Mathematical logic Questions in English

(N/A) The component statements are:
$p: \sqrt{2}$ is a rational number.
$q: \sqrt{2}$ is an irrational number.
The first statement $p$ is false,as $\sqrt{2}$ cannot be expressed in the form $\frac{p}{q}$ where $p, q$ are integers and $q \neq 0$.
The second statement $q$ is true.
Since the connecting word is 'or',the compound statement is true if at least one of the component statements is true. Therefore,the compound statement is true.

(N/A) The component statements are:
$p: 24$ is a multiple of $2$.
$q: 24$ is a multiple of $4$.
$r: 24$ is a multiple of $8$.
All three statements are true. Here,the connecting word is 'and'. Thus,we observe that compound statements are made up of two or more statements connected by words like 'and','or',etc. These words have special meanings in mathematics.

(N/A) The negation of the statement is: Chennai is not the capital of Tamil Nadu.

(N/A) The negation of a statement $p$ is denoted as $\sim p$.
Given the statement $p$: $\sqrt{2}$ is not a complex number.
The negation $\sim p$ is: $\sqrt{2}$ is a complex number.

(N/A) The negation of the statement "All triangles are not equilateral triangles" is "There exists at least one triangle which is an equilateral triangle."

(B) The negation of a statement $P$ is denoted by $\sim P.$
Given statement $P$: The number $2$ is greater than $7.$
Therefore,the negation $\sim P$ is: The number $2$ is not greater than $7.$

(N/A) The negation of the statement "Every natural number is an integer" is "There exists at least one natural number which is not an integer."

(A) The negation of the first statement,'$x$ is not a rational number',is '$x$ is a rational number'.
Since the set of real numbers consists of rational and irrational numbers,the statement '$x$ is not an irrational number' is logically equivalent to '$x$ is a rational number'.
Therefore,the two given statements are negations of each other.

(A) The negation of the first statement,'$x$ is a rational number',is '$x$ is not a rational number'.
Since a real number that is not rational is by definition an irrational number,the statement '$x$ is not a rational number' is equivalent to '$x$ is an irrational number'.
Therefore,the given pairs of statements are indeed negations of each other.

(N/A) The component statements are as follows:
$p: \text{Number } 3 \text{ is prime.}$
$q: \text{Number } 3 \text{ is odd.}$
Since $3$ is a prime number,statement $p$ is true.
Since $3$ is an odd number,statement $q$ is true.
Therefore,both component statements are true.

(N/A) The component statements are as follows:
$p:$ All integers are positive.
$q:$ All integers are negative.
Both the statements are false because $0$ is an integer which is neither positive nor negative.

The component statements are as follows:
$p: 100$ is divisible by $3$.
$q: 100$ is divisible by $11$.
$r: 100$ is divisible by $5$.
Checking the truth values:
$100$ is not divisible by $3$,so $p$ is false.
$100$ is not divisible by $11$,so $q$ is false.
$100$ is divisible by $5$,so $r$ is true.

(N/A) The component statements are:
$p:$ $A$ line is straight.
$q:$ $A$ line extends indefinitely in both directions.
Both these statements are true,therefore,the compound statement is true.

(N/A) The component statements are:
$p: 0$ is less than every positive integer.
$q: 0$ is less than every negative integer.
Since $0$ is greater than every negative integer,the statement $q$ is false.
Since the compound statement is connected by 'and',it is true only if both component statements are true.
Therefore,the compound statement is false.

(B) The two component statements are:
$p:$ All living things have two legs.
$q:$ All living things have two eyes.
Both these statements are false because many living things (like fish,birds,or insects) do not have two legs or two eyes. Therefore,the compound statement is false.

(A) In this statement,the "Or" is inclusive. This is because a person can possess both a passport and a voter registration card simultaneously,and either or both documents are sufficient to enter the country.

(A) In this statement,the "Or" is inclusive. The school remains closed if it is a holiday,if it is a Sunday,or if it is both (e.g.,a Sunday that is also a holiday). Since the condition is satisfied in all these cases,it is an inclusive "Or".

(B) The "Or" used here is exclusive.
Reason: It is impossible for two lines in a plane to both intersect at a point and be parallel simultaneously. Since both conditions cannot be true at the same time,the "Or" is exclusive.

(B) In this statement,the "Or" is exclusive.
This is because a student is typically required to choose only one language as their third language and cannot take both French and Sanskrit simultaneously.

