# Tag Info

2

Learn by playing with it. Take a piece of paper and start drawing! Put a $1$ at the bottom and just start drawing arrows. You put an arrow whenever one number evenly divides into another, for example$$1\rightarrow 2\rightarrow 4 \rightarrow 8 \rightarrow 16$$ should be in your drawing. If you manage to draw all the possible arrows you have completely ...

0

I have a robot arm with a gripper. I know the gripper pose (relative to the robot base coordinate system) at any moment. $T_{\text{base}:\text{gripper}}$ At startup, I record the pose of the gripper and set this as the original pose O $T_{\text{base}:\text{gripper}_O}$ Then, the gripper moves to its new pose N, again in the robot base ...

2

A relation $R$ is said to be transitive if for all $a,b,c$ in its domain, $aRb$ and $bRc \Rightarrow aRc$ In your case, the relation would be transitive if, for all $x,y,z \in \mathbb N$, $x+y=10$ and $y+z=10$ implied $x+z=10$, which is clearly not the case.

0

The relation E={{1,2},{3,4}} already disproves the first point, it seems, since {1} and {2,3} are related by F and {2,3} and {4} are related by F but {1} and {4} are not.

2

It is only a preorder. If $u$ is any non-trivial unit, and $x$ is any non-zero element, then $x|ux|x$, but $x \neq ux$.

0

Let $F:X\to Y$ be constant function: $\forall x\in X, F(x)=c$, where $c$ is a constant element of $Y$. Let $A$ and $B$ be subsets of $X$, such that $A-B$ is not empty. Then $$F(A)−F(B)= \varnothing,$$ while, $$F(A-B)=\{c\}$$

0

Hint: A pair $(x,y)$ is in $R_1\circ R_2$ precisely if you can find a $z$ with $(x,z)$ in $R_2$ and $(z,y)$ in $R_1$ (there may be more than one such "enabling" $z$ for a given $(x,y)$, but that doesn't matter--there just has to be at least one). For example, $(\boxed{a},\boxed{c})$ is in $R_1\circ R_2$ because $(\boxed{a},b)$ is in $R_2$ and ...

0

I find much easier to reason this kind of problems with a drawing of the relations in hand. Here the arrows $x\to y$ means $(x,y)\in R$ From the digraph, it is clear that the trichotomy property holds, since every pair of nodes has one and only one edge between them. Transitivity also holds, since every time we have (x,y) and (y,z) we also have (x,z) ...

4

It's in order to maintain consistency with function composition and application. For functions $g=S, f=R$, the notation for composition of relations $S, R$ should be the same as that of composition of $g,f$: $(g\circ f)(x) = g(f(x))$. If composition were defined so that the first function $f$ applied (or relation $R$) appears "leftmost", then we would want ...

0

An easier way to deal with this relation is to define $g(x)=x^3+x$ and say that $xPy\iff g(x)\ge g(y)$. Which immediately shows that $P$ is indeed transitive but not symmetric.

1

Your resolution is unclear and incorrect in some places. For reflexivity, that's not what you have to prove (which is also wrong! as, if $x=0.5$ then $x^3\not\ge x$), you have to show that $x^3-x\ge x^3 -x$ which is obviously true (as the expressions are equal). Also, saying "$xRy$ does not equal $yRx$" doesn't make much sense, you should use a word like ...

3

Note that if $x=0$ then, as we're given $y=7x$, so we get $$y=7\times 0=0$$ Therefore we have an initial point on the plane: $A=(0,0)$. The same story can be done for $x=15$ to find another point which is $B=(15,7\times 15)=(15,105)$. Now draw the line $y=7x$ first and then bold the line between $A$ and $B$.

1

$x\cap B$ is a subset of $B$ and $B$ has $8$ subsets. For each of these subsets there is one equivalence class. So the required number is $8$.

0

This sound's a lot like linear algebra. I am assuming that you want an approximate function f(x,y) = z(where x,y are the first numbers and z is the thired). If you where looking for the connection between tow numbers you would do something like assuming f is polynomial of whatever degree. To force your situation to work with the one dimensional problem you ...

0

Though the other answers give a good impression of why $x\sim y\iff 4|(5x+3y)$ defines an equivalence relation, it needs to be said that this is not exactly what is asked in the OP. The OP literally asks: Show that $R = \{ (x,y) \in \mathbb{Z} \times \mathbb{Z} : 4 | (5x+3y)\}$ is an equivalence relation. Then one should say: $R$ is not an ...

1

In general, let $a$ and $b$ be integers and $m$ a positive integer. Then, $$R:=\big\{(x,y)\in\mathbb{Z}\times\mathbb{Z}\,|\,ax+by\text{ is divisible by }m\big\}$$ is an equivalence relation on $\mathbb{Z}$ if and only if $m$ divides $a+b$. Furthermore, if $m$ divides $a+b$, then the set of equivalence classes of $\mathbb{Z}$ under $R$ is precisely ...

