Is it correct to say that   "A function or a map or a mapping is a binary relation such that ..."
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I wonder if there is any difference between mapping and a function. Somebody told me that the only difference is that mapping can be from any set to any set, but function must be from $\mathbb R$ to $\mathbb R$. But I am not ok with this answer. I need a simple way to explain the differences between mapping and function to a lay man together with some illustration (if possible).  Thanks for any help.
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May somebody explain this in other words?   I know this is an akward question, but I really want to know the answer, it's just curiosity, I'm learning all that stuff of low level programming and learn binary system is one of the first steps to accomplish the objective.
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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...
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I want to be able to prove that for all functions f, that $a = b \to f(a) = f(b)$.  That this is true is obvious, but I'm not sure how to formally prove it using only the rules of inference in first-order logic (using natural deduction).  I'm guessing I need to reference a formal definition for a function, where $f \subset A \times B$ and satisfies:  $$ (\forall x \in A)(\exists y \in B)((x,y) \in f \wedge (\forall z \in B)((x,z) \in f\to y=z))$$  However, I don't have any clue where to begin with this.  Can anyone help?  Please only give hints/answers using natural deduction as the deductive system.
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A binary operation $*$ is defined on $\mathbb{Q}$ such that $a*b=ab$.  So, it is just simple multiplication.   I found out that this is a group with identity $=1$ and inverse is reciprocal of element "$a$" in $G$.  But, my answer key says that this is not a group because axiom 3 (inverse) fails.  I can't see why it is not a group.  Is my answer key wrong or am I wrong?
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$\def\p{\mathrel p}$If $\p$ is a relation on a set $A$, define $\p^2$ by $a \mathrel{\p^2} b$ if and only if there exists $c$ with $a \p c$ and $c \p b$.   If $p$ is reflexive/symmetric/transitive does $p^2$ have the same properties?  I'm not even sure how to start this, I assume I would need to use the $a$ related to $c$, $c$ related to $b$ somehow?
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For sets $A,B$, let $|A|\leq^*|B|$ say that there exists an onto map $f:B\rightarrow A$ or $A=\emptyset$.   My question is, is $$\forall A,B(|A|\leq^*|B|\longrightarrow |A|\leq|B|)$$ equivalent to the axiom of choice?  Thanks
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Please guide me in this problem. I am confused about whether its asking that having the relation $z&gt;0$ does not satisfy the order axioms.  Any help would be really appreciated.  Thanks!
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Let $r$ be a relation on $A \times A$ such that $(a,b) r (c,d) \iff ad = bc.$ How can I show that this relation is transitive, ie. $(a, b)r(c,d)$ and $(c,d)r(e, f) \implies (a,b)r(e,f)$?  I tried to say that $(a,b) r (c,d)$ means that $c=ka$ and $d=kb,$ $k$ a coefficient, such that $ad = bc \iff akb = bka,$ and going on from there, but I'm not sure this is valid for all values of $a, b, c,$ and $d$. How can I show the transitivity of the relation?
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I am very new to proofs so please excuse any trivial errors.  In lecture we were told that:     A set $\mathbb S$ is called finite if there exists a one-to-one    mapping (bijective mapping) between $\mathbb S$ and the set $\mathbb  N_n=\{0,1,2,3,...,n-1\}$ for some natural number $n$. A set is called    countable infinite if there exists a bijective mapping between    $\mathbb S$ and the set $\mathbb N=\{0,1,2,...\} $ of all natural    numbers   I want to answer the following questions:  1. Show that the set $\mathbb S=\{a,b,c\}$ is countable  Does countable mean countable infinite? An attempt from another post on this site: For all triples $(a,b,c)$ there exists an according sum $S=a+b+c \implies ...$ But wouldn't the sum be a surjection between $\mathbb S$ and $\mathbb N$ and not a bijection?  2. Show that the set of integer numbers $\mathbb Z=\{0,\pm1,\pm 2,...\}$ is countable infinite  Before I continue with this proof I have a question about mapping. The definition says that "if there exists a bijective mapping...". In this context, can a function $f(x)$ be considered a mapping? If yes, then can't I just define myself some map $$f:\mathbb Z\to \mathbb N$$  and let $\space f$ be: $$x \mapsto \lvert x \rvert$$  which would then automatically imply that $\mathbb Z$ is countable infinite? (or wouldn't it)?  3. Show that $\mathbb S=\mathbb N \times \mathbb N=\{x |x=(p,q),p,q \in \mathbb N\}$ is countable infinite.   I am not sure about this one but if my "intuition" from 2) turns out to be correct I can maybe define some sort of norm and show that it is a bijection between $\mathbb S$ and $\mathbb N$
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"Consider a language with a single object constant a, a single unary function constant s, and two unary relation constants p and q. We start with the premises shown below. We know that p is true of s(a) and only s(a). We know that q is also true of s(a), but we do not know whether it is true of anything else.  ¬p(a) p(s(a)) ∀x:¬p(s(s(x))) q(s(a)) Prove ∀x.(p(x) ⇒ q(x)). Hint: Break the problem into two parts - first prove the result for s(x), and then use that intermediate conclusion to prove the overall result."  I know I have to reach AX:(p(s(X)) => q(s(X))) and p(a) => q(a) first, but I'm struggling getting started. I also don't know if my first steps are correct. Any help? Thank you.
