Mathematical logic Questions in English

(N/A) If the banana trees stay warm for a month,then they will bloom.

(N/A) If the diagonals of a quadrilateral bisect each other,then it is a parallelogram.

(N/A) If you want to get an $A^{+}$ in the class,then you must do all the exercises of the book.

(N/A) The given statement is of the form "If $P$,then $Q$",where $P$ is "you live in Delhi" and $Q$ is "you have winter clothes".
$(i)$ The statement "If you do not have winter clothes,then you do not live in Delhi" is of the form "If not $Q$,then not $P$",which is the contrapositive of the given statement.
$(ii)$ The statement "If you have winter clothes,then you live in Delhi" is of the form "If $Q$,then $P$",which is the converse of the given statement.

(N/A) Let $P$ be the statement "$A$ quadrilateral is a parallelogram" and $Q$ be the statement "Its diagonals bisect each other". The given statement is $P \implies Q$.
$(i)$ The statement "If the diagonals of a quadrilateral do not bisect each other,then the quadrilateral is not a parallelogram" is of the form $\neg Q \implies \neg P$,which is the contrapositive of $P \implies Q$.
$(ii)$ The statement "If the diagonals of a quadrilateral bisect each other,then it is a parallelogram" is of the form $Q \implies P$,which is the converse of $P \implies Q$.

(A) Let $p: x, y \in \mathbb{Z}$ such that $x$ and $y$ are odd.
Let $q: xy$ is odd.
To check the validity of the given statement,we assume that if $p$ is true,then $q$ is true.
If $p$ is true,then $x$ and $y$ are odd integers.
We can write $x = 2m + 1$ and $y = 2n + 1$ for some integers $m, n \in \mathbb{Z}$.
Then,$xy = (2m + 1)(2n + 1) = 4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1$.
Since $2mn + m + n$ is an integer,$xy$ is of the form $2k + 1$,which means $xy$ is odd.
Therefore,the statement is true.

(A) Let the statements be defined as follows:
$p: xy$ is odd.
$q:$ Both $x$ and $y$ are odd.
We need to verify the implication $p \Rightarrow q$ by checking its contrapositive statement,which is $\sim q \Rightarrow \sim p$.
$\sim q:$ It is false that both $x$ and $y$ are odd. This means at least one of $x$ or $y$ is even.
Let $x = 2n$ for some integer $n \in \mathbb{Z}$.
Then $xy = (2n)y = 2(ny)$,which is an even number.
Thus,$\sim p$ is true (i.e.,$xy$ is even).
Since $\sim q \Rightarrow \sim p$ is true,the original statement $p \Rightarrow q$ is true.

(N/A) In this method,we assume that the given statement is false. That is,we assume that $\sqrt{7}$ is rational.
This means that there exist positive integers $a$ and $b$ such that $\sqrt{7} = \frac{a}{b}$,where $a$ and $b$ have no common factors.
Squaring the equation,we get $7 = \frac{a^2}{b^2} \Rightarrow a^2 = 7b^2$.
This implies that $7$ divides $a^2$,and since $7$ is prime,$7$ must divide $a$.
Therefore,there exists an integer $c$ such that $a = 7c$.
Substituting this into $a^2 = 7b^2$,we get $(7c)^2 = 7b^2$ $\Rightarrow 49c^2 = 7b^2$ $\Rightarrow b^2 = 7c^2$.
This implies that $7$ divides $b^2$,and consequently,$7$ must divide $b$.
Thus,$7$ is a common factor of both $a$ and $b$,which contradicts our earlier assumption that $a$ and $b$ have no common factors.
This shows that the assumption that $\sqrt{7}$ is rational is false.
Hence,the statement $p: \sqrt{7}$ is irrational is true.

(N/A) The given statement is in the form "If $p$,then $q$". To show that this statement is false,we need to find a counterexample where $p$ is true but $q$ is false.
Here,$p$ is "$n$ is an odd integer" and $q$ is "$n$ is prime".
We look for an odd integer $n$ that is not a prime number.
Consider $n = 9$.
$9$ is an odd integer,so $p$ is true.
However,$9$ is a composite number $(9 = 3 \times 3)$,so $q$ is false.
Since we found an odd integer that is not prime,the original statement is false.

