\displaystyle \textbf{Question 1: }\text{Check the validity of the following statements.}
\displaystyle \text{Answer:}

\displaystyle \text{(i) }p:\ 100\text{ is a multiple of }4\text{ and }5.
\displaystyle \text{Since }100\text{ is a multiple of both }4\text{ and }5,\text{ both component statements are true.}
\displaystyle \text{The connective is ``AND''. Hence, the compound statement is true and is a valid statement.}

\displaystyle \text{(ii) }q:\ 125\text{ is a multiple of }5\text{ and }7.
\displaystyle \text{Since }125\text{ is a multiple of }5\text{ but not of }7,\text{ one component statement is false.}
\displaystyle \text{The connective is ``AND''. Hence, the compound statement is false and is not a valid statement.}

\displaystyle \text{(iii) }r:\ 60\text{ is a multiple of }3\text{ or }5.
\displaystyle \text{Since }60\text{ is a multiple of both }3\text{ and }5,\text{ both component statements are true.}
\displaystyle \text{The connective is ``OR''. Hence, the compound statement is true and is a valid statement.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Check whether the following statements are true or not:}
\displaystyle \text{(i) }p:\text{ If }x\text{ and }y\text{ are odd integers, then }x+y\text{ is an even integer.}
\displaystyle \text{(ii) }q:\text{ If }x,y\text{ are integers such that }xy\text{ is even, then at least one of}
\displaystyle x\text{ and }y\text{ is an even integer.}
\displaystyle \text{Answer:}

\displaystyle \text{(i) Let the component statements be:}
\displaystyle p_1:x\text{ and }y\text{ are odd integers.}
\displaystyle q_1:x+y\text{ is an even integer.}
\displaystyle \text{The given statement is }p_1\Rightarrow q_1.
\displaystyle \text{Let }p_1\text{ be true. Then }x\text{ and }y\text{ are odd integers.}
\displaystyle \therefore x=2m+1\text{ and }y=2n+1\text{ for some integers }m\text{ and }n.
\displaystyle x+y=(2m+1)+(2n+1)
\displaystyle =2m+2n+2
\displaystyle =2(m+n+1).
\displaystyle \text{Since }m+n+1\text{ is an integer, }x+y\text{ is an even integer.}
\displaystyle \therefore q_1\text{ is true whenever }p_1\text{ is true.}
\displaystyle \therefore \text{The given statement is true.}

\displaystyle \text{(ii) Let the component statements be:}
\displaystyle p_2:xy\text{ is an even integer.}
\displaystyle q_2:\text{At least one of }x\text{ and }y\text{ is an even integer.}
\displaystyle \text{The given statement is }p_2\Rightarrow q_2.
\displaystyle \text{We prove it by the contrapositive method.}
\displaystyle \text{Let }q_2\text{ be false. Then neither }x\text{ nor }y\text{ is even.}
\displaystyle \therefore x\text{ and }y\text{ are both odd integers.}
\displaystyle \therefore x=2m+1\text{ and }y=2n+1\text{ for some integers }m\text{ and }n.
\displaystyle xy=(2m+1)(2n+1)
\displaystyle =4mn+2m+2n+1
\displaystyle =2(2mn+m+n)+1.
\displaystyle \text{Hence, }xy\text{ is an odd integer, so }p_2\text{ is false.}
\displaystyle \therefore \sim q_2\Rightarrow\sim p_2.
\displaystyle \therefore p_2\Rightarrow q_2\text{ is true.}
\displaystyle \therefore \text{The given statement is true.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Show that the statement }p:\text{ If }x\text{ is a real number such that}
\displaystyle x^3+x=0,\text{ then }x=0,\text{ is true by:}
\displaystyle \text{(i) direct method}\qquad\text{(ii) method of contrapositive}\qquad\text{(iii) method of contradiction.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the component statements be:}
\displaystyle q:x\text{ is a real number such that }x^3+x=0.
\displaystyle r:x=0.
\displaystyle \text{The given statement is }q\Rightarrow r.

