Question paper, Set 8

Sol: Every integer can be written with $q = 1$, and $1 \ne 0$.

$\displaystyle 8 = \frac{8}{1}$

$\displaystyle 18 = \frac{18}{1}$

Answer: $\displaystyle 8 = \frac{8}{1}$, $\displaystyle 18 = \frac{18}{1}$

Sol: (i) $\displaystyle \sqrt{9} = 3 = \frac{3}{1}$, so it is rational.

(ii) $\displaystyle \frac{21}{4}$ is in the form $\displaystyle \frac{p}{q}$ with $q \ne 0$, so it is rational.

(iii) $\sqrt{61} = 7.8102\ldots$ is a non terminating non recurring decimal, so it is irrational.

(iv) $2.\overline{5}$ is a non terminating recurring decimal, so it is rational.

Answer: (i) rational (ii) rational (iii) irrational (iv) rational

Sol: $\pi$ is irrational and $\displaystyle \frac{22}{7}$ is rational, so they are not equal.

$\displaystyle \pi \cong \frac{22}{7}$ ($\pi$ is approximately equal to $\displaystyle \frac{22}{7}$).

Answer: False

Sol: $4\sqrt{2} - 6\sqrt{3} - 11\sqrt{2} + 3\sqrt{3}$

$= (4 - 11)\sqrt{2} + (-6 + 3)\sqrt{3}$

$= -7\sqrt{2} - 3\sqrt{3}$

Answer: $-7\sqrt{2} - 3\sqrt{3}$

Sol: $(-11\sqrt{6} - 9\sqrt{13}) - (10\sqrt{6} - 7\sqrt{13})$

$= -11\sqrt{6} - 9\sqrt{13} - 10\sqrt{6} + 7\sqrt{13}$

$= (-11 - 10)\sqrt{6} + (-9 + 7)\sqrt{13}$

$= -21\sqrt{6} - 2\sqrt{13}$

Answer: $-21\sqrt{6} - 2\sqrt{13}$

Sol: $\sqrt{15} \times \sqrt{7} = \sqrt{15 \times 7}$

$= \sqrt{105}$

Answer: $\sqrt{105}$

Sol: $5\sqrt{2} \times 6\sqrt{2} = 5 \times 6 \times \sqrt{2} \times \sqrt{2}$

$= 30 \times 2$

$= 60$

Answer: $60$

Sol: (i) Positive terms: $93607\sqrt{2} + 930105\sqrt{2} = 1023712\sqrt{2}$

Negative terms: $24222\sqrt{2} + 60468\sqrt{2} = 84690\sqrt{2}$

$93607\sqrt{2} + 930105\sqrt{2} - 24222\sqrt{2} - 60468\sqrt{2} = 1023712\sqrt{2} - 84690\sqrt{2}$

$= (1023712 - 84690)\sqrt{2} = 939022\sqrt{2}$

(ii) Positive terms: $6236\sqrt{15}$

Negative terms: $62063\sqrt{15} + 17518\sqrt{15} = 79581\sqrt{15}$

$-62063\sqrt{15} - 17518\sqrt{15} + 6236\sqrt{15} = 6236\sqrt{15} - 79581\sqrt{15}$

$= -(79581 - 6236)\sqrt{15} = -73345\sqrt{15}$

Answer: (i) $939022\sqrt{2}$ (ii) $-73345\sqrt{15}$

Sol: (i) $9\sqrt{13} \times 4\sqrt{13} = 9 \times 4 \times \sqrt{13} \times \sqrt{13}$

$= 36 \times 13 = 468$ (since $\sqrt{a} \times \sqrt{a} = a$)

(ii) $3\sqrt{7} \times (-6\sqrt{28}) = 3 \times (-6) \times \sqrt{7 \times 28}$

$= -18 \times \sqrt{196}$

$\sqrt{196} = 14$

$= -18 \times 14 = -252$

(iii) $4\sqrt{3} \times 4\sqrt{147} = 4 \times 4 \times \sqrt{3 \times 147}$

$= 16 \times \sqrt{441}$

Long division: $2 \times 2 = 4$, remainder $0$; bring down $41$ to get $41$; $41 \times 1 = 41$. So $\sqrt{441} = 21$

$= 16 \times 21 = 336$

Answer: (i) $468$ (ii) $-252$ (iii) $336$

Sol: $\displaystyle \frac{4 \times \sqrt{441}}{\sqrt{21}} = 4 \times \frac{\sqrt{441}}{\sqrt{21}}$

$\displaystyle = 4 \times \sqrt{\frac{441}{21}} = 4 \times \sqrt{21}$

$= 4\sqrt{21}$

Answer: $4\sqrt{21}$

Sol: $\sqrt{208} = \sqrt{16 \times 13} = \sqrt{16} \times \sqrt{13} = 4\sqrt{13}$