(N/A) The statement $p$ is: "If $x$ is a real number such that $x^{3}+4x=0$,then $x$ is $0$."
Let $q$ be the statement: "$x$ is a real number such that $x^{3}+4x=0$."
Let $r$ be the statement: "$x$ is $0$."
To prove $p$ is true by the direct method,we assume $q$ is true and show that $r$ must be true.
Assume $q$ is true,so $x^{3}+4x=0$.
Factoring the equation,we get $x(x^{2}+4)=0$.
This implies $x=0$ or $x^{2}+4=0$.
Since $x$ is a real number,$x^{2} \geq 0$,which means $x^{2}+4 \geq 4$.
Therefore,$x^{2}+4$ cannot be $0$ for any real number $x$.
Thus,the only possibility is $x=0$.
Since we have shown that $q$ implies $r$,the statement $p$ is true.

(N/A) Let $p$ be the statement: 'If $x$ is a real number such that $x^{3}+4x=0$,then $x$ is $0$.'
To prove $p$ by contradiction,we assume that $p$ is false.
The negation of the statement 'If $q$,then $r$' is '$q$ and not $r$'.
Thus,we assume that $x$ is a real number such that $x^{3}+4x=0$ and $x \neq 0$.
Given $x^{3}+4x=0$,we can factor this as $x(x^{2}+4)=0$.
This implies $x=0$ or $x^{2}+4=0$.
Since we assumed $x \neq 0$,we must have $x^{2}+4=0$,which means $x^{2}=-4$.
However,for any real number $x$,$x^{2} \geq 0$.
Therefore,$x^{2}=-4$ has no real solution.
This contradicts our assumption that $x$ is a real number.
Hence,our assumption that $p$ is false must be incorrect.
Therefore,the statement $p$ is true.

(A) The statement $p$ is of the form 'If $q$,then $r$',where:
$q: x$ is a real number such that $x^{3}+4x=0$
$r: x=0$
To prove $p$ is true by the contrapositive method,we must show that $\sim r \Rightarrow \sim q$.
Assume $\sim r$ is true,which means $x \neq 0$.
We need to show that $\sim q$ is true,i.e.,$x^{3}+4x \neq 0$.
Since $x \neq 0$,we have $x^{2} > 0$.
Therefore,$x^{2}+4 > 4$,which implies $x^{2}+4 > 0$.
Since $x \neq 0$ and $(x^{2}+4) > 0$,their product $x(x^{2}+4)$ must be non-zero.
Thus,$x^{3}+4x \neq 0$,which is $\sim q$.
Since $\sim r \Rightarrow \sim q$ is true,the original statement $p$ is true.

(A) The given statement can be written in the form of "if-then" as follows:
If $a$ and $b$ are real numbers such that $a^{2}=b^{2}$,then $a=b$.
Let $p$ be the statement: $a$ and $b$ are real numbers such that $a^{2}=b^{2}$.
Let $q$ be the statement: $a=b$.
To prove the statement is false,we need to find a case where $p$ is true but $q$ is false (i.e.,$a^{2}=b^{2}$ but $a \neq b$).
Let $a=1$ and $b=-1$.
Then $a^{2}=(1)^{2}=1$ and $b^{2}=(-1)^{2}=1$.
Thus,$a^{2}=b^{2}$ holds true.
However,$a=1$ and $b=-1$,so $a \neq b$.
Since we found a case where $a^{2}=b^{2}$ but $a \neq b$,the statement is false.

