Theory of Positive Integers : Logic

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Example Questions

Example Question #1 : Logic

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

For all \displaystyle B\ \epsilon\ S which \displaystyle P(B) is true? 

Possible Answers:

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

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

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

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

Correct answer:

Explanation:

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

Looking at what is given,

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

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

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

Thus the final solution is,

Example Question #1 : Theory Of Positive Integers

Negate the following statement.

\displaystyle P: 17 is a prime number.

Possible Answers:

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

\displaystyle \sim P:17 is an odd number

\displaystyle \sim P:17 is a prime number

\displaystyle \sim P:17 is an even number

\displaystyle P:17 is not a prime number

Correct answer:

\displaystyle \sim P:17 is not a prime 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,

\displaystyle P: 17 is a prime number.

contains to components.

Component one: \displaystyle P:17

Component two: "is a prime number"

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

\displaystyle \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.

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

Example Question #3 : Theory Of Positive Integers

Determine which statement is true giving the following information.

\displaystyle P: 17 is a prime number \displaystyle Q:50 is odd

Possible Answers:

\displaystyle P\wedge Q

\displaystyle \sim (P\vee Q)

\displaystyle \sim P

\displaystyle P\vee Q

None of the answers.

Correct answer:

\displaystyle P\vee Q

Explanation:

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

The first component of this statement is:

\displaystyle P: 17 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:

\displaystyle Q:50 is odd

This statement is false since \displaystyle 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,

\displaystyle P\vee Q

Example Question #4 : Theory Of Positive Integers

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

For all \displaystyle B\ \epsilon\ S which \displaystyle P(B) is true? 

Possible Answers:

\displaystyle B\ \epsilon\ \begin{Bmatrix} \O \end{Bmatrix}

\displaystyle B\ \epsilon\ \begin{Bmatrix} 10,23 \end{Bmatrix}

\displaystyle B\ \epsilon\ \begin{Bmatrix} 10 \end{Bmatrix}

\displaystyle B\ \epsilon\ \begin{Bmatrix} \O, 10,20 \end{Bmatrix}

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

Correct answer:

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

Explanation:

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

Looking at what is given,

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

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

it is seen that only ten lives in \displaystyle B and \displaystyle S therefore both these elements will be in the partition of \displaystyle B. Another element that also exists in both sets is the empty set.

Thus the final solution is,

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

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