The purpose of restricting the domain to integers $&gt;1$ is to remove obvious counterexamples that arise from equations like $x\uparrow (a\cdot z) = (x\uparrow a)\uparrow z\ \ $ or $\ \ 2\uparrow^y 2=4\uparrow^{y'} 1\ \ $ or $\ \ 1\uparrow^y z=1\uparrow^{y'} z'.$)  Proof:  Suppose, for the sake of contradiction, that there is a counterexample to $(1)$, in which the LHS of the implication holds but not the RHS. Then the LHS is a nontrivial equation for some hyperoperation(s) beyond exponentiation, which is impossible as per $(*)$. QED
12861
I wonder about the foundations of set theory and my question can be stated in some related forms:   If we base Zermelo–Fraenkel set theory on first order logic, does that mean first order logic is not allowed to contain the notion of sets? The axioms of Zermelo–Fraenkel set theory seem to already expect the notion of a set to be defined. Is there are pre-definition of what we are dealing with? And where? In set theory, if a function is defined as a set using tuples, why or how does first order logic and the axioms of Zermelo–Fraenkel set theory contain parameter dependend properties $\psi(u_1,u_2,q,...)$, which basically are functions?
59560
Is it possible to figure out if a number $z$ is the addition of two triangular number without recursion or finding the values to $x$ and $y$?  $$\frac{x(x+1)}{2} + \frac{y(y+1)}{2} = z$$  An example of this would be following:   $$\frac{345(345+1)}{2} + \frac{234(234+1)}{2} = 87180$$  Is it possible to determine if $87180$ originated from the addition of two triangular numbers without recursively going back and plugging every possibility for $x$ and $y$?
62075
I know perfectly well the Peano axioms, but if they were sufficient for defining $\mathbb N$, there would be no controversy whether $0$ is a member of $\mathbb N$ or not because $\mathbb N$ is isomorphic to a host of sub- and supersets of $\mathbb Z$.  So it seems that mathematicians, when specifying $\mathbb N$, rarely mean the set of natural numbers according to its definition but rather some particular embedding into sets defined by different axiomatic systems.  Doesn't this sort of misstate then the role of the Peano axioms as the defining characteristics of $\mathbb N$?  It seems like most of the time the Peano axioms are not more than a retrofit to a subset of $\mathbb Z$ or other sets defined by a different axiomatic system with one or several axioms incompatible with the Peano axioms.
48792
I was wondering why do some people use redundant axioms in definitions?  If you just expand $(a+1)(b+1)=(a+1)b+a+1=ab+b+a+1$ $(a+1)(b+1)=a(b+1)+b+1=ab+a+b+1$. Hence, $ab+a+b+1=ab+b+a+1$, then cancel ab and 1. Then, you get it's commutative for free.   Why do we then assume it?   Also, is it even stronger. If we drop the condition that 1 is in R, then can we still deduce a+b=b+a? As clearly the argument just done falls apart if you haven't got 1 in R.
84369
I do not even know if the question makes sense.  The point is rather simply. If I have the sum of all natural numbers,  $$\sum_{n\in \mathbb{N}}n$$  this is clearly "equal to infinity".  But since almost a century ago, we know that there are (a lot of) different "infinities".  So, is this sum equal to something countable or something bigger?  I tried to look for references, but couldn't find anything and, since I am not an expert in Logic, Set Theory or Foundations of Mathematics, I thought that it would be good to ask here.  Thanks in advance, Davide  PS: This question is about sum of cardinals. It is not about Calculus.
74710
For each $x,y,z∈\mathbb{N}$, if $x&lt;y$ then $x+z&lt;y+z$  I know how to solve this if it was $x=y$ then $x+z=y+z$ (which my professor said I need to do) Is there a definition that says I can change the $&lt;$ into a $=$ and then change it back to $&lt;$ at the end?  I have a paper full of Peano's axioms that I'm able to build off of but I don't know what one I should be using.  Would For each $x,y∈\mathbb{N}$, $x&lt;y$ or $x=y$ or $x&gt;y$  be able to be used to change the $&lt;$ to $=$?
4721
So I answered this question:  I believe my answer is correct, however one thing I couldn't explain fully, and which is bugging me, is why the rationals $\mathbb Q$, integers $\mathbb Z$ and naturals $\mathbb N$ all have cardinality $| \mathbb Q |=| \mathbb Z | = |\mathbb N| = \aleph_0$, when $\mathbb N \subsetneq \mathbb Z \subsetneq \mathbb Q$.  The basic proof of the other question, from my answer, is that for any $r$ in $\mathbb Q$ there is a single function $f(r)$ such that $f(r) \in \mathbb Z$, and the same holds between $\mathbb Z$ and $\mathbb N$. Because there is this 1:1 transformation possible, there must be the same number of numbers in the three sets, because otherwise there would be a number in one set for which the bijection could not produce a number of the other set, and this is not so.  Now, Belgi explained why $1&lt;2$ in the other question by defining the value 0 as the cardinality of the empty set $\emptyset$, 1 as the cardinality of a set of sets containing only the empty set, and 2 as the set of sets containing the empty set and a set containing the empty set, then proceeding as follows:     Now that we have defined the natural numbers we can define when one number   is smaller than another. The definition is $x&lt;y$ is $x\subset y$   and $x\neq y$ .      Clearly $\{\emptyset\}\neq\{\emptyset,\{\emptyset\}\}$ and $\{\emptyset\}\subset\neq\{\emptyset,\{\emptyset\}\}$   So this a proof, by definition, of why $1&lt;2$.   ... however, by the same definition, because $\mathbb N \subsetneq \mathbb Z \subsetneq \mathbb Q$, then $| \mathbb N | &lt; | \mathbb Z | &lt; |\mathbb Q|$ and thus a maximum of one of these quantities can be the ordinal quantity $\aleph_0$.