(N/A) The statement $p$ is: If $x$ is an integer and $x^{2}$ is even,then $x$ is even.
To prove this by the method of contrapositive,we assume the negation of the conclusion and prove the negation of the hypothesis.
Let $q: x$ is an integer and $x^{2}$ is even.
Let $r: x$ is even.
The contrapositive of the statement $q \implies r$ is $\neg r \implies \neg q$.
Assume $\neg r$: $x$ is not even,i.e.,$x$ is odd.
If $x$ is odd,then $x = 2k + 1$ for some integer $k$.
Then $x^{2} = (2k + 1)^{2} = 4k^{2} + 4k + 1 = 2(2k^{2} + 2k) + 1$.
Since $2(2k^{2} + 2k) + 1$ is of the form $2m + 1$,$x^{2}$ is odd.
Thus,$\neg q$ is true: $x^{2}$ is not even.
Since $\neg r \implies \neg q$ is true,the original statement $p$ is true.

(N/A) The given statement is of the form 'if $q$ then $r$'.
$q:$ All the angles of a triangle are equal.
$r:$ The triangle is an obtuse-angled triangle.
The statement $p$ is false if we can find a case where $q$ is true but $r$ is false.
In an equilateral triangle,all three angles are equal to $60^{\circ}$.
Since $60^{\circ} < 90^{\circ}$,an equilateral triangle is an acute-angled triangle,not an obtuse-angled triangle.
Thus,the statement $p$ is false because we have found a counterexample (the equilateral triangle) where the premise is true but the conclusion is false.

(N/A) The given statement is: $q:$ The equation $x^{2}-1=0$ does not have a root lying between $0$ and $2$.
To prove this statement false,we need a counterexample.
Consider the equation $x^{2}-1=0$.
Solving for $x$:
$x^{2} = 1$
$x = \pm 1$
The roots of the equation are $x = 1$ and $x = -1$.
Since the root $x = 1$ lies between $0$ and $2$,the statement that there is no root between $0$ and $2$ is false.

(A) Given the inequality $x > y$.
According to the rules of inequality,if we multiply both sides of an inequality by a negative number,the direction of the inequality sign reverses.
Multiplying both sides by $-1$,we get:
$(-1) \times x < (-1) \times y$
$-x < -y$
Thus,the given statement $s$ is true.

(N/A) The "Or" used in the given statement is inclusive because it is possible that it rains and you are in the river simultaneously.
The component statements of the given statement are:
$p:$ You are wet when it rains.
$q:$ You are wet when you are in a river.
Since both component statements $p$ and $q$ can be true,the compound statement $t$ is true.

(N/A) The negation of a statement involving the quantifier 'for every' is formed by replacing it with 'there exists' and negating the condition.
The negation of $p$ is:
$\sim p:$ There exists a real number $x$ such that $x^{2} \leq x.$

(N/A) The negation of $q$ is denoted by $\sim q$,which means "it is false that $q$".
$\sim q:$ There does not exist a rational number $x$ such that $x^{2} = 2$.
This statement can also be expressed as:
$\sim q:$ For all rational numbers $x, x^{2} \neq 2$.

(N/A) The negation of the statement is:
$\sim r:$ There exists at least one bird which does not have wings.

(N/A) The negation of the given statement is $\sim s$: There exists at least one student who does not study mathematics at the elementary level.

(N/A) The statement "The integer $n$ is odd if and only if $n^{2}$ is odd" can be rewritten as: "The condition that the integer $n$ is odd is necessary and sufficient for $n^{2}$ to be odd."
Let $p$ be the statement: "The integer $n$ is odd."
Let $q$ be the statement: "$n^{2}$ is odd."
To check the validity of $p \iff q$,we examine both implications:
$1$. If $p$ is true,then $n = 2k + 1$ for some integer $k$. Then $n^{2} = (2k + 1)^{2} = 4k^{2} + 4k + 1 = 2(2k^{2} + 2k) + 1$,which is odd. Thus,$p \implies q$ is true.
$2$. If $q$ is true,we use the contrapositive: if $n$ is even,then $n = 2k$. Then $n^{2} = (2k)^{2} = 4k^{2} = 2(2k^{2})$,which is even. Since the contrapositive is true,$q \implies p$ is true.
Since both $p \implies q$ and $q \implies p$ are true,the original statement is true.