1

Your proofs for reflexivity and stymmetry are somewhat correct. I say "somewhat" because you mean to say the right thing, but I'm inclined to say you natation is incorrect. More on the notation at the end of this answer. To prove transitivity need to do something like this: $$(x,y)\in \rho\wedge(y,z)\in\rho\implies x-y=3k_1\wedge y-z=3k_2\implies ... 1 A relation is simply any declaration that a R b. A relationship is a collection of possible relations. They can be as "small" as "no term is related to anything else" or as large as "everything is related to everything else", which is the same as "2 R 2, 2R3, 3R2, 3R3. As claiming a R b is really just another way of stating an ordered pair (a,b) ... 1 If f:\{2,3\}\to \{2,3\} is a function then f(2) can equal 2 or 3. For each choice of f(2) there are 2 choices for f(3). Thus 2x2=4 functions.There are 16 subsets of A=\{2,3\}\times \{2,3\}. Subsets of A are what binary relations on \{2,3\} are. 0 Let X be any finite set. Then a relation is a subset of X \times X and it is reflexive if it contains \Delta_X = \{(x,x) \:|\: x \in X\}. Thus we are looking for the number of subsets of X \times X containing \Delta_X and this is the same as looking for the number of subsets of (X \times X) - \Delta_X. This set has |X|^2 - |X| elements and so ... 2 Let d(n) be the number of digits of a positive integer n. Then$$d(n)= 1+ \lfloor \log_{10}(n) \rfloor$$This uses the floor function and the base 10 logarithm. The slow growth rate you mentioned is because of the fact that logarithms, which are the inverse of exponents, grow very slowly. You should read that article on logarithms - they are extremely ... 4 Because of R, we must have 1=2=4=5=6, 7=8, and 3=3. So there are at most three equivalence classes. You can also combine them in various ways, e.g. 1=2=4=5=6 and 3=7=8. 1 Why would, for example (1,3) need to be in the relation? The definition is: R is symmetric if (and only if) it holds for all a and b that if (a,b)\in R then (b,a)\in R which we can unfold to R is symmetric if (and only if) all of the following are true: If (1,1)\in R then (1,1)\in R If (1,2)\in R then (2,1)\in R ... 3 No, the number of equivalence classes is finite, because there are only finitely many propositional variables, namely p,q,r. Any propositional formula in P represents (or induces) a truth function — a function from n tuples of truth values to truth values. The truth table of a formula defines this truth function. The formulas of P define 3-ary truth ... 2 Hint: in this example, you can test to see if two propositions are equivalent by computing their truth tables. If the final answers are identical (intermediate working being irrelevant), then they are equivalent; if the final answers are different, they are not. So the question is in effect the same as asking: how many possible final answers are there for ... 0 Assuming that x \in \mathbb{Z}. The binomial coefficient \binom{7-x}{x} is defined for 0 \leq x \leq 7 and x \leq 7-x. Thus x \in \{0,1,2,3\}. Now you can find possible values of this coefficient. And this gives the size of set A. Any reflexive relation R should have pairs of the form (x,x) for all x \in A. So if |A|=n, then R should ... 1 Towards showing that S/\sigma^* is a group: can you show that (the equivalence class corresponding to) aa^{-1} is an identity in S/\sigma^*? More precisely: that baa^{-1}\sigma b for any b? Towards showing that \sigma^* is the least group congruence: suppose S/\rho is a group. Then show that if a\sigma b, the equivalence class corresponding ... 1 Let us consider induction on n of a slightly stronger statement. If R is a transitive binary relation that contains a cycle of length n, then xRy for any x and y in the cycle. In this case we will say R is full with respect to the cycle. Base case n=1 and n=2 are trivial. Inductive step. Assume that whenever there is a cycle of length ... 0 If this is an exercise, I would say no, you need to give more details. It was a good idea to start with the second version, but now writing the full solution takes less space that writing your question... The interesting part is transitivity. You have to consider three functions f, g and h such that f \mathop{R} g, g \mathop{R} h. Thus there ... 0 It's much easier if you consider the function f\colon \mathbb{R}^2\to\{-1,0,1\} defined by$$ f(x,y)=\operatorname{sgn}(y+\pi x) $$Then you can rewrite your relation by stating that$$ (x,y)\mathrel{\rho}(a,b) \quad\text{if and only if}\quad f(x,y)=f(a,b)  whereby proving that it's an equivalence relation is very simple. Moreover, the equivalence ...

0

reflection : (x,y)~(x,y) ; it is obvious because of "=" sign. symmetry: if (x,y)~(a,b) than it has to be (a,b)~(x,y) you don't change the position of x,y because you work with ordered pair in R2. transitivity: if (x,y)~(a,b) and (a,b)~(c,d) than it has to be (x,y)~(c,d).This is obvious again because of "=" sign. Function sgn() gives you 1/0/-1 so it is ...

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