Q: Let \$N\$ denote the set of all natural numbers and \$R\$ be the relation on \$N\times N\$ defined by \$(a, b) R (c, d)\Longleftrightarrow ad (b+c)=bc(a+d)\$. Check weather \$R\$ is an equivalence relation.

Let R be defined on \$N\times N\$ as \$(a, b) R (c, d)\Longleftrightarrow ad (b+c)=bc(a+d)\$. ....(1)Reflexivity:We can write \$ab(b+a)= ba(a+b)\$ for all \$a, b\in N\$Since, sum and product of natural numbers obeys commutative propertyHence, by def (1), we can write \$(a,b) R (a,b)\$ for all \$(a,b) \in N\times N\$Hence, \$R\$ is reflexive.Symmmetry :Let \$(a,b) R (c,d)\$ \$\Rightarrow ad(b+c)=bc(a+d)\$\$\Rightarrow da(c+b)=cb(d+a)\$ (Since, sum and product of natural numbers obeys commutative property)or \$cb(d+a)=da(c+b)\$\$\Rightarrow (c,d) R(a,b)\$Hence, \$R\$ is symmetric Transitivity :Let \$(a,b) ,(c,d) , (e,f) \in N\times N\$Let \$(a,b)R (c,d)\$ and \$(c,d)R(e,f)\$\$ad(b+c)=bc(a+d)\$ and \$cf(d+e)=de(c+f)\$\$\Rightarrow \dfrac{ab}{a-b} =\dfrac{cd}{c-d}\$ and \$\dfrac{cd}{c-d}=\dfrac{ef}{e-f}\$\$\Rightarrow \dfrac{ab}{a-b}=\dfrac{ef}{e-f}\$\$\Rightarrow af(b+e) =be(a+f)\$\$\Rightarrow (a,b)R(e,f)\$Hence ,\$R\$ is transitive\$\therefore R\$ is Equivalence Relation.

Q: Find how many three digits natural numbers are divisible by \$5\$.

Three digit number are \$100,105.........995\$a\$= 100\$ (a= first term)d\$= 105-100=5\$ (d=common difference)\$\begin{array}{l} \Rightarrow { a_{ n } }=a+(n-1)d \\ \Rightarrow 995=100+(n-1)5 \\ \Rightarrow 995-100=(n-1)5 \\ \Rightarrow \frac { { 895 } }{ 5 } =n-1 \\ \Rightarrow 179=n-1 \\ \Rightarrow n=179+1 \\ \Rightarrow n=180 \end{array}\$Hence,\$180\$ three digit natural number are divisible by 5.

Q: Find the reduced form of the ratio of the first quantity to second quantity.\$5\ litre,\ 2500\ ml\$

\$\frac{5litre}{2500ml}=\frac{5\times 1000ml}{2500ml}=\frac{50}{25}=2\$

Q: Find an irrational number between \$3\$ and \$4\$

\$\textbf{Step - 1 : Find an irrational number between 3 and 4.}\$ \$\text{We know,}{ \sqrt {ab} } \text{ is an irrational number between a and b.}\$ \$\text{Let, }a= 3 \text{ and } b= 4 \$ \$\therefore\sqrt {ab} = \sqrt {3\times 4}\$ \$= 2\sqrt 3\$\$\textbf{Hence, an irrational number between 3 and 4 is }\bf{2\sqrt 3.}\$

Q: State true or false:Every whole number is a natural number.

Natural numbers \$N =\$ Integers from \$1\$ to infinityWhole numbers \$W =\$ Integers from \$0\$ to infinityHence, every natural number is a whole number, but, every whole number is not a natural number.

Q: Is there any natural number that has no predecessor?

Yes, it will be \$1\$.Predecessor of \$1\$ is \$1-1=0\$.\$0\$ is a whole number but not a natural number.Hence the statement is true.

Question 1

Let $N$ denote the set of all natural numbers and $R$ be the relation on $N\times N$ defined by $(a, b) R (c, d)\Longleftrightarrow  ad (b+c)=bc(a+d)$. Check weather $R$ is an equivalence relation.

