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48

Here's a non-mathematical one: "is the father of". You are not your own father. You are not your father's father. Your father's father is not your father.


14

Think of three points $u, v, w$ with relation $R = \{(u, v), (v, w) \}$. So $u$ is related to $v$ and $v$ is related to $w$. This is not reflexive since $(u,u) \notin R$, not symmetric because $(v, u) \notin R$ and not transitive because $(u, w) \notin R$.


6

A simple one is : Define $R$ on $\mathbb{Z}$ by $(x,y)\in R$ if and only if $x-y=10$.


5

On $\mathbb{N}$, consider $$a \sim b \iff a +2b = 5,$$ then $1\nsim 1$, $3\sim 1$ but $1 \nsim 3$, $3\sim 1, 1 \sim 2$ but $3 \nsim 2$.


5

What beats what in Roshambo or "Rock, Paper, Scissors" is such a relation. not reflexive: rock does not beat rock. not symmetric: rock beats scissors, but scissors does not beat rock. not transitive: rock beats scissors and scissors beats paper, but rock does not beat paper. The same is true of "Rock, Paper, Scissors, Lizard, Spock".


5

Daoud is not richer than Fatma.


5

Let $A=B=C=\{\emptyset\}$. Then $$ (A\times B)\times C=\{((\emptyset,\emptyset),\emptyset)\}$$ whereas $$ A\times (B\times C)=\{(\emptyset,(\emptyset,\emptyset))\}$$ This looks different. If we use the definition $(x,y):=\{\{x\},\{x,y\}\}$, we see that indeed $$\begin{align} (\emptyset,\emptyset)&=\{\{\emptyset\}\},&\text{ hence}\\ (A\times ...


5

Define $$R = \{(m, n)\mid m, n\in \mathbb N \;\text{ and }\;m, n \text{ are both even}\}$$ Then, for every odd integer $t\in \mathbb N$, $(t, t)\notin R$, hence reflexivity fails. Recall that for reflexivity to hold, it must be the case that for every $n\in \mathbb N$, $(n, n) \in \mathbb N$. No exceptions. However, it is easy to verify that $R$ is both ...


4

$\varnothing$. It's transitive and symmetric by vacuous arguments. But it is not reflexive since $(n,n)\notin\varnothing$ for all $n\in\Bbb N$!


4

An $n$-ary (or $n$-place) function symbol is an expression which combines with $n$ terms to form another term. (If you like, think of the symbol coming with $n$ slots to be filled in; and when the slots are filled, we then get a complete term.) For example, in arithmetic the function symbol '$+(\ldots,\ldots)$' combines with the two numerals '$2$' and '$4$' ...


4

A relation over geographic regions: New York State is within the bounds of the United States. New York City is within the bounds of New York State. Manhattan is within the bounds of New York City. Manhattan is within the bounds of Manhattan; where else would it be? Because it is within New York City, it must be within the bounds of New York State, and ...


4

How about: "is the square of", defined on the set of positive integers? In other words, $$a \sim b \iff a=b^2$$ This relation is not reflexive (most numbers are not their own square), not symmetric (if $a$ is the square of $b$ then in most cases $b$ is not the square of $a$) and not transitive (if $a$ is the square of $b$ and $b$ is the square of $c$ then ...


4

Your proof of reflexivity is incorrect. To show that $xRx$, show that $x-x$ is rational (and it is because $0 \in \mathbb Q$). Your proof of symmetry and transitivity should make some reference to the fact that $\mathbb Q$ is closed under negatives and addition, respectively. Now, your discussion of the equivalence classes of $0$, $1$, and $\sqrt{2}$ are ...


3

Your "reflexive" proof is wrong. It should read $$\forall x\in\mathbb{R}, \quad x-x=0\in\mathbb{Q}\implies xRx$$ I'm not sure if this was implied, but for transitivity the correct argument is $$\begin{align}xRy, \;yRz&\implies x-y,y-z\in\mathbb{Q} \\ &\implies (x-y)-(y-z)\in\mathbb{Q} \\ &\implies x-z\in \mathbb{Q} \\ &\implies ...


3

Minimal example: $X = \{0,1,2\}$ $R = \{(0,1),(1,2)\}$


3

Here's a somewhat artificial one, but how about: $$ x\sim y \iff y-x = 0 \text{ or } y-x = 1 $$ Note $1\sim 2$ but $2 \not \sim 1$, and $1 \not\sim 3$ even though $1 \sim 2$ and $2 \sim 3$ .


