chillywilly12a

2021-08-13

Discrete Math Question

Consider the relation R on Z defined by the rule that$(a,b)\in R$ if and only if $a+2b$ is even. Briefly justify your responses to the following.

a) Is this relation reflexive?

b) Is this relation symmetric?

c) Is this relation transitive?

Consider the relation R on Z defined by the rule that

a) Is this relation reflexive?

b) Is this relation symmetric?

c) Is this relation transitive?

Latisha Oneil

Skilled2021-08-14Added 100 answers

Step 1

a) Let$a=b=1$

Then$a+2b=1+2\left(1\right)=1+2=3$ , which is not even, so $(1,1)\mathrm{\neg}\{\in \}R$

Therefore, for$a\in Z,(a,a)\mathrm{\neg}\{\in \}R$

Hence, R is not reflexive.

Step 2

b) Let$a=2,b=1$

Then$a+2b=2+2\left(1\right)=4$ , which is even, so $(2,1)\in R$

Now, take$a=1,b=2$

Then$1+2b=1+2\left(2\right)=5$ , which is not even, so $(1,2)\mathrm{\neg}\{\in \}R$

Therefore, for$a,b\in Z,(a,b)\in R$ but $(b,a)\mathrm{\neg}\{\in \}R$

Hence, R is not symmetric.

Step 3

c) If$(a,b)\in R$ and $(b,c)\in R$ then a and b must be even.

Then the expression$a+2c$ will clearly be even.

Hence, R is transitive.

a) Let

Then

Therefore, for

Hence, R is not reflexive.

Step 2

b) Let

Then

Now, take

Then

Therefore, for

Hence, R is not symmetric.

Step 3

c) If

Then the expression

Hence, R is transitive.

Show that the sequence $a}_{n$ is an solution of the recurrence relation [a^n = a^n−1 + 2a^n−2 + 2n − 9 if a] if

a^n = −n + 2

Why does a magnet always attract iron, not wood?

Why is $f(n)={n}^{3}$ not an onto function?

I was doing an example in a book where it asked which of these functions are one to one, the answer in the back said for $f(n)={n}^{3}$ that it is a one to one function. Then it asked which of the functions from the previous example are onto and $f(n)={n}^{3}$ was not included in the list of onto functions.

In a later example, a question asked which of these functions is a bijection, the answer included $f(x)={x}^{3}$

This is confusing because doesn't a function have to be both an onto and one to one to be a bijection? Why would the book say it was not a onto in a previous example yet declare it to be a bijection? Is the book wrong?What is the meaning of l.h.s and r.h.s?

Showing that if n is a natural number larger than 3, then $n!>{2}^{n}$

Showing that if n is a natural number larger than $3$, then $n!>{2}^{n}$

My try:

Base Case:

If $n=4$, then $4!>{2}^{4}$

$24>16$

So, the base case is true.

Assuming $P(k)$ is true.

$k!>{2}^{k}$

Now we need to show that $P(k+1)$ is true.

$(k+1)!={2}^{k+1}$

Proof:

$(k+1)!>(k+1)k!$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}(k+1){2}^{k}$

After this I have no idea how to solve further.

Can anyone explain how to continue.What type of material allows electrons to flow freely?

Conductor

Resistor

Hestitation

RegulatorA set X with cardinality X has how many elements in its power set (the set of all subsets of a set)?

a. $|x{|}^{2}$

b. $|X\times X|$

c. ${2}^{|x|}$

d. $|X\cup X|$Why is the value of $\pi $ is $\frac{22}{7}$?

Can a dot product of a permutation of n (1,-1) with a sequence of primes generate unique numbers?

According to Fundamental Theorem of Arithmetic any positive whole number is the product of primes. Therefore, I can create unique numbers by multiplying n primes.

If I have a list of length n generated by a finite sequence in the set 1,-1, e.g., (1,1,-1,-1,1,...), and I do a dot product with a sequence of n primes starting with 3, e.g., (3,5,7,11,13,...), do I have any guarantee that I will generate unique numbers by doing different permutations of 1s and -1s?

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Given {1,…,2n}, how do you pair off the elements such that: when you multiply within each of the n pairs, the resulting products have maximal sum?

Here is an example: For $n=2$, this would mean pairing off elements of {1,2,3,4}, taking the product within each pair, and then adding all of those products up: where the goal is to maximize the resulting sum.