(N/A) Let $p$ and $q$ denote the statements:
$p:$ You drive over $80 \, km/h$.
$q:$ You will get a fine.
The implication $p \implies q$ indicates that $p$ is a sufficient condition for $q$. That is,driving over $80 \, km/h$ is sufficient to get a fine.
Therefore,the sufficient condition is "driving over $80 \, km/h$".
Similarly,the implication $p \implies q$ indicates that $q$ is a necessary condition for $p$. That is,to drive over $80 \, km/h$,it is necessary to get a fine.
Therefore,the necessary condition is "getting a fine".

(N/A) The negation of a statement involving the quantifier 'for every' is formed by replacing it with 'there exists' and negating the predicate.
The negation of statement $p$ is:
There exists a positive real number $x$ such that $x-1$ is not positive.

(N/A) The negation of the statement $q$ is:
$\sim q:$ There exists at least one cat that does not scratch.

(N/A) The negation of a statement involving the quantifier 'for every' is 'there exists'.
The negation of 'either $P$ or $Q$' is 'neither $P$ nor $Q$',which is equivalent to 'not $P$ and not $Q$'.
Therefore,the negation of statement $r$ is:
There exists a real number $x$ such that $x \leq 1$ and $x \geq 1$.

(D) The negation of a statement involving the existential quantifier 'There exists' is formed by replacing it with the universal quantifier 'For all' and negating the condition.
The original statement is: 'There exists a number $x$ such that $0 < x < 1.$'
The negation is: 'For all numbers $x$,it is not the case that $0 < x < 1.$'
This simplifies to: 'For all numbers $x$,$x \le 0$ or $x \ge 1.$'

(N/A) The statement $p$ can be written in the form 'If $q$,then $r$':
If a positive integer is prime,then it has no divisors other than $1$ and itself.
The converse of the statement is:
If a positive integer has no divisors other than $1$ and itself,then it is prime.
The contrapositive of the statement is:
If a positive integer has divisors other than $1$ and itself,then it is not prime.

(N/A) The given statement can be written in the form $p \implies q$ as: If it is a sunny day,then $I$ go to a beach.
The converse of the statement $p \implies q$ is $q \implies p$:
If $I$ go to a beach,then it is a sunny day.
The contrapositive of the statement $p \implies q$ is $\sim q \implies \sim p$:
If $I$ do not go to a beach,then it is not a sunny day.

(N/A) The converse of the statement $r$ is: If you feel thirsty,then it is hot outside.
The contrapositive of the statement $r$ is: If you do not feel thirsty,then it is not hot outside.

(N/A) The statement can be rewritten in the form "if $p$,then $q$" as follows:
If you log on to the server,then you must have a password.

(N/A) The statement "There is a traffic jam whenever it rains" means that if it rains,then there is a traffic jam.
Thus,the statement in the form "if $p$,then $q$" is: If it rains,then there is a traffic jam.

(N/A) The statement $r$ is of the form "$p$ only if $q$",which is logically equivalent to "if $p$,then $q$".
Here,$p$ is "You can access the website" and $q$ is "You pay a subscription fee".
Therefore,the statement in the form "if $p$,then $q$" is: "If you can access the website,then you pay a subscription fee."

(N/A) The statement "If you watch television,then your mind is free" and "If your mind is free,then you watch television" is equivalent to "You watch television if and only if your mind is free."

(A) You get an $A$ grade if and only if you do all the homework regularly.

(N/A) quadrilateral is equiangular if and only if it is a rectangle.

(N/A) The compound statement with 'And' is: "$25$ is a multiple of $5$ and $8$."
This is a false statement,since $25$ is not a multiple of $8$.
The compound statement with 'Or' is: "$25$ is a multiple of $5$ or $8$."
This is a true statement,since $25$ is a multiple of $5$ (even though it is not a multiple of $8$,the 'Or' condition is satisfied if at least one statement is true).