Answer: $4\sqrt{13}$

Sol: $\sqrt{294} = \sqrt{49 \times 6} = \sqrt{49} \times \sqrt{6} = 7\sqrt{6}$

$\sqrt{320} = \sqrt{64 \times 5} = \sqrt{64} \times \sqrt{5} = 8\sqrt{5}$

$\sqrt{80} = \sqrt{16 \times 5} = \sqrt{16} \times \sqrt{5} = 4\sqrt{5}$, so $3\sqrt{80} = 3 \times 4\sqrt{5} = 12\sqrt{5}$

$\sqrt{96} = \sqrt{16 \times 6} = \sqrt{16} \times \sqrt{6} = 4\sqrt{6}$, so $3\sqrt{96} = 3 \times 4\sqrt{6} = 12\sqrt{6}$

$\therefore \sqrt{294} - \sqrt{320} - 3\sqrt{80} - 3\sqrt{96} = 7\sqrt{6} - 8\sqrt{5} - 12\sqrt{5} - 12\sqrt{6}$

$= (7 - 12)\sqrt{6} + (-8 - 12)\sqrt{5}$

$= -5\sqrt{6} - 20\sqrt{5}$

Answer: $-5\sqrt{6} - 20\sqrt{5}$

Sol: $a = 34 \in R$ and $b = 0 \in R$

$\displaystyle \frac{a}{b} = \frac{34}{0}$ is not defined.

$\displaystyle \therefore \frac{34}{0} \notin R$, so real numbers are not closed under division.

Answer: $34, 0 \in R$ but $\displaystyle \frac{34}{0} \notin R$

Sol: $a + b = 6\sqrt{10} + (-5\sqrt{10}) = \sqrt{10}$

$a - b = 6\sqrt{10} - (-5\sqrt{10}) = 11\sqrt{10}$

$a \times b = 6\sqrt{10} \times (-5\sqrt{10}) = -30 \times 10 = -300$

All three are real numbers, so $a + b, a - b, a \times b \in R$.

Answer: $a + b = \sqrt{10}$, $a - b = 11\sqrt{10}$, $a \times b = -300$; each $\in R$

Sol: The order of the two numbers is changed and the answer is the same.

This is the commutative property under multiplication.

Answer: Commutative property under multiplication

Sol: $a + b = 8\sqrt{17} + 7\sqrt{17} = 15\sqrt{17}$

$b + a = 7\sqrt{17} + 8\sqrt{17} = 15\sqrt{17}$

$a \times b = 8\sqrt{17} \times 7\sqrt{17} = 56 \times 17 = 952$

$b \times a = 7\sqrt{17} \times 8\sqrt{17} = 56 \times 17 = 952$

So $a + b = b + a$ and $a \times b = b \times a$.

Answer: $a + b = b + a = 15\sqrt{17}$, $a \times b = b \times a = 952$

Sol: (i) LHS $= 54 + (18 + 3) = 54 + 21 = 75$, RHS $= (54 + 18) + 3 = 72 + 3 = 75$

LHS $=$ RHS

(ii) LHS $= 54 - (18 - 3) = 54 - 15 = 39$, RHS $= (54 - 18) - 3 = 36 - 3 = 33$

LHS $\ne$ RHS

(iii) LHS $= 54 \times (18 \times 3) = 54 \times 54 = 2916$, RHS $= (54 \times 18) \times 3 = 972 \times 3 = 2916$

LHS $=$ RHS

(iv) LHS $= (54 \div 18) \div 3 = 3 \div 3 = 1$, RHS $= 54 \div (18 \div 3) = 54 \div 6 = 9$

LHS $\ne$ RHS

Answer: (i) $75 = 75$ (ii) $39 \ne 33$ (iii) $2916 = 2916$ (iv) $1 \ne 9$

Sol: $a + 0 = a$, where $0$ is the additive identity.

$\therefore 5 + 0 = 5$

Answer: $0$

Sol: (i) Additive inverse of $a$ is $-a$, since $a + (-a) = 0$. So it is $-8\sqrt{2}$.

(ii) Multiplicative inverse of $a$ is $\displaystyle \frac{1}{a}$, since $\displaystyle a \times \frac{1}{a} = 1$. So it is $\displaystyle -\frac{4}{7}$.

Answer: (i) $-8\sqrt{2}$ (ii) $\displaystyle -\frac{4}{7}$

Sol: Additive inverse of $26$ is $-26$.

Multiplicative inverse of $26$ is $\displaystyle \frac{1}{26}$.

Check: $26 + (-26) = 0$

Check: $\displaystyle 26 \times \frac{1}{26} = 1$

Answer: Additive inverse $-26$, multiplicative inverse $\displaystyle \frac{1}{26}$