(B) The component statements are:
$p: \sqrt{2}$ is a rational number.
$q: \sqrt{2}$ is an irrational number.
Here,the first statement $p$ is false and the second statement $q$ is true.
Since $\sqrt{2}$ cannot be both rational and irrational simultaneously,the "Or" used here is exclusive.
In an exclusive "Or" statement,if one component is true and the other is false,the compound statement is true.
Therefore,the compound statement is true.

(N/A) The component statements are:
$p:$ To get into a public library,children need an identity card.
$q:$ To get into a public library,children need a letter from the school authorities.
Children can enter the library if they have either of the two,an identity card or the letter,as well as when they have both. Therefore,it is an inclusive "Or". The compound statement is true if a child has either an identity card or a letter,or both.

(N/A) The statement is composed of two component statements:
$p$: $A$ rectangle is a quadrilateral.
$q$: $A$ rectangle is a $5$-sided polygon.
Here,the "Or" used is exclusive because a rectangle cannot be both a quadrilateral and a $5$-sided polygon simultaneously.
Since $p$ is true and $q$ is false,the disjunction $p \lor q$ is true.

(N/A) The given compound statement is: 'All rational numbers are real and all real numbers are not complex.'
$1$. The connecting word is 'and'.
$2$. The component statements are:
$p:$ All rational numbers are real.
$q:$ All real numbers are not complex.

(N/A) The connecting word is "Or".
The component statements are as follows:
$p:$ The square of an integer is positive.
$q:$ The square of an integer is negative.

(N/A) Here,the connecting word is 'and'.
The component statements are as follows:
$p:$ The sand heats up quickly in the sun.
$q:$ The sand does not cool down fast at night.

(N/A) The connecting word is 'and'.
The component statements are:
$p: x = 2$ is a root of the equation $3x^{2} - x - 10 = 0$.
$q: x = 3$ is a root of the equation $3x^{2} - x - 10 = 0$.

(N/A) The quantifier in the statement is 'There exists'.
The negation of the statement 'There exists a number which is equal to its square' is 'There does not exist a number which is equal to its square'.

(N/A) The quantifier is 'For every'.
The negation of the statement is: There exists a real number $x$ such that $x$ is not less than $x+1.$

(N/A) The statement is: 'There exists a capital for every state in India.'
$1$. The quantifier is 'There exists'.
$2$. The negation of the statement is: 'There exists a state in India which does not have a capital.'

(A) Let $P$ be the statement: 'There exists real numbers $x$ and $y$ for which $x+y=y+x.$'
This is an existential statement of the form 'There exists $x, y$ such that $P(x, y)$'.
The negation of an existential statement 'There exists $x$ such that $P(x)$' is 'For all $x$,not $P(x)$'.
Therefore,the negation of $P$ is: 'For all real numbers $x$ and $y$,$x+y \neq y+x.$'
Since statement $II$ is exactly this negation,the given pair of statements are indeed negations of each other.

(A) In the statement "Sun rises or Moon sets",the "or" is exclusive.
This is because the two events,the Sun rising and the Moon setting,are distinct astronomical phenomena that cannot occur simultaneously in the same context.

(B) In this statement,the word "or" is inclusive.
This is because a person can possess both a ration card and a passport to apply for a driving licence.
Since the condition is satisfied if a person has either one or both documents,it is an inclusive "or".

(A) In the statement "All integers are positive or negative",the "or" is exclusive.
This is because an integer cannot be both positive and negative simultaneously. Since both conditions cannot be true at the same time,it is an exclusive "or".

(C) The contrapositive of a statement of the form $P \implies Q$ is $\neg Q \implies \neg P.$
Here,$P$ is "a number is divisible by $9$" and $Q$ is "a number is divisible by $3.$"
Therefore,the contrapositive is: "If a number is not divisible by $3,$ then it is not divisible by $9.$"

(B) The contrapositive of a conditional statement of the form $P \implies Q$ is given by $\neg Q \implies \neg P$.
Here,$P$ is "You are born in India" and $Q$ is "You are a citizen of India".
Therefore,$\neg Q$ is "You are not a citizen of India" and $\neg P$ is "You are not born in India".
Thus,the contrapositive is: "If you are not a citizen of India,then you are not born in India."