(A) The given statement is: $p:$ The sum of an irrational number and a rational number is irrational.
To check the validity by the contradiction method,we assume the negation of the statement is true.
Let us assume that the sum of an irrational number $x$ and a rational number $y$ is a rational number $z$.
So,$x + y = z$,where $x$ is irrational and $y, z$ are rational.
This implies $x = z - y$.
Since the difference of two rational numbers $(z - y)$ is always a rational number,this implies that $x$ is a rational number.
This contradicts our initial premise that $x$ is an irrational number.
Therefore,our assumption that the sum is rational must be false.
Hence,the sum of an irrational number and a rational number is irrational. The statement $p$ is true.

(N/A) The given statement $q$ is: If $n$ is a real number with $n > 3$,then $n^{2} > 9$.
To prove this by contradiction,we assume the negation of the statement.
Let us assume that $n$ is a real number with $n > 3$,but $n^{2} \leq 9$.
Since $n > 3$ and $n$ is a real number,we can square both sides of the inequality $n > 3$ without changing the inequality sign.
$n^{2} > 3^{2}$
$n^{2} > 9$
This contradicts our assumption that $n^{2} \leq 9$.
Since the assumption leads to a contradiction,the original statement must be true. Thus,if $n$ is a real number with $n > 3$,then $n^{2} > 9$.

(N/A) The given statement can be written in five different ways as follows:
$(i)$ $A$ triangle being equiangular implies that it is an obtuse-angled triangle.
$(ii)$ $A$ triangle is equiangular only if it is an obtuse-angled triangle.
$(iii)$ For a triangle to be equiangular,it is necessary that the triangle is an obtuse-angled triangle.
$(iv)$ For a triangle to be an obtuse-angled triangle,it is sufficient that the triangle is equiangular.
$(v)$ If a triangle is not an obtuse-angled triangle,then the triangle is not equiangular.

(D) tautology is a statement that is true for all possible truth values of its components. We evaluate the truth table for the expression $(p \wedge q)$ $\rightarrow (p$ $\rightarrow q)$:

$p$	$q$	$p \wedge q$	$p \rightarrow q$	$(p \wedge q)$ $\rightarrow (p$ $\rightarrow q)$
$T$	$T$	$T$	$T$	$T$
$T$	$F$	$F$	$F$	$T$
$F$	$T$	$F$	$T$	$T$
$F$	$F$	$F$	$T$	$T$

Since the final column contains only $T$ (True),the expression $(p \wedge q)$ $\rightarrow (p$ $\rightarrow q)$ is a tautology.

(A) Given the expression $(p \Rightarrow q) \Leftrightarrow (q * (\sim p))$ is a tautology.
We know that $p \Rightarrow q \equiv \sim p \vee q$.
Comparing $(p \Rightarrow q)$ with $(q * (\sim p))$,we observe that the operator $*$ corresponds to the disjunction operator $\vee$.
Thus,$p * (\sim q) \equiv p \vee (\sim q)$.
Using the implication rule $\sim a \vee b \equiv a \Rightarrow b$,we get $p \vee (\sim q) \equiv \sim (\sim p) \vee (\sim q) \equiv \sim p \Rightarrow \sim q$,which is not directly listed.
Alternatively,$p \vee (\sim q) \equiv \sim q \vee p \equiv q \Rightarrow p$.
Therefore,$p * (\sim q) \equiv q \Rightarrow p$.

(D) In mathematical logic,a statement is a declarative sentence that is either true or false.
Since the sentence "What is your name?" is an interrogative sentence (a question),it cannot be classified as true or false.
Therefore,it is not a statement.

(C) Let $p$ and $q$ be the statements:
$p: \text{I reach the station in time.}$
$q: \text{I will catch the train.}$
The given statement is of the form $p \rightarrow q$.
The contrapositive of the implication $p \rightarrow q$ is defined as $\sim q \rightarrow \sim p$.
Here,$\sim q$ is "$I$ will not catch the train" and $\sim p$ is "$I$ do not reach the station in time."
Therefore,the contrapositive is: "If $I$ will not catch the train,then $I$ do not reach the station in time."
This matches option $C$.