\displaystyle \text{(i) Direct method}
\displaystyle \text{Let }q\text{ be true. Then }x\in\mathbb{R}\text{ and }x^3+x=0.
\displaystyle x(x^2+1)=0.
\displaystyle \text{For every real number }x,\ x^2+1>0.
\displaystyle \therefore x^2+1\neq0.
\displaystyle \therefore x=0.
\displaystyle \therefore r\text{ is true whenever }q\text{ is true.}
\displaystyle \therefore q\Rightarrow r\text{ is true.}
\displaystyle \text{Hence, the given statement is true by the direct method.}

\displaystyle \text{(ii) Method of contrapositive}
\displaystyle \text{The contrapositive of }q\Rightarrow r\text{ is }\sim r\Rightarrow\sim q.
\displaystyle \text{Let }r\text{ be false. Then }x\neq0.
\displaystyle \text{Also, since }x\in\mathbb{R},\ x^2+1>0.
\displaystyle \therefore x\neq0\text{ and }x^2+1\neq0.
\displaystyle \therefore x(x^2+1)\neq0.
\displaystyle \therefore x^3+x\neq0.
\displaystyle \therefore q\text{ is false.}
\displaystyle \therefore \sim r\Rightarrow\sim q\text{ is true.}
\displaystyle \therefore q\Rightarrow r\text{ is true.}
\displaystyle \text{Hence, the given statement is true by the method of contrapositive.}

\displaystyle \text{(iii) Method of contradiction}
\displaystyle \text{Assume, if possible, that }q\Rightarrow r\text{ is false.}
\displaystyle \therefore q\text{ is true and }r\text{ is false.}
\displaystyle \therefore x^3+x=0\text{ and }x\neq0.
\displaystyle x(x^2+1)=0.
\displaystyle \text{Since }x^2+1>0,\text{ we must have }x=0.
\displaystyle \text{Thus, }x=0\text{ and }x\neq0,\text{ which is a contradiction.}
\displaystyle \therefore q\Rightarrow r\text{ is true.}
\displaystyle \text{Hence, the given statement is true by the method of contradiction.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Show that the following statement is true by the method of contrapositive:}
\displaystyle p:\text{ If }x\text{ is an integer and }x^2\text{ is odd, then }x\text{ is also odd.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the component statements be:}
\displaystyle q:x\text{ is an integer and }x^2\text{ is odd.}
\displaystyle r:x\text{ is an odd integer.}
\displaystyle \text{The given statement is }q\Rightarrow r.
\displaystyle \text{Its contrapositive is }\sim r\Rightarrow\sim q.
\displaystyle \text{Let }r\text{ be false. Then }x\text{ is not an odd integer.}
\displaystyle \therefore x\text{ is an even integer.}
\displaystyle \therefore x=2n\text{ for some integer }n.
\displaystyle x^2=(2n)^2=4n^2=2(2n^2).
\displaystyle \therefore x^2\text{ is an even integer.}
\displaystyle \therefore q\text{ is false.}
\displaystyle \therefore \sim r\Rightarrow\sim q\text{ is true.}
\displaystyle \therefore q\Rightarrow r\text{ is true.}
\displaystyle \therefore \text{The given statement is true by the method of contrapositive.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Show that the following statement is true:}
\displaystyle \text{``The integer }n\text{ is even if and only if }n^2\text{ is even.''}
\displaystyle \text{Answer:}
\displaystyle \text{Let the statements be:}
\displaystyle p:n\text{ is an even integer.}
\displaystyle q:n^2\text{ is an even integer.}
\displaystyle \text{To prove }p\Leftrightarrow q,\text{ we prove both }p\Rightarrow q\text{ and }q\Rightarrow p.

\displaystyle \text{(i) Proof of }p\Rightarrow q:
\displaystyle \text{Let }p\text{ be true. Then }n=2k\text{ for some integer }k.
\displaystyle n^2=(2k)^2=4k^2=2(2k^2).
\displaystyle \therefore n^2\text{ is an even integer.}
\displaystyle \therefore q\text{ is true whenever }p\text{ is true.}
\displaystyle \therefore p\Rightarrow q\text{ is true.}

