On the Wikibooks page , the author included $0 \neq 1$ as the axiom that relates addition, and multiplication.   But, what I see the most is just the distributive law. What does $0 \neq 1$ mean? And how it is related to addition and multiplication? Why is it included in this list of axioms?
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The title pretty much says it all.  Of course, one answer (IMO unsatisfactory) to such questions goes something like "a definition is a definition, period."  In my experience, mathematical definitions are rarely completely arbitrary.  Therefore I figure there must be a good reason for insisting that a field be a non-trivial ring (among other things), but it's not obvious to me.  I realize that a trivial ring would make for a very boring field, but it is also a rather boring ring, and yet it is not disallowed as such.  EDIT: I found a hint of a rationale in the statement that "the zero ring ... does not behave like a finite field," but I could not find in the  for this assertion exactly how the zero ring fails to behave like a finite field.
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I have no idea how to prove it by using only what the book has talked about so far. Can anyone help? The proof shouldn't be using set theory as set theory is only mentioned in the following chapter. The proof should only make use of the addition of natural numbers, order properties of natural numbers, the trichotomy of order for natural numbers and principle of induction.  Proposition 2.2.14 (Strong principle of induction). Let $m_0$ be a natural number, and let $P(m)$ be a property pertaining to an arbitrary natural number $m$. Suppose that for each $m ≥ m_0$, we have the following implication: if $P(m')$ is true for all natural numbers $m_0 ≤ m' &lt; m$, then $P(m)$ is also true. (In particular, this means that $P(m_0)$ is true, since in this case the hypothesis is vacuous.) Then we can conclude that $P(m)$ is true for all natural numbers $m ≥ m_0$. Exercise 2.2.5. Prove Proposition 2.2.14. (Hint: define $Q(n)$ to be the property that $P(m)$ is true for all $m_0 ≤ m &lt; n$; note that $Q(n)$ is vacuously true when $n &lt; m_0$.)  Thanks!
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a+b and ab are the usual integer addition and multiplication. You can assume that this new operation forms a ring, say R is the set of integers with these operations. Then does R have zero-divisors? And Are R and Z (the integers) isomorphic?  Any help will be appreciated!
109990
Let's say we are talking about addition defined in the real numbers. Then, by induction we define $\sum_{i=0}^{0}a_i=a_0$ and $\sum_{i=0}^{n}a_i=\sum_{i=0}^{n-1}a_i+a_n$ for $n&gt; 1.\:$  Now, how do you define general associativity? I know that this has something to do with the fact that $\sum_{i=0}^{n}=\sum_{i=0}^{k}+\sum_{i=k+1}^{n}$ for $0\leq k&lt;n$, being $\sum_{i=k}^{n}a_i=\sum_{i=0}^{n-k}a_{i+k}$ by definition. But the thing is that this doesn't quite define the notion of different ways of arranging brackets, like for example $(a_0+(a_1+a_2))+((a_3+a_4)+(a_5+a_6))$.   So my question is how they formally define this process of bracketing. Think of the case when someone just tell you to prove the general associativity for the real numbers. How do they actually define this property in order to prove it? is it necessary to have one?  Look at for example this , specifically at the point where professor M. Zuker says: "Let us now assume that any bracketing of $a_1, a_2,...,a_k$ equals the standard form for $1\leq k\leq n-1$, where $n&gt;3$". But, then again my question: what is the definition of bracketing? is it actually necessary to have a definition of bracketing or this proof works whatever the definition of bracketing is?  Also I have found this paper by William P. Wardlow - A generalized general associative law-  that contains different proofs of this general associativity law. The first one, the one that he suggests as his favorite, is done by Nathan Jacobson in his book "Lectures of Abstract Algebra" Vol. 1 page 20. Looking at this proof there is one point where he says "Consider now any product associated with $(a_1, a_2,..., a_n)$...", which means "any bracketing associated with...". Then again the same question.      I hope you understand my point. If not, please fell free of asking anything related to my question.    Edit:   For clarification let's say we are talking about sum in the real numbers. Then,   1.- $(...(((a_0+a_1)+a_2)+a_3)...+a_n)$ is a representation of the formal definition by recursion, meaning $\sum_{i=0}^{n}a_i$ just as defined above:  $\sum_{i=0}^{0}a_i=a_0$ and $\sum_{i=0}^{n}a_i=\sum_{i=0}^{n-1}a_i+a_n$ for $n&gt; 1$.  2.- What is the formal definition of $a_0+(a_1+(a_2+...+(a_{n-1}+a_n)...)$ ?  3.- What is the formal definition of something like $(a_0+((a_1+a_2)+a_3))+(((a_4+a_5)+a_6)+....+(a_{n-1}+a_n))$?
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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 $\aleph_0$.
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My question is whether $x/x$ is always equal to $1$. I am mostly intersted in real numbers and particularly wonder whether $x/x$ is defined at $x=0$.  On one hand the division should simplify to $1$, on the other hand you should not be allowed to divide by zero.   I have been trying to find whether the simplification 'goes first' or whether the division causes trouble first, but it has proven impossible for me to find usefull search terms.     Note that this question arose after reading  and its first comment.