In this example, you could use:

$\{1,2\},\{3,4\}\to 1\cdot 2+3\cdot 4=2+12=14$

$\{1,3\},\{2,4\}\to 1\cdot 3+2\cdot 4=3+8=11$

$\{1,4\},\{2,3\}\to 1\cdot 4+2\cdot 3=4+6=10$

Since the pairwise product sums in this case are 14,11, and 10, we find that the maximum is 14. In particular, the maximum occurs when we pair off consecutive numbers. I suspect that the consecutive pairing may always be optimal (and that pairing first with last, second with second to last, etc may always result in the minimal pairwise product sum).

I'm not sure how to prove (or disprove!) this, and would appreciate a proof or a pointer to one. I suspect this problem has already been considered, so I include a ref-req tag as there may be terminology unfamiliar to me. If there are generalizations of this problem (e.g., starting with a set other than the first consecutive 2n positive integers) then related references will be most welcome, too!Compute summation of modules expression?

In particular, what I want to look at is the sum

$\sum _{k=1}^{n}(pk\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}q))$

where $p,q\in {\mathbb{Z}}_{\ge 1}$ can be assumed to be coprime but it would be best if solved in the fullest generality. In the above expression, n is a variable, p,q are fixed, and a(modb) means taking the representative set $\{0,1,2,...,b-1\}$. For example, $7\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}3)=1$ is the only value we agree upon and $7\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}3)\ne -2$.

The problem with this is that the list of representatives are permuted by p and hence the methods presented in the initial link are no longer valid.

It would be nice if we can come up with a closed form, but a really tight upper bound of the expression also works.Counting words of length n from k-sized alphabet with no substring of k consecutive distinct letters

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Other than that, no restrictions apply: any amount of distinct letters may be used throughout the entire word, and any letter can be used as many times as we like, as long as every substring inside the length n word complies with the rules above.

There is the obvious case of $k=2$ which results in 2 words, for every n, because you can only start with either letter, and they alternate.

For the larger case, I have come up with a recursive formula:

$C(n,k)=f(0,0,k)$

$f(i,d,k)=\{\begin{array}{ll}{\displaystyle (k-d)\cdot f(i+1,d+1,k)+\sum _{c=1}^{d}f(i+1,c,k)}& i<n,d<k\\ 1& i=n,d<k\\ 0& i>n\\ 0& d\ge k\end{array}$

With d being the current length of consecutive distinct letters, and i the current word length.

At every step, we can either use a letter other than the previous d letters, in which case the length is increased by one, and the chain-length d of distinct consecutive letters is also increased by one. In this case, there are $(k-d)$ such letters we can use at the stage, each subsequently resulting in the same contribution.

Or, we can use a letter already in the last d letters. In this case, the position of the letter matters. If we use the last of the d letters, a whole new chain begins. If instead we use the second last d letter, then a chain of length 2 of consecutive distinct letters begins. The 3rd last would result in a chain of length 3 and so on.

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Problem: Define a relation S on $\mathbb{N}\to \{0,1\}$ as follows:$\u27e8f,g\u27e9\in S\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}$ there exists a bijection $h:\mathbb{N}\to \mathbb{N}$ s.t. $f=g\circ h$.

S is an equivalence relation on $\mathbb{N}\to \{0,1\}$ (no need to prove this). Write a Representative set for the relation S. There's no need to prove that the relation you wrote is indeed a Representative set.

Reminder: Suppose $T\subseteq X\times X$ is an equivalence relation over X. $\text{}\text{}A\subseteq X$ will be called a Representative set of T, if it occurs that: $\mathrm{\forall}x\in X.|[x{]}_{T}\cap A|=1$.

Attempt: I don't really know what Representative set to define. It seems to me I'm missing something simple here. I tried to look at the functions: ${f}_{1}(n)=0,{f}_{2}(n)=1,{f}_{3}(n)=\{\begin{array}{ll}0& \text{n=0}\\ 1& \text{else}\end{array}$, ${f}_{4}(n)=\{\begin{array}{ll}0& n\in {\mathbb{N}}_{even}\\ 1& n\in {\mathbb{N}}_{odd}\end{array},\mathrm{\forall}n\in \mathbb{N}$. None of these functions relate through relation S since there does not exist a bijection between them. I feel lost, do you have any idea what to do?