Tamara Bryan
2022-07-17
Answered

Suppose the following two statements $(x=a)\vee (x\ne a)$ is it true or false? I know that its false when only both operators are false, but I don't know in this case if it can be the case

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Quenchingof

Answered 2022-07-18
Author has **14** answers

Step 1

For any sentence $P,P\vee \mathrm{\neg}P$ is a tautology and is therefore true.

Step 2

Consider the two possible truth values of P. Normally your sentence would have an implied quatification of $\mathrm{\forall}x,a$ before it to bind the variables.

For any sentence $P,P\vee \mathrm{\neg}P$ is a tautology and is therefore true.

Step 2

Consider the two possible truth values of P. Normally your sentence would have an implied quatification of $\mathrm{\forall}x,a$ before it to bind the variables.

Jadon Melendez

Answered 2022-07-19
Author has **4** answers

Step 1

You can have $x=a$ or $x\ne a$. If $x=a$, then $x=a\text{}\vee \text{}x\ne a$ is true since $x=a$ is true; otherwise, if $x\ne a$, then $x=a\text{}\vee \text{}x\ne a$ is true.

Step 2

In general, given a property $P\vee \mathrm{\neg}P$ is always true; similarly, $P\wedge \mathrm{\neg}P$ is always false.

You can have $x=a$ or $x\ne a$. If $x=a$, then $x=a\text{}\vee \text{}x\ne a$ is true since $x=a$ is true; otherwise, if $x\ne a$, then $x=a\text{}\vee \text{}x\ne a$ is true.

Step 2

In general, given a property $P\vee \mathrm{\neg}P$ is always true; similarly, $P\wedge \mathrm{\neg}P$ is always false.

asked 2020-11-09

Use proof by Contradiction to prove that the sum of an irrational number and a rational number is irrational.

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Discrete Mathematics Basics

1) Find out if the relation R is transitive, symmetric, antisymmetric, or reflexive on the set of all web pages.where $(a,b)\in R$ if and only if

I)Web page a has been accessed by everyone who has also accessed Web page b.

II) Both Web page a and Web page b lack any shared links.

III) Web pages a and b both have at least one shared link.

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

asked 2021-08-02

Suppose that A is the set of sophomores at your school and B is the set of students taking 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: $\cap \cup$

asked 2022-09-04

Prove that at least one student solved all the problems using PHP

In a physics exam, 5 problems were given to a class of N students. Suppose every two of these problems were solved by more than $\frac{3N}{5}$ students.Prove that at least one student solved all the problems

I have used PHP here. We know if at least $(k.n+1)$ objects are distributed among n boxes, then one of the boxes must contain at least $(k+1)$ objects. So according to this problem $(k.n+1)$ = 2 |where $$n=\frac{3N}{5}$$. But unable to prove at least one student solved all the problems. I have to prove basically $(k+1)$ $$=5$$. It will be helpful for me if someone helps me with this.

In a physics exam, 5 problems were given to a class of N students. Suppose every two of these problems were solved by more than $\frac{3N}{5}$ students.Prove that at least one student solved all the problems

I have used PHP here. We know if at least $(k.n+1)$ objects are distributed among n boxes, then one of the boxes must contain at least $(k+1)$ objects. So according to this problem $(k.n+1)$ = 2 |where $$n=\frac{3N}{5}$$. But unable to prove at least one student solved all the problems. I have to prove basically $(k+1)$ $$=5$$. It will be helpful for me if someone helps me with this.

asked 2022-07-14

Linear congruences - system of two equations with two variables - solutions don't satisfy

$9x+27y\equiv 3\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}102)$

$16x+4y\equiv 2\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}102)$

Here's what I did. I subtracted the first equation from the second to get

$9x+27y\equiv 3\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}102)$

$7x-23y\equiv -1\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}102)$

After that, I multiplied the first equation by 7. Next, I added to it the second equation multiplied by -9 to eliminate x from the first equation:

$I=II\cdot -9+I$

The first equation now looks like this:

$396y\equiv 30\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}102)$

I checked the solutions online, and this congruence has six solutions less than m which are: $y=6,23,40,57,74,91$

$9x+27y\equiv 3\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}102)$

$16x+4y\equiv 2\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}102)$

Here's what I did. I subtracted the first equation from the second to get

$9x+27y\equiv 3\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}102)$

$7x-23y\equiv -1\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}102)$

After that, I multiplied the first equation by 7. Next, I added to it the second equation multiplied by -9 to eliminate x from the first equation:

$I=II\cdot -9+I$

The first equation now looks like this:

$396y\equiv 30\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}102)$

I checked the solutions online, and this congruence has six solutions less than m which are: $y=6,23,40,57,74,91$

asked 2021-08-11

Consider the following statement:

” A is a subset of B. Therefore A is a subset of P(B).”

This statement is incorrect as written. Assuming the first sentence is true, what is incorrect about the second sentence? State the second sentence correctly.

” A is a subset of B. Therefore A is a subset of P(B).”

This statement is incorrect as written. Assuming the first sentence is true, what is incorrect about the second sentence? State the second sentence correctly.