(A) The contrapositive of a statement of the form $P \implies Q$ is $\neg Q \implies \neg P$.
Here,$P$ is "a triangle is equilateral" and $Q$ is "a triangle is isosceles".
Therefore,the contrapositive is "If a triangle is not isosceles,then it is not equilateral."

(N/A) The converse of the statement "If $p$,then $q$" is "If $q$,then $p$.
Given statement: If $n$ is even,then $n^{2}$ is even.
Therefore,the converse is: If $n^{2}$ is even,then $n$ is even.

(N/A) The converse of a conditional statement of the form "If $P$,then $Q$" is "If $Q$,then $P$.
Given statement: If you do all the exercises in the book $(P)$,then you get an $A$ grade in the class $(Q)$.
Therefore,the converse is: If you get an $A$ grade in the class,then you have done all the exercises in the book.

(N/A) The converse of a statement of the form "If $P$,then $Q$" is "If $Q$,then $P$.
Here,$P$ is "two integers $a$ and $b$ are such that $a > b$" and $Q$ is "$a - b$ is always a positive integer".
Therefore,the converse is: "If $a - b$ is always a positive integer,then two integers $a$ and $b$ are such that $a > b$."

(N/A) The component statements are given by:
$p: \text{Triangle } ABC \text{ is equilateral.}$
$q: \text{Triangle } ABC \text{ is isosceles.}$
Since every equilateral triangle is also an isosceles triangle,the given compound statement is true.

(N/A) The component statements are given by:
$p: a$ and $b$ are integers.
$q: ab$ is a rational number.
Since the product of two integers is always an integer,and every integer is a rational number,the compound statement is true.

(N/A) rectangle is a square if and only if all its four sides are equal.

(N/A) number is divisible by $3$ if and only if the sum of its digits is divisible by $3.$

(N/A) The given statement can be expressed in five different ways as follows:
$(i)$ $A$ natural number being odd implies that its square is odd.
$(ii)$ $A$ natural number is odd only if its square is odd.
$(iii)$ For a natural number to be odd,it is necessary that its square is odd.
$(iv)$ For the square of a natural number to be odd,it is sufficient that the number is odd.
$(v)$ If the square of a natural number is not odd,then the natural number is not odd.

(N/A) The given statement is of the form $P \implies Q$,where $P$ is '$x$ is a prime number' and $Q$ is '$x$ is odd'.
The contrapositive of $P \implies Q$ is $\neg Q \implies \neg P$.
Thus,the contrapositive is: If $x$ is not odd,then $x$ is not a prime number.
The converse of $P \implies Q$ is $Q \implies P$.
Thus,the converse is: If $x$ is odd,then $x$ is a prime number.

(N/A) Let $p$ be the statement: "Two lines are parallel".
Let $q$ be the statement: "They do not intersect in the same plane".
The given statement is of the form "If $p$,then $q$ $(p \implies q)$".
The contrapositive is "If not $q$,then not $p$ $(
eg q \implies \neg p)$":
If two lines intersect in the same plane,then they are not parallel.
The converse is "If $q$,then $p$ $(q \implies p)$":
If two lines do not intersect in the same plane,then they are parallel.

(N/A) Let $p$ be the statement 'Something is cold' and $q$ be the statement 'It has low temperature'.
The given statement is of the form $p \implies q$.
The contrapositive of $p \implies q$ is $\neg q \implies \neg p$.
Thus,the contrapositive is: 'If something does not have low temperature,then it is not cold.'
The converse of $p \implies q$ is $q \implies p$.
Thus,the converse is: 'If something has low temperature,then it is cold.'

(N/A) The given statement is of the form "If $p$,then $q$",where $p$ is "You do not know how to reason deductively" and $q$ is "You cannot comprehend geometry".
The contrapositive of "If $p$,then $q$" is "If not $q$,then not $p$".
Here,not $q$ is "You can comprehend geometry" and not $p$ is "You know how to reason deductively".
Thus,the contrapositive is: "If you can comprehend geometry,then you know how to reason deductively."
The converse of "If $p$,then $q$" is "If $q$,then $p$".
Thus,the converse is: "If you cannot comprehend geometry,then you do not know how to reason deductively."

(N/A) The given statement is: If $x$ is an even number,then $x$ is divisible by $4.$
$1$. The contrapositive is: If $x$ is not divisible by $4,$ then $x$ is not an even number.
$2$. The converse is: If $x$ is divisible by $4,$ then $x$ is an even number.

(N/A) If you get a job,then your credentials are good.