Accepted Answer

Let R be defined on $N\times N$ as $(a, b) R (c, d)\Longleftrightarrow  ad (b+c)=bc(a+d)$.     ....(1)Reflexivity:We can write $ab(b+a)= ba(a+b)$ for all $a, b\in N$Since, sum and product of natural numbers obeys commutative propertyHence, by def (1), we can write $(a,b) R (a,b)$   for all $(a,b) \in N\times N$Hence, $R$ is reflexive.Symmmetry :Let $(a,b) R (c,d)$ $\Rightarrow ad(b+c)=bc(a+d)$$\Rightarrow da(c+b)=cb(d+a)$    (Since, sum and product of natural numbers obeys commutative property)or $cb(d+a)=da(c+b)$$\Rightarrow (c,d) R(a,b)$Hence, $R$ is symmetric Transitivity :Let $(a,b) ,(c,d) , (e,f) \in N\times N$Let $(a,b)R (c,d)$ and $(c,d)R(e,f)$$ad(b+c)=bc(a+d)$ and $cf(d+e)=de(c+f)$$\Rightarrow \dfrac{ab}{a-b} =\dfrac{cd}{c-d}$ and $\dfrac{cd}{c-d}=\dfrac{ef}{e-f}$$\Rightarrow \dfrac{ab}{a-b}=\dfrac{ef}{e-f}$$\Rightarrow af(b+e) =be(a+f)$$\Rightarrow (a,b)R(e,f)$Hence ,$R$ is transitive$\therefore R$ is Equivalence Relation.

Question 2

The property represented by $a \times  (b + c) = a \times b + a \times  c$ is

Accepted Answer

Distributive property say that multiplication an division can be distributed over parenthesis.a x (b+c) = a x b + a x cSo correct answer will be option C

Question 3

Which of the following statements is incorrect?

Accepted Answer

The difference of two whole numbers is not necessarily a whole number. The statement whole numbers are closed under subtraction is an incorrect statement.

Question 4

Find how many three digits natural numbers are divisible by $5$.

Accepted Answer

Three digit number are $100,105.........995$a$= 100$                            (a= first term)d$= 105-100=5$         (d=common difference)$\begin{array}{l} \Rightarrow { a_{ n } }=a+(n-1)d \ \Rightarrow 995=100+(n-1)5 \ \Rightarrow 995-100=(n-1)5 \ \Rightarrow \frac { { 895 } }{ 5 } =n-1 \ \Rightarrow 179=n-1 \ \Rightarrow n=179+1 \ \Rightarrow n=180 \end{array}$Hence,$180$ three digit natural number are divisible by 5.

Question 5

Find the reduced form of the ratio of the first quantity to second quantity.$5\ litre,\ 2500\ ml$

Accepted Answer

$\frac{5litre}{2500ml}=\frac{5\times 1000ml}{2500ml}=\frac{50}{25}=2$

Question 6

Find an irrational number between $3$ and $4$

Accepted Answer

$\textbf{Step - 1 : Find an irrational number between 3 and 4.}$                    $\text{We know,}{ \sqrt {ab} } \text{ is an irrational number between a and b.}$                    $\text{Let, }a= 3 \text{ and } b= 4 $                    $\therefore\sqrt {ab} = \sqrt {3\times 4}$                                 $= 2\sqrt 3$$\textbf{Hence, an irrational number between 3 and 4 is }\bf{2\sqrt 3.}$

Question 7

How are the quantum numbers $n$, $1$ and $m_{1}$ arrived at? Explain the significance of these quantum numbers.?

Accepted Answer

Quantum numbers are a way to describe the discrete states that a particle can hold. These can then be used for solutions to the Schrodinger equation(above).Principal Quantum Number (n)&#8212;this designates the size of the orbital, also called the shell. The larger the value for $n$, the greater the average distance of an electron in the orbital from the nucleus and therefore the larger the orbital. (I could always remember this trend by relating it back to Coulomb&#8217;s Law of Attractive Forces.) From the equation that defines the line spectrum of Hydrogen, we can also know that the value for n represents energy levels.Angular Momentum Quantum Number (l)&#8212;this describes the shape of the orbital, but again, you can relate it back to the periodic table. The values of $l$ are integers that depend on the value of the principal quantum number, $n$. For any given value of $n$, the possible range of values for $l$ go from $0$ to $n - 1$. If $n = 1$, there&#8217;s only one possible value of $l$; that is, $0$ ($n - 1$ where $n = 1$). If $n = 2$, there&#8217;s two values for $l$ : $0$ and $1$. If $n = 3$, $l$ has three values: $0$, $1$, and $2$.Magnetic Quantum Number (ml)&#8212;this describes the orientation of the orbital in space, or the direction along the axes in which it faces. Within a sub-shell, the value of this depends on the value of $l$. For a certain value of $l$, there are $(2l + 1)$ integral values, or $-l, &#8230; 0, &#8230; +l$

Question 8

State true or false:Every whole number is a natural number.

Accepted Answer

Natural numbers $N =$ Integers from $1$ to infinityWhole numbers $W =$ Integers from $0$ to infinityHence, every natural number is a whole number, but, every whole number is not a natural number.

Question 9

Is there any natural number which has no successor? Is there a last number number?

Accepted Answer

Every natural number has successor.There are infinite natural numbers & hence there is no last natural number.Hence, every natural number has successor.

Question 10

Is there any natural number that has no predecessor?

Accepted Answer

Yes, it will be $1$.Predecessor of $1$ is $1-1=0$.$0$ is a whole number but not a natural number.Hence the statement is true.