54806
In Peano arithmetic (first order), we first define natural numbers using a successor function and Peano axioms, then we define addition (and multiplication), and then, we define inequality, as:  $a\leq b\leftrightarrow\exists c\left(a+c=b\right)$   Is there any way to define inequality first, directly from the successor function and Peano axioms? (I mean, if I don't need addition for my purpose, why define it?).
27580
Reading  Wikipedia page I found this definition:     A real number $a$ is first-order definable in the language of set   theory, without parameters, if there is a formula $\phi$ in the   language of set theory, with one free variable, such that $a$ is the   unique real number such that $\phi(a)$ holds in the standard model of   set theory.   A few lines later we find the statement:     Assuming they form a set, the definable numbers form a field....   But, since they are a subset of the set of real numbers, why shouldn't they be a set?   Coming back from this question to the definition, I've another doubt: if ZFC is consistent this does not means that every set-theoretic object (and so any real number) is definable in some model?  Reading the whole article does not lessen my confusion .... and the ''talk'' is too difficult for me and it does not help.  More generally, this Wikipedia article is "disputed" ad has many "!" So I doubt that it is not reliable.  A brief surf on the web give me many pages on this subject but I've  found nothing that I can understand and give a response to the question: we can well define what is a definable real number?
40663
I realize this is possibly a copy of another question (The Axiom of Choice and definability) but I would appreciate an explanation with less set theoretical explanations if that is at all possible.  I recently took the opportunity to look up the axiom of choice (AC). Wikipedia has both a formal and an informal definition.  Based on my shortcomings in set theory I have to rely on the informal version: “given any collection of bins, each containing at least one object, it is possible to make a selection of exactly one object from each bin”.  It seems that this can be used to define an endless series of decimals such as an arbitrary number x between 0 and 1, in term corresponding to an arbitrary positive real number through such a function as f(x)=1/x-1, i.e. it can be used to define any real number (assuming a real number is an infinite decimal series).  A beauty spot is the cases of 0.99999… and 1.00000… - discussed in a number of questions on the site. It points to several issues:  I argue, albeit in a roundabout way, that for a choice involving the AC to have any value then the number received through the choice must also allow for existing mathematical operations, i.e. 2*0.11111…=0.22222…  as discussed in another question. (“You're also assuming that you can multiply 0.9999... by 10 and get 9.9999... (or, rather, that arithmetic with infinite decimals works normally), which is not at all unreasonable to assume. – Isaac Jul 20 '10 at 20:59”) i)  0.99999…must be equal to 1.00000… for the same reason. ii) The fact that 0.99999…=1.00000…can challenge the one-to-one correlation between the real numbers, as we view as being without these doublets. However it can be disregarded as   I was assured in a recent question, in that removing a countable set such as 0,99999…, between the endless decimal series and 0.099999…and 0.0099999…  etc. from a uncountable set would not change the one-to-one correlation to numbers without this anomaly. (I was referred to i.a. the Cantor–Bernstein–Schroeder theorem).  Reasoning in this manner I perceive the AC to be identical of postulating - or at least allowing for the definition of – real numbers. My question is whether or not this is the case.  Edit much later: I should add that I primarily am looking at the problems of defing invividual real numbers, so called pointwise definition, rather than definition of the whole set of reals. I didn't realize at the time that these are two different issues.
83486
Besides the axioms of extensionality and regularity, all of the axioms of ZFC either postulate the existence of a set or give a method for generating new sets from existing sets.  Extensionality then gives an equivalence relation on sets which allow us to confirm our definitions we create are well defined.  Regularity then gives a rule that all sets must satisfy.   From all the other axioms (besides regularity) we can start generating sets and decide which generated sets are equivalent. For all generated sets they either will satisfy the proposition in the axiom of regularity or they will not.  If all sets generated do satisfy the axiom of regularity then what is the point of the axiom of regularity? It adds no additional structure to the sets of ZFC.  If there exists a generated set that does not satisfy the axiom of regularity then aren't the axioms inconsistant?  I believe my misunderstanding stems from a misunderstanding with how logic/formal systems work. Any clarification would be greatly appreciated.
117817
I am curious about why the existence of inaccessible cardinals cannot be proven in ZFC.My intuitive proof is:  Suppose we can prove there is an inaccessible cardinal ϰ,then by definition we must show that for any cardinals b&amp;c which satisfying b&lt;ϰ and c&lt;ϰ,then b^c&lt;ϰ.Then if we consider the set S which is the union of all the cardinals less than ϰ and consider the power set of S--P(S).It follows that |P(S)|>=ϰ.This implies we can proof the existence of inaccessible cardinals under ZFC iff we can disprove it.Therefore we cannot prove the existence using ZFC.  Is this the intuitive idea behind the proof?  I only have learnt a little bit in set theory by self-studying.But I don't know any thing about mathematical logic.So I don't know the thing such as the second incompleteness theorem.This statement was found in my set theory book but without prrof it.So any comment and improvement of my work is welcome!
27882
For natural numbers $a$ and $b$, show that $a \Bbb Z + b \Bbb Z = \gcd(a, b)\Bbb Z $  I just basically said that the gcd of a and b { written as C } obviously divides aZ and bZ, therefore it can be rewritten as  CdZ + CeZ -> C[dZ + eZ] -> CZ[d+e] where Cd=a and Ce=b.   if a and b and C are natural numbers, then  so are d and e. the sum of d and e is an integer, that when multiplied by an integer produces an integer. therefore  CZ[d+e] = CZ, so aZ + bZ = GCD(a,b)Z  { Where C is the gcd(a,b) }