Theory of Positive Integers

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Which of the following is a classification of a set whose cardinality is infinite?

Uncountably

Positively

Negatively

Infinitely

None of the answers

Explanation

If a set is stated to have infinite cardinality then it will fall one of the following categories,

I. Countably

II. Uncountably

Countably infinite sets are those that the elements within the set are able to be counted.

For example, the set of natural numbers

$\mathbb{N}:\begin{Bmatrix} 0,1,2,3,3,...,n \end{Bmatrix}$

is a countably infinite set.

Uncountably infinite sets are those that the elements cannot be counted.

For example, the set of real numbers,

$\mathbb{R}:\begin{Bmatrix} ...,-1,-\frac{2}{3},0.4567786,5,\frac{145}{2},... \end{Bmatrix}$

The reason the set of real numbers is not countable is because there are infinitely many elements between each element. This differs from the natural numbers because in the natural numbers there are no elements between elements.

Therefore the correct solution is "uncountably".

Which of the following is a property of a relation?

Transitive Property

Non-symmetric Property

Equivalency Property

Partition Property

All are properties of relations.

Explanation

For a relation to exist there must be a non empty set present. If a non empty set is present then there are three relation properties.

These properties are:

I. Reflexive Property

$x\approx x$

II. Symmetric Property

$x\approx y\rightarrow y\approx x$

III. Transitive Property

$x\approx y, y\approx z \rightarrow x\approx z$

When all three properties represent a specific set, then that set is known to have an equivalence relation.

$B\subseteq \begin{Bmatrix} {3,4,7,9,11} \end{Bmatrix}$ over the domain $S=P(\begin{Bmatrix} {4,7,10,12} \end{Bmatrix})$

For all $B\ \epsilon\ S$ which is true?

$B\ \epsilon\ \begin{Bmatrix} \O, 4,7 \end{Bmatrix}$

$B\ \epsilon\ \begin{Bmatrix} 4,7 \end{Bmatrix}$

$B\ \epsilon\ \begin{Bmatrix} 4,7,11 \end{Bmatrix}$

$B\ \epsilon\ \begin{Bmatrix} 3,7,10 \end{Bmatrix}$

$B\ \epsilon\ \begin{Bmatrix} 4,12 \end{Bmatrix}$

Explanation

This question is giving a subset who lives in the domain and it is asking for the partition or group of elements that live in both and .

Looking at what is given,

$B\subseteq \begin{Bmatrix} {3,4,7,9,11} \end{Bmatrix}$

$S=P(\begin{Bmatrix} {4,7,10,12} \end{Bmatrix})$

it is seen that both four and seven live in and therefore both these elements will be in the partition of . Another element that also exists in both sets is the empty set.

Thus the final solution is,

$B\ \epsilon\ \begin{Bmatrix} \O, 4,7 \end{Bmatrix}$

Let

$S:\begin{Bmatrix} -10,-8,-6,-4,-2,0,2,4 \end{Bmatrix}$

if is some condition of such that it can be described as $\begin{Bmatrix} x\ \epsilon\ S: p(x) \end{Bmatrix}$ what is when $B=\begin{Bmatrix} 2,4 \end{Bmatrix}$ ?

$B=\begin{Bmatrix} x\ \epsilon\ S: x>0 \end{Bmatrix}$

$B=\begin{Bmatrix} x\ \epsilon\ S: x\geq0 \end{Bmatrix}$

$B=\begin{Bmatrix} x\ \epsilon\ S: x<0 \end{Bmatrix}$

$B=\begin{Bmatrix} x\ \epsilon\ S: x\leq0 \end{Bmatrix}$

None of the answers

Explanation

First, identify what is given.

$S:\begin{Bmatrix} -10,-8,-6,-4,-2,0,2,4 \end{Bmatrix}$

$B=\begin{Bmatrix} 2,4 \end{Bmatrix}$

and can be described in the following format

$B=\begin{Bmatrix} x\ \epsilon\ S: p(x) \end{Bmatrix}$

Since contains the elements in that are greater than zero, can be written as follows.

$B=\begin{Bmatrix} x\ \epsilon\ S: x>0 \end{Bmatrix}$

Let

$S:\begin{Bmatrix} -10,-8,-6,-4,-2,0,2,4 \end{Bmatrix}$

if is some condition of such that it can be described as $\begin{Bmatrix} x\ \epsilon\ S: p(x) \end{Bmatrix}$ what is when $B=\begin{Bmatrix} 2,4 \end{Bmatrix}$ ?

$B=\begin{Bmatrix} x\ \epsilon\ S: x>0 \end{Bmatrix}$

$B=\begin{Bmatrix} x\ \epsilon\ S: x\geq0 \end{Bmatrix}$

$B=\begin{Bmatrix} x\ \epsilon\ S: x<0 \end{Bmatrix}$

$B=\begin{Bmatrix} x\ \epsilon\ S: x\leq0 \end{Bmatrix}$

None of the answers

Explanation

First, identify what is given.

