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Ronsly Yassi
2022-07-09

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asked 2021-08-15

How many elements are in the set
{ 0, { { 0 } }?

asked 2021-08-18

Discrete Mathematics Basics

1) Determine whether the relation R on the set of all Web pages is reflexive, symmetric, antisymmetric, and/or transitive, where$(a,b)\in R$ if and only if

I) everyone who has visited Web page a has also visited Web page b.

II) there are no common links found on both Web page a and Web page b.

III) there is at least one common link on Web page a and Web page b.

1) Determine whether the relation R on the set of all Web pages is reflexive, symmetric, antisymmetric, and/or transitive, where

I) everyone who has visited Web page a has also visited Web page b.

II) there are no common links found on both Web page a and Web page b.

III) there is at least one common link on Web page a and Web page b.

asked 2021-07-28

Let A, B, and C be sets. Show that

asked 2021-08-02

Suppose that A is the set of sophomores at your school and B is the set of students in discrete mathematics at your school. Express each of these sets in terms of A and B.

a) the set of sophomores taking discrete mathematics in your school

b) the set of sophomores at your school who are not taking discrete mathematics

c) the set of students at your school who either are sophomores or are taking discrete mathematics

Use these symbols:$\cap \cup$

a) the set of sophomores taking discrete mathematics in your school

b) the set of sophomores at your school who are not taking discrete mathematics

c) the set of students at your school who either are sophomores or are taking discrete mathematics

Use these symbols:

asked 2022-07-07

How is the summation being expanded?

I am trying to understand summations by solving some example problems, but I could not understand how is the second to last line being expanded? I would really appreciate if you could explain me how is it being expanded.

$\begin{array}{rl}& \sum _{i=1}^{n-1}\sum _{j=i+1}^{n}\sum _{k=1}^{j}1=\\ & \sum _{i=1}^{n-1}\sum _{j=i+1}^{n}j=\\ & \sum _{i=1}^{n-1}(\sum _{j=1}^{n}j-\sum _{j=1}^{i}j)=\\ & \sum _{i=1}^{n-1}(\frac{n(n+1)}{2}-\frac{i(i+1)}{2})=\\ & \frac{1}{2}\sum _{i=1}^{n-1}{n}^{2}+n-{i}^{2}-i=\\ & \frac{1}{2}((n-1){n}^{2}+(n-1)n-(\frac{n(n+1)(2n+1)}{6}-{n}^{2})-(\frac{n(n+1)}{2}-n))=\\ & f(n)=\frac{n(n(n+1))}{2}-\frac{n(n+1)(2n+1)}{12}-\frac{n(n+1)}{4}\end{array}$

I am trying to understand summations by solving some example problems, but I could not understand how is the second to last line being expanded? I would really appreciate if you could explain me how is it being expanded.

$\begin{array}{rl}& \sum _{i=1}^{n-1}\sum _{j=i+1}^{n}\sum _{k=1}^{j}1=\\ & \sum _{i=1}^{n-1}\sum _{j=i+1}^{n}j=\\ & \sum _{i=1}^{n-1}(\sum _{j=1}^{n}j-\sum _{j=1}^{i}j)=\\ & \sum _{i=1}^{n-1}(\frac{n(n+1)}{2}-\frac{i(i+1)}{2})=\\ & \frac{1}{2}\sum _{i=1}^{n-1}{n}^{2}+n-{i}^{2}-i=\\ & \frac{1}{2}((n-1){n}^{2}+(n-1)n-(\frac{n(n+1)(2n+1)}{6}-{n}^{2})-(\frac{n(n+1)}{2}-n))=\\ & f(n)=\frac{n(n(n+1))}{2}-\frac{n(n+1)(2n+1)}{12}-\frac{n(n+1)}{4}\end{array}$

asked 2022-07-13

I'm self studying How to Prove book and have been working out the following problem in which I have to analyze it to logical form:

Nobody in the calculus class is smarter than everybody in the discrete math class

Now, this is how, I started solving it:

¬(Somebody in the calculus class is smarter than everybody in the discrete math class) ¬(If x is in calculus class then x is smartert than everybody in the discrete maths class)

$C(x)=x$ is in calculus class. $D(y)=y$ is in discrete class. $S(x,y)=x$ is smarter than y

$\neg \mathrm{\exists}x(C(x)\to \mathrm{\forall}y(D(y)\wedge S(x,y)))$

But this is the solution given in the Velleman's book:

$\neg \mathrm{\exists}x[C(x)\wedge \mathrm{\forall}y(D(y)\to S(x,y))]$

I cannot understand how that answer is correct. Can someone explain the thing I'm missing there ?

Nobody in the calculus class is smarter than everybody in the discrete math class

Now, this is how, I started solving it:

¬(Somebody in the calculus class is smarter than everybody in the discrete math class) ¬(If x is in calculus class then x is smartert than everybody in the discrete maths class)

$C(x)=x$ is in calculus class. $D(y)=y$ is in discrete class. $S(x,y)=x$ is smarter than y

$\neg \mathrm{\exists}x(C(x)\to \mathrm{\forall}y(D(y)\wedge S(x,y)))$

But this is the solution given in the Velleman's book:

$\neg \mathrm{\exists}x[C(x)\wedge \mathrm{\forall}y(D(y)\to S(x,y))]$

I cannot understand how that answer is correct. Can someone explain the thing I'm missing there ?

asked 2021-08-21

What is the coefficeint of $a}^{2}{b}^{3}{c}^{4$ in the expansion of

$(a+2b+3c)}^{9$ ?