Write down the converse of following statements.
(i) If a rectangle ' $R$ ' is a square, then $R$ is a rhombus.
(ii) If today is Monday, then tomorrow is Tuesday.
(iii) If you go to Agra, then you must visit Taj Mahal.
(iv) If sum of squares of two sides of a triangle is equal to the square of third side of a triangle, then the triangle is right angled.
(v) If all three angles of a triangle are equal, then the triangle is equilateral.
(vi) If $x: y=3: 2$, then $2 x=3 y$.
(vii) If $S$ is a cyclic quadrilateral, then the opposite angles of $S$ are supplementary.
(viii) If $x$ is zero, then $x$ is neither positive nor negative.
(ix) If two triangles are similar, then the ratio of their corresponding sides are equal.
(i) If thes rectangle ' $R$ ' is rhombus, then it is square.
(ii) If tomorrow is Tuesday, then today is Monday.
(iii) If you must visit Taj Mahal, you go to Agra.
(iv) If the triangle is right angle, then sum of squares of two sides of a triangle is equal to the square of third side.
(v) If the triangle is equilateral, then all three angles of triangle are equal.
(vi) If $2 x=3 y$, then $x: y=3: 2$
(vii) If the opposite angles of a quadrilateral are supplementary, then $S$ is cyclic.
(viii) If $x$ is neither positive nor negative, then $x$ is 0 .
(ix) If the ratio of corresponding sides of two triangles are equal, then triangles are similar.
Identify the quantifiers in the following statements.
(i) There exists a triangle which is not equilateral.
(ii) For all real numbers $x$ and $y, x y=y x$.
(iii) There exists a real number which is not a rational number.
(iv) For every natural number $x, x+1$ is also a natural number.
(v) For all real numbers $x$ with $x>3, x^2$ is greater than 9 .
(vi) There exists a triangle which is not an isosceles triangle.
(vii) For all negative integers $x, x^3$ is also a negative integers.
(viii) There exists a statement in above statements which is not true.
(ix) There exists an even prime number other than 2.
(x) There exists a real number $x$ such that $x^2+1=0$.
Quantifier are the phrases like 'There exist' and 'For every', 'For all' etc.
(i) There exists
(ii) For all
(iii) There exists
(iv) For every
(v) For all
(vi) There exists
(vii) For all
(viii) There exists
(ix) There exists
(x) There exists
Prove by direct method that for any integer ' $n$ ', $n^3-n$ is always even.
Here, two cases arise
Case I When $n$ is even,
$$\begin{aligned} & \text { Let } \quad n=2 K, K \in N \\ & \Rightarrow \quad n^3-n=(2 K)^3-(2 K)=2 K\left(4 K^2-1\right) \\ & =2 \lambda \text {, where } \lambda=K\left(4 K^2-1\right) \end{aligned}$$
Thus, $\left(n^3-n\right)$ is even when $n$ is even.
Case II When $n$ is odd,
$$\begin{aligned} \text{Let}\quad n & =2 K+1, K \in N \\ \Rightarrow \quad & n^3-n & =(2 K+1)^3-(2 K+1) \\ & =(2 K+1)\left[(2 K+1)^2-1\right] \\ & =(2 K+1)\left[4 K^2+1+4 K-1\right] \\ & =(2 K+1)\left(4 K^2+4 K\right) \\ & =4 K(2 K+1)(K+1) \\ & =2 \propto, \text { when } \alpha=2 K(K+1)(2 K+1) \end{aligned}$$
Then, $n^3-n$ is even when $n$ is odd.
So, $n^3-n$ is always even.
Check validity of the following statement.
(i) $p: 125$ is divisible by 5 and 7.
(ii) $q: 131$ is a multiple of 3 or 11 .
(i) $p: 125$ is divisible by 5 and 7 .
Let $q: 125$ is divisible by 5 .
$r: 125$ is divisible by 7 .
$q$ is true, $r$ is false.
$\Rightarrow q \wedge r$ is false.
[since, $p \wedge q$ has the truth value $F$ (false) whenever either $p$ or $q$ or both have the truth value F.]
Hence, $p$ is not valid.
(ii) $p: 131$ is a multiple of 3 or 11 .
Let $q: 131$ is multiple of 3 .
$r: 131$ is a multiple of 11 .
$p$ is true, $r$ is false.
$\Rightarrow \quad p \vee r$ is true. [since, $p \vee q$ has the truth value $T$ (true) whenever either $p$ or $q$ or both have the truth value T]
Hence, $q$ is valid.
Prove the following statement by contradiction method
$p$ : The sum of an irrational number and a rational number is irrational.
Let $p$ is false i.e., sum of an irrational and a rational number is rational.
Let $\sqrt{m}$ is irrational and $n$ is rational number.
$\Rightarrow\quad\sqrt{m}+n=r\quad \text{[rational]}$
$\Rightarrow\quad\sqrt{m}=r-n$
$\sqrt{\mathrm{m}}$ is irrational, where as $(r-n)$ is rational. This is contradiction.
Then, our supposition is wrong.
Hence, $p$ is true.