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Assume that my definition of a Boolean algebras is the following one: I have a set $B$ with two binary operations $\vee$ and $\wedge$ which both satisfy the commutative, associative and distributive laws. Both the operations have an identity element ($0$ and $1$) and for each element $b\in B$ there exists a complement (i.e. an element $b{'}\in B$ such that $b\vee b{'}=1$ and $b\wedge b{'}=0$).  Now I found the following definition of Boolean Algebra. Let $B$ be a set on which we define a binary operation $\cap$ and an unary operation ${'}$. We also require that those operations satisfy the following rules:  1) $x\cap y=y\cap x$ for each $x,\, y\in B$  2) $(x\cap y)\cap z=x\cap (y\cap z)$  3) $x\cap y{'}=z\cap z{'}$ iff $x\cap y=x$  Are those equivalent definitions? I was trying to work the thing out in the following way.  a) First we may put $0:=z\cap z'$ by property 3).  b) We may define $x\cup y:=(x{'}\cap y{'}){'}$ and put $0{'}=:1$ and we may prove the commutative law for $\cup$.  Anyway I cannot prove the associativity of $\cup$ or either the distributivity or even the fact the $0$ and $1$ are identity elements.
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For nonempty sets $A$ and $B$ and functions $ f: A \to B$ and $g: B \to A$, suppose that $ g \circ f = i_A$. Prove that if $f$ is onto, then $g$ is $1-1$.   Here is what I started, I don't think it's correct.   Suppose $f$ is onto  i.e for each element $b \in B$ there is an element of $a\in A$, $f(a)=b$. (definition of onto).   Let $y,\ z$ be any two elements of $B$.Then $y=f(a)$ and $z=f(b)$ for some $a,b \in A$.   If $g(y)=g(z)$, then $g(f(a)) = g(f(b))$,which implies that $a=b$, so $f(a)=f(b)$, so $y=z$.   Thus, whenever $g$ maps $y$ and $z$ to the same element in $A$, $y=z$, so $g$ is one-to-one.   PS: New to proof writing, any constructive criticism is much appreciated.
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The map $\phi: \mathbb{Z} \to \mathbb{Z}$ defined by $\phi (n) = n + 1$ for $n \in \mathbb{Z}$ is one to one and onto $\mathbb{Z}$. Give the definition of a binary operation $*$ on $\mathbb{Z}$ such that $\phi$ is an isomorphism mapping   a) $\langle\mathbb{Z},+\rangle$ onto $\langle\mathbb{Z}, *\rangle$    I absolutely do not understand this. What is the solution doing? Why are we starting out with $m*n$? And why do we have $\phi(m-1) * \phi(n-1)$?
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Possible Duplicate:          I'm studying the Atiyah's commutative algebra book and I realized that in the beginning of the book, the author says as one of the conditions to a map be a homomorphism is $f(1)=1$, I would like to know if I can left this condition behind, i. e., if $f(x+y)=f(x)+f(y)$ and $f(xy)=f(x)f(y)$ we can have $f(1)=1$.  Thanks
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For the surjectivity part, I showed that if $(a,b)\in I$ then $(a,c)\in R_{1}$ and $(c,b)\in R_{1}$ and $(a,c)\in R_{2}$ and $(c,b)\in R_{2}$ for some $c\in S$. Now for arbituary $(x,y)\in R_{1}$ we have $(x,z)\in I$ which implies that $(y,z)\in R_{2}$. But also $(y,z)\in R_{1}$, so $R_{1}$ is surjective. Similar argument for $R_{2}$  And for Injectivity, I've tried to show that if $aR_{1};R_{2}c$ and $bR_{2};R_{1}c$, then $a=b$ but I am not entirely sure what $a=b$ means in this context. Do I have to show that there is an equivalence relation containing (a,b)?  edit: $R_{1};R_{2}$ is a composition of two relations and $I=R_{1};R_{2}$
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I just began the topic of functions in my Mathematics textbook and in the first paragraph itself, two conditions about a relation being a function were mentioned.  They were :   1) For every a∈A, there exists b∈B such that (a,b)∈f     I think  that is another way of saying that both the input and output must lie in the given sets 2) If (a,b)∈f and (a,c)∈f, then b = c     And I think that this is another way of saying that an input has a unique output   Let me know if I have misunderstood these statements.  Now, let's think of the square root function...  The square root of 4 is ±2, which means that there are two outputs 2 and -2 for one input 4, so there are two images of the same input, so it defies the second condition that states that a function must have a unique output/image of some input  Let R represent the set of real numbers, then if square root is a function, it means that both the input and output must lie in the set R. In the case of square root of -1, the output will be i (iota) which is not a member of the set R but is a member of C (the set of complex numbers), this defies the first condition that makes a relation a function.  Please let me know if I have misunderstood some concept related to functions and if square root is really a function or not.  Thank You...