Logic - Theory of Positive Integers

Card 0 of 16

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

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

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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,

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

Negate the following statement.

is a prime number.

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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.

Determine which statement is true giving the following information.

is a prime number is odd

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To determine which statement is true first state what is known.

The first component of this statement is:

is a prime number

This is a true statement since only one and seventeen are factors of seventeen.

The second component of this statement is:

is odd

This statement is false since $50 \mod 2=0$ .

Therefore, the only true statement is the one that uses the "or" operator because only one component is true.

Thus the correct answer is,

$P\vee Q$

$B\subseteq \begin{Bmatrix} {2,10,11,23} \end{Bmatrix}$ over the domain $S=P(\begin{Bmatrix} {10,20,30} \end{Bmatrix})$

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

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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,

$B\subseteq \begin{Bmatrix} {2,10,11,23} \end{Bmatrix}$

$S=P(\begin{Bmatrix} {10,20,30} \end{Bmatrix})$

it is seen that only ten lives 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,

$B\ \epsilon\ \begin{Bmatrix} \O, 10 \end{Bmatrix}$

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

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

Tap to see back →

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,

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

Negate the following statement.

is a prime number.

Tap to see back →

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.

Determine which statement is true giving the following information.

is a prime number is odd

Tap to see back →

To determine which statement is true first state what is known.

The first component of this statement is:

is a prime number

This is a true statement since only one and seventeen are factors of seventeen.

The second component of this statement is:

is odd

This statement is false since $50 \mod 2=0$ .

Therefore, the only true statement is the one that uses the "or" operator because only one component is true.

Thus the correct answer is,

$P\vee Q$

$B\subseteq \begin{Bmatrix} {2,10,11,23} \end{Bmatrix}$ over the domain $S=P(\begin{Bmatrix} {10,20,30} \end{Bmatrix})$

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

Tap to see back →

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,

$B\subseteq \begin{Bmatrix} {2,10,11,23} \end{Bmatrix}$

$S=P(\begin{Bmatrix} {10,20,30} \end{Bmatrix})$

it is seen that only ten lives 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,

$B\ \epsilon\ \begin{Bmatrix} \O, 10 \end{Bmatrix}$

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

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

Tap to see back →

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,

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

Negate the following statement.

is a prime number.

Tap to see back →

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.

Determine which statement is true giving the following information.

is a prime number is odd

Tap to see back →

To determine which statement is true first state what is known.

The first component of this statement is:

is a prime number

This is a true statement since only one and seventeen are factors of seventeen.

The second component of this statement is:

is odd

This statement is false since $50 \mod 2=0$ .

Therefore, the only true statement is the one that uses the "or" operator because only one component is true.

Thus the correct answer is,

$P\vee Q$

$B\subseteq \begin{Bmatrix} {2,10,11,23} \end{Bmatrix}$ over the domain $S=P(\begin{Bmatrix} {10,20,30} \end{Bmatrix})$

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

Tap to see back →

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,

$B\subseteq \begin{Bmatrix} {2,10,11,23} \end{Bmatrix}$

$S=P(\begin{Bmatrix} {10,20,30} \end{Bmatrix})$

it is seen that only ten lives 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,

$B\ \epsilon\ \begin{Bmatrix} \O, 10 \end{Bmatrix}$

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

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

Tap to see back →

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,

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

Negate the following statement.

is a prime number.

Tap to see back →

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.

Determine which statement is true giving the following information.

is a prime number is odd

Tap to see back →

To determine which statement is true first state what is known.

The first component of this statement is:

is a prime number

This is a true statement since only one and seventeen are factors of seventeen.

The second component of this statement is:

is odd

This statement is false since $50 \mod 2=0$ .

Therefore, the only true statement is the one that uses the "or" operator because only one component is true.

Thus the correct answer is,

$P\vee Q$

$B\subseteq \begin{Bmatrix} {2,10,11,23} \end{Bmatrix}$ over the domain $S=P(\begin{Bmatrix} {10,20,30} \end{Bmatrix})$

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

Tap to see back →

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,

$B\subseteq \begin{Bmatrix} {2,10,11,23} \end{Bmatrix}$

$S=P(\begin{Bmatrix} {10,20,30} \end{Bmatrix})$

it is seen that only ten lives 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,

$B\ \epsilon\ \begin{Bmatrix} \O, 10 \end{Bmatrix}$