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Context: Rings.     Are $x \cdot 0 = 0$ and $x \cdot 1 = x$ and $-(-x) = x$ axioms?   Arguably three questions in one, but since they all are properties of the multiplication, I'll try my luck...
102547
Prove if $x, y ∈ R$ and $xy &gt; 0$ then either $x &gt; 0$ and $y &gt; 0$, or, $x &lt; 0$ and $y &lt; 0$ using only the field axioms.  These include the Field axioms for addition, multiplication, distribution, order axioms (trichotomy and transitivity), ordered field axioms. I'm stuck on this and not sure how to approach it. Would proof by contrapositive be a good approach in this case?
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This is my attempt to prove that, if $a &gt; 0$, then $\frac{1}{a} &gt; 0$, using just field and order axioms.  $$a &gt; 0 \implies a \cdot \left(\frac{1}{a}\right)^2 &gt; 0 \cdot \left(\frac{1}{a}\right)^2 \implies a \cdot \left(\frac{1}{a}\right)^2 &gt; 0 \implies a \cdot \left(\left(\frac{1}{a}\right) \cdot \left(\frac{1}{a}\right)\right) &gt; 0$$  By associativity of multiplication:  $$\left(a \cdot \left(\frac{1}{a}\right)\right) \cdot \left(\frac{1}{a}\right) &gt; 0$$  We know that $a \cdot \left(\frac{1}{a}\right) = 1$, because one is the multiplication inverse of the other.   So we have:  $$1 \cdot \left(\frac{1}{a}\right) &gt; 0 \implies \frac{1}{a} &gt; 0 $$  Which is what we wanted to prove.   My question are:    In the field axioms, there's no axiom that says that $0$ times another number is $0$, but I have used this property. How would you translate it to field and order axioms? I have assumed that $\left(\frac{1}{a}\right)^2$ is positive, have you noticed? Note that I need a positive number in order not to change the $&gt;$ sign. But this is not present in the field or order axioms. How would you explain this?
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I know books have slightly different ways of presenting the axioms, but I think they tend to go like this: Group is a set with a law of composition with closure that satisfies the following properties (i) associativity, i.e. a(bc) = (ab)c (ii) identity, i.e. 1a = a1 = a (iii) inverse, i.e. every element a has an inverse such that ab = ba = 1.  Michael Artin's "Algebra" made no mention of this, but Fraleigh's "A First Course in Abstract Algebra" implied that it was important that the axioms were in that specific order, and wouldn't work any other way.  Why is this?  Thanks in advance.
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Some very wise people here have just told me that  $\aleph_0 = \mathbb{N}$, i.e. that the cardinality of the set of natural numbers is just the set of natural numbers itself. Is this now the general consensus in mathematics, real analysis in particular? Or have I got completely the wrong end of the stick as usual?
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Excerpt from a mathematical proof that serves to prove by contradiction, using the Well Ordered Principle, that predicate P(n) is true for all nonzero integers n.     If n belongs to set C i.e. n >= c, then P(n) is false.      This implies the contrapositive,      If P(n) is true, n has to not be a part of set C i.e n &lt; c.   I know that the above implication is valid but why so? How should I know that n cannot take on 2 values simultaneously -- n being both >= c and &lt; c, thus P(n) being both true and false simultaneously? (Yes I am thinking Dialetheism, maybe even quantum logic but I know nothing other than that quantum logic defies certain aspects of propositional logic.) What is the basis behind a predicate (or any proposition) being either true or false? Or would the simultaneity be possible just that I wouldn't be dealing with a proposition to begin with?  Also, if you don't mind, please do let me know if you feel my understanding of concepts and terminologies (i.e. discrete, predicate, contrapositive, assume) are erroneous. I am extremely new to discrete mathematics and could really use the extra guidance.  Thank you.
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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.
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We all know that multiplication is the inverse of division, and therefore $x\div{x}\cdot{x}=x$ But what if $x=0$? $0\div0$ is undefined so $0\div0\cdot0$ should be too, but whatever happens when we divide that first $0$ by $0$ should be reversed when we multiply it by $0$ again, so what is the right answer? 0, undefined, or something else entirely?
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I know $[0,1]$ denotes the interval between 0 and 1 with the boundary. And $(0,1)$ denotes the interval between 0 and 1 without the boundary. Today, I encounter some expressions as $]0,1]$, $]-\infty,+\infty]$. What does $]0,1]$ mean? Thanks.
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The book is Foundations of Analysis by E. Landau.  I delve into analysis for the first time. Before I would encounter merely a definition of something in a textbook, but now there is something more. In Theorem 4 the author gives a definition of addition. Then he first proves that A) for any fixed $x$ there is at most one possibility to define $x + y$. Then he proves that B) for each $x$ it is possible to define $x + y$.  My question is why does he prove B) that it is possible to define $x + y$? He has already made a definition, so he brought it into existence.  Why does he need proofs here at all? It is a definition, you can define it however you want.
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From Putnam and Beyond:    The Solution is:    The only part I do NOT understand is how: $a_k + b_k = 1$ for every $k$? The problem just specifies nonnegative numbers?