\displaystyle \text{(ii) Proof of }q\Rightarrow p:
\displaystyle \text{We prove the contrapositive. Assume }p\text{ is false. Then }n\text{ is odd.}
\displaystyle \therefore n=2k+1\text{ for some integer }k.
\displaystyle n^2=(2k+1)^2=4k^2+4k+1.
\displaystyle =2(2k^2+2k)+1,
\displaystyle \text{which is an odd integer.}
\displaystyle \therefore q\text{ is false.}
\displaystyle \therefore \sim p\Rightarrow\sim q\text{ is true.}
\displaystyle \therefore q\Rightarrow p\text{ is true.}
\displaystyle \therefore p\Leftrightarrow q\text{ is true.}
\displaystyle \therefore \text{The integer }n\text{ is even if and only if }n^2\text{ is even.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{By giving a counterexample, show that the following statement is not true.}
\displaystyle p:\text{ ``If all the angles of a triangle are equal, then the triangle is an obtuse-angled triangle.''}
\displaystyle \text{Answer:}
\displaystyle \text{Consider an equilateral triangle }PQR.
\displaystyle \therefore \angle P=\angle Q=\angle R=60^\circ.
\displaystyle \text{Since each angle is }60^\circ,\text{ no angle is greater than }90^\circ.
\displaystyle \therefore \triangle PQR\text{ is not an obtuse-angled triangle.}
\displaystyle \text{Thus, the given statement is not true.}
\displaystyle \text{Hence, }\triangle PQR\text{ is a counterexample to the given statement.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Which of the following statements are true and which are false?}
\displaystyle \text{In each case, give a valid reason for saying so.}
\displaystyle \text{Answer:}

\displaystyle \text{(i) }p:\text{ Each radius of a circle is a chord of the circle.}
\displaystyle \text{The given statement is false.}
\displaystyle \text{A chord is a line segment whose two endpoints lie on the circle, whereas a radius has}
\displaystyle \text{one endpoint at the centre and the other endpoint on the circle.}
\displaystyle \therefore \text{A radius is not a chord of the circle.}

\displaystyle \text{(ii) }q:\text{ The centre of a circle bisects each chord of the circle.}
\displaystyle \text{The given statement is false.}
\displaystyle \text{The centre does not lie on every chord and hence cannot bisect every chord.}
\displaystyle \text{Only a chord passing through the centre, that is, a diameter, is bisected by the centre.}
\displaystyle \text{In general, the perpendicular drawn from the centre to a chord bisects the chord.}

\displaystyle \text{(iii) }r:\text{ A circle is a particular case of an ellipse.}
\displaystyle \text{The given statement is true.}
\displaystyle \text{The standard equation of an ellipse is }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.
\displaystyle \text{If }a=b=r,\text{ then}
\displaystyle \frac{x^2}{r^2}+\frac{y^2}{r^2}=1
\displaystyle \Rightarrow x^2+y^2=r^2,
\displaystyle \text{which is the equation of a circle with centre at the origin and radius }r.
\displaystyle \therefore \text{A circle is a particular case of an ellipse.}

\displaystyle \text{(iv) }s:\text{ If }x\text{ and }y\text{ are integers such that }x>y,\text{ then }-x<-y.
\displaystyle \text{The given statement is true.}
\displaystyle x>y
\displaystyle \Rightarrow -x<-y.
\displaystyle \text{The sign of an inequality is reversed when both sides are multiplied by }-1.

\displaystyle \text{(v) }t:\sqrt{11}\text{ is a rational number.}
\displaystyle \text{The given statement is false.}
\displaystyle \text{Since }11\text{ is not a perfect square, }\sqrt{11}\text{ is an irrational number.}
\displaystyle \therefore \sqrt{11}\text{ is not a rational number.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Determine whether the argument used to check the validity of the following statement is correct:}
\displaystyle p:\text{ ``If }x^2\text{ is irrational, then }x\text{ is rational.''}
\displaystyle \text{The statement is true because }x^2=\pi^2\text{ is irrational; therefore }x=\pi\text{ is irrational.}
\displaystyle \text{Answer:}
\displaystyle \text{Argument used: }x^2=\pi^2\text{ is irrational and }x=\pi\text{ is irrational.}
\displaystyle \text{Here, }x=\pi\text{ is irrational and }x^2=\pi^2\text{ is also irrational.}
\displaystyle \text{Thus, the hypothesis }(x^2\text{ is irrational})\text{ is true, but the conclusion}
\displaystyle (x\text{ is rational})\text{ is false.}
\displaystyle \text{Hence, the implication is false.}
\displaystyle \text{Therefore, the given argument correctly shows that the statement is not true.}
\displaystyle \text{Thus, the argument used is correct.}
\displaystyle \\

 

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