$S:\begin{Bmatrix} -10,-8,-6,-4,-2,0,2,4 \end{Bmatrix}$

$B=\begin{Bmatrix} 2,4 \end{Bmatrix}$

and can be described in the following format

$B=\begin{Bmatrix} x\ \epsilon\ S: p(x) \end{Bmatrix}$

Since contains the elements in that are greater than zero, can be written as follows.

$B=\begin{Bmatrix} x\ \epsilon\ S: x>0 \end{Bmatrix}$

$B\subseteq \begin{Bmatrix} {3,4,7,9,11} \end{Bmatrix}$ over the domain $S=P(\begin{Bmatrix} {4,7,10,12} \end{Bmatrix})$

For all $B\ \epsilon\ S$ which is true?

$B\ \epsilon\ \begin{Bmatrix} \O, 4,7 \end{Bmatrix}$

$B\ \epsilon\ \begin{Bmatrix} 4,7 \end{Bmatrix}$

$B\ \epsilon\ \begin{Bmatrix} 4,7,11 \end{Bmatrix}$

$B\ \epsilon\ \begin{Bmatrix} 3,7,10 \end{Bmatrix}$

$B\ \epsilon\ \begin{Bmatrix} 4,12 \end{Bmatrix}$

Explanation

This question is giving a subset who lives in the domain and it is asking for the partition or group of elements that live in both and .

Looking at what is given,

$B\subseteq \begin{Bmatrix} {3,4,7,9,11} \end{Bmatrix}$

$S=P(\begin{Bmatrix} {4,7,10,12} \end{Bmatrix})$

it is seen that both four and seven live in and therefore both these elements will be in the partition of . Another element that also exists in both sets is the empty set.

Thus the final solution is,

$B\ \epsilon\ \begin{Bmatrix} \O, 4,7 \end{Bmatrix}$

Which of the following is a classification of a set whose cardinality is infinite?

Uncountably

Positively

Negatively

Infinitely

None of the answers

Explanation

If a set is stated to have infinite cardinality then it will fall one of the following categories,

I. Countably

II. Uncountably

Countably infinite sets are those that the elements within the set are able to be counted.

For example, the set of natural numbers

$\mathbb{N}:\begin{Bmatrix} 0,1,2,3,3,...,n \end{Bmatrix}$

is a countably infinite set.

Uncountably infinite sets are those that the elements cannot be counted.

For example, the set of real numbers,

$\mathbb{R}:\begin{Bmatrix} ...,-1,-\frac{2}{3},0.4567786,5,\frac{145}{2},... \end{Bmatrix}$

Therefore the correct solution is "uncountably".

Which of the following is a property of a relation?

Transitive Property

Non-symmetric Property

Equivalency Property

Partition Property

All are properties of relations.

Explanation

For a relation to exist there must be a non empty set present. If a non empty set is present then there are three relation properties.

These properties are:

I. Reflexive Property

$x\approx x$

II. Symmetric Property

$x\approx y\rightarrow y\approx x$

III. Transitive Property

$x\approx y, y\approx z \rightarrow x\approx z$

When all three properties represent a specific set, then that set is known to have an equivalence relation.

Negate the following statement.

is a prime number.

$\sim P:17$ is not a prime number

$\sim P:17$ is a prime number

is not a prime number

$\sim P:17$ is an even number

$\sim P:17$ is an odd number

Explanation

Negating a statement means to take the opposite of it.

To negate a statement completely, each component of the statement needs to be negated.

The given statement,

is a prime number.

contains to components.

Component one:

Component two: "is a prime number"

To negate component one, simply take the compliment of it. In mathematical terms this looks as follows,

$\sim P:17$

To negate component two, simply add a "not" before the phrase "a prime number".

Now, combine these two components back together for the complete negation.

$\sim P:17$ is not a prime number.

Which of the following is a classification of a set whose cardinality is infinite?

Countably

Positively

Negatively

Continuously

None of the answers

Explanation

If a set is stated to have infinite cardinality then it will fall one of the following categories,

I. Countably

II. Uncountably

Countably infinite sets are those that the elements within the set are able to be counted.

For example, the set of natural numbers

$\mathbb{N}:\begin{Bmatrix} 0,1,2,3,3,...,n \end{Bmatrix}$

is a countably infinite set.

Uncountably infinite sets are those that the elements cannot be counted.

For example, the set of real numbers,

$\mathbb{R}:\begin{Bmatrix} ...,-1,-\frac{2}{3},0.4567786,5,\frac{145}{2},... \end{Bmatrix}$

Therefore the correct solution is "countably".

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