3

Assuming you have $\ge$ defined (along with its total order), I would proceed as follows (outline/sketch): Antisymmetric $$(a\le b) \wedge (b\le a)$$ $$(b-a)\ge 0 \wedge (a-b) \ge 0$$ $$(b-a)\ge 0 \wedge -(b-a)\ge 0$$ Now, we know that $x\ge 0 \wedge -x \ge 0 \implies x = 0$ (I'm assuming this can be worked out from whatever definition of $\ge$ you're ...


3

The diagram is incomplete, the picture of me close to the top of the ladder has been left out. I should have known better than to use a $13$ foot ladder. Let $x=x(t)$ be the distance of the foot of the ladder from the foot of the wall. Let $y=y(t)$ be the distance of the top of the ladder from the ground. By the Pythagorean Theorem, we have ...


3

The partitioning of $A$ into subsets/cells induces (and is induced by) the following relation $$R = \{(1, 1), (2, 2), (3,3), (4, 4), (5, 5), (6, 6), (1, 3), (3, 1), (1, 5), (5, 1), \\ (3, 5), (5, 3), (4, 6), (6, 4) \}$$ such that for any $a, b\in A$, $(a, b) \in R$ if and only if $a$ and $b$ are in the same subset/cell of the partition. N.B. ...


3

As $|A| = 6$, $|\mathcal P(A)| = 2^6 = 64$, so there are at most 64 distinct sums. Now, note that $12 + 11 + 10 + 9 + 8 + 7 = 57$, and so no subset of $A$ can sum to more than $57$, by distinctness of the elements of $A$. Further, as each element of $A$ is positive, no subset of $A$ can sum to less than $0$. Therefore, the possible sums for subsets of $A$ ...


3

I would prefer to speak about a "functional relation" here rather than a "function", because the latter wording implies that the only thing you're planning to do with it is to apply it to inputs, and here you're doing something quite different. That being said, here are some comments on your observations: $f(x)=−x$ is total and symmetric, In general a ...


3

$f(x)=x^2\sin x$ and $g(x)=x.$


2

There are $2^6=64$ possible subsets of $A$, the sums of all of them is less than $12+11+10+9+8+7=57$. There is not enough room for the sums to be all distinct.


2

Let $$f(x)= \begin{cases} x^2 & 2n\le x<2n+1 \\ 1 & 2n+1\le x < 2n+2\end{cases}$$ and $g(x) = x$.


2

Use the Cantor-Bernstein theorem. Then you first have to prove that there are at least $\mathfrak c$ countably infinite subsets of $\mathbb R$, and then that there at most $\mathfrak c$ of them. The first one is easy -- the sets $\mathbb{Z}\cup\{x\}$ for $x\in(0,1)$ are $\mathfrak c$ different countably infinite subsets. For the second one, you need a ...


2

A relation on $A$ is a subset of $A\times A$. A reflexive relation, in this case, is any relation $R$ such that $(1, 1) \in R$ and $(b, b)\in R$ and $(\varnothing, \varnothing) \in R$. Any subset of $A\times A$ containing those three ordered pairs is reflexive. One such relation is given by $R_1 = \{(1, 1), (b, b), (\varnothing, \varnothing)\}$. Another ...


2

Good observation skills! Let's chase definitions and then be happy :) Let me write $n = 10, m = 3$ and let your matrix be denoted $A$. Then we calculate! $$ \begin{align} f &= \|A\|_F \;\; \text{ your defn} \\ & = \sqrt{\sum_{i=1}^n\sum_{j=1}^m |A_{ij}|^2}\;\; \text{ defn of Frobenius norm} \\ & = \sqrt{\sum_{i=1}^n\sqrt{\sum_{j=1}^m ...


2

We have $\def\R{\mathrel R}\def\S{\mathrel S}$ $$ R \circ S = \{(x,z) \mid \exists y. \, x\S y\land y \R z\} $$ hence $$ R \circ S = \{(3,6), (6,3), (9,2)\} $$


2

You know that $x^2+y^2=13^3.$ Now, if the ladder moves, we $x$ and $y$ are functions of time. Taking derivatives in the equality above we have $2x(t)\frac{dx}{dt}+2y(t)\frac{dy}{dt}=0,$ or, equivalently $x(t)\frac{dx}{dt}+y(t)\frac{dy}{dt}=0.$ If you parametrize $x(t)=12+3t$ ($12$ is the initial distance from the wall and $3$ the velocity) you get that ...


2

Given a number $x$, we have $yRx \iff y=x+n$ for some integer $n$. So the equivalence class of $x$ is $x+\mathbb{Z}$, that is, {$...,x-2,x-1,x,x+1,x+2,...$}. The set of all such equivalence classes is the partition formed on $\mathbb{R}$. Now, $xRy$ iff $x$ and $y$ have the same decimal part. For example, $1.23R5.23$. So the distinct equivalence ...



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