(A) To determine which expression is a tautology,we evaluate the truth table or simplify the logical expressions.
For option $A$: $(\sim p \wedge (p \vee q)) \rightarrow q$
Using the distributive law: $(\sim p \wedge p) \vee (\sim p \wedge q) \rightarrow q$
Since $(\sim p \wedge p) \equiv F$ (Contradiction),we have: $F \vee (\sim p \wedge q) \rightarrow q$
This simplifies to: $(\sim p \wedge q) \rightarrow q$
Using the implication law $a \rightarrow b \equiv \sim a \vee b$: $\sim (\sim p \wedge q) \vee q$
Applying De Morgan's law: $(p \vee \sim q) \vee q$
By associativity: $p \vee (\sim q \vee q)$
Since $(\sim q \vee q) \equiv T$ (Tautology): $p \vee T \equiv T$
Thus,option $A$ is a tautology.

(A) Given the proposition: $p \rightarrow \sim( p \wedge \sim q )$
Using the implication law $A \rightarrow B \equiv \sim A \vee B$,we get:
$\sim p \vee \sim( p \wedge \sim q )$
Applying De Morgan's Law $\sim( A \wedge B ) \equiv \sim A \vee \sim B$:
$\sim p \vee (\sim p \vee \sim(\sim q))$
Since $\sim(\sim q) \equiv q$:
$\sim p \vee \sim p \vee q$
Using the idempotent law $\sim p \vee \sim p \equiv \sim p$:
$\sim p \vee q$
Thus,the proposition is equivalent to $(\sim p) \vee q$.

(C) Let $p$ be the statement: 'Function $f$ is differentiable at $a$'.
Let $q$ be the statement: 'Function $f$ is continuous at $a$'.
The given statement is $p \rightarrow q$.
The contrapositive of $p \rightarrow q$ is defined as $\sim q \rightarrow \sim p$.
Here,$\sim q$ is: 'Function $f$ is not continuous at $a$'.
And $\sim p$ is: 'Function $f$ is not differentiable at $a$'.
Therefore,the contrapositive is: 'If a function $f$ is not continuous at $a$,then it is not differentiable at $a$'.

(C) For $(S_{1}): (q \vee p) \rightarrow (p \leftrightarrow \sim q)$
If $p = T$ and $q = T$,then $(T \vee T)$ $\rightarrow (T \leftrightarrow F)$ $\Rightarrow T$ $\rightarrow F = F$. Since it is not true for all truth values,$(S_{1})$ is not a tautology.
For $(S_{2}): \sim q \wedge (\sim p \leftrightarrow q)$
If $p = F$ and $q = F$,then $\sim F \wedge (\sim F \leftrightarrow F)$ $\Rightarrow T \wedge (T \leftrightarrow F)$ $\Rightarrow T \wedge F = F$.
If $p = T$ and $q = F$,then $\sim F \wedge (\sim T \leftrightarrow F)$ $\Rightarrow T \wedge (F \leftrightarrow F)$ $\Rightarrow T \wedge T = T$.
Since there exists a case where the truth value is $T$,$(S_{2})$ is not a fallacy (contradiction).
Thus,both $(S_{1})$ and $(S_{2})$ are incorrect.

(C) To determine if the statement is a tautology,we construct a truth table for the expression $(p$ $\rightarrow (q$ $\rightarrow p))$ $\rightarrow (p$ $\rightarrow (p \vee q))$.

$p$	$q$	$q \rightarrow p$	$p$ $\rightarrow (q$ $\rightarrow p)$	$p \vee q$	$p \rightarrow (p \vee q)$	$(p$ $\rightarrow (q$ $\rightarrow p))$ $\rightarrow (p$ $\rightarrow (p \vee q))$
$T$	$T$	$T$	$T$	$T$	$T$	$T$
$T$	$F$	$T$	$T$	$T$	$T$	$T$
$F$	$T$	$F$	$T$	$T$	$T$	$T$
$F$	$F$	$T$	$T$	$F$	$T$	$T$

Since the final column contains only $T$ (True) values for all possible combinations of $p$ and $q$,the statement is a tautology.