The World of Numbers Class 9 CBSE Important Questions with Answers

Answer: (c) $\sqrt{7}$√7.

$\sqrt{16}=4$√16=4 is a whole number, $0.\overline{3}=\tfrac13$0.3=1/3 is rational, and $\tfrac{22}{7}$22/7 is a ratio of integers. Only $\sqrt7$√7 is not a perfect square, giving a non-terminating, non-recurring decimal — it is irrational.

Answer: (b) terminating or recurring.

Every rational number $\tfrac{p}{q}$p/q has a decimal expansion that either terminates or repeats a block of digits. Non-terminating, non-recurring decimals belong to irrational numbers.

Answer: (a) $\dfrac{3}{8}$3/8.

$\dfrac38=0.375$3/8=0.375 terminates because its denominator $8=2^3$8=2³ has no prime factor other than $2$2 or $5$5. $\dfrac13=0.\overline{3}$1/3=0.3, $\dfrac27=0.\overline{285714}$2/7=0.285714 and $\dfrac56=0.8\overline{3}$5/6=0.83 all recur.

Answer: (d) Infinitely many.

Rational numbers are dense: between any two of them there is another (for example their average), and between that one and either end there is yet another, and so on without end.

Answer: (c) A is true but R is false. $\sqrt2 + (-\sqrt2) = 0$√2 + (-√2) = 0, which is rational, so A is true. But the sum of two irrationals is not always irrational (this example itself gives a rational $0$0), so R is false.

Write both with denominator $24$24: $\dfrac13=\dfrac{8}{24}$1/3=8/24 and $\dfrac12=\dfrac{12}{24}$1/2=12/24.

Any of $\dfrac{9}{24},\ \dfrac{10}{24},\ \dfrac{11}{24}$9/24, 10/24, 11/24 lies between them. Two of them are $\dfrac{9}{24}=\dfrac38$9/24=3/8 and $\dfrac{10}{24}=\dfrac{5}{12}$10/24=5/12.

Multiply numerator and denominator by $4$4 (one more than the number of rationals needed): $\dfrac35=\dfrac{12}{20}$3/5=12/20 and $\dfrac45=\dfrac{16}{20}$4/5=16/20.

So $\dfrac{13}{20},\ \dfrac{14}{20}=\dfrac{7}{10}$13/20, 14/20=7/10 and $\dfrac{15}{20}=\dfrac34$15/20=3/4 lie between them.

Let $x = 0.\overline{36} = 0.363636\ldots$x = 0.36 = 0.363636…

Since two digits repeat, multiply by $100$100: $100x = 36.3636\ldots$100x = 36.3636…

Subtract: $100x - x = 36.3636\ldots - 0.3636\ldots$100x - x = 36.3636… - 0.3636…, so $99x = 36$99x = 36.

Thus $x = \dfrac{36}{99} = \dfrac{4}{11}$x = 36/99 = 4/11 (dividing top and bottom by $9$9).

Hence $0.\overline{36} = \dfrac{4}{11}$0.36 = 4/11.

Suppose, on the contrary, that $\sqrt2$√2 is rational. Then $\sqrt2=\dfrac{p}{q}$√2=p/q, where $p$p and $q$q are integers with no common factor other than $1$1, and $q\neq0$q≠0.

Squaring, $2=\dfrac{p^2}{q^2}$2=p²/q², so $p^2=2q^2$p²=2q². Hence $p^2$p² is even, and so $p$p is even (the square of an odd number is odd). Write $p=2m$p=2m.

Then $4m^2=2q^2$4m²=2q², so $q^2=2m^2$q²=2m². Hence $q^2$q² is even, and so $q$q is even.

Now $p$p and $q$q are both even, so they have the common factor $2$2. This contradicts our assumption. Therefore $\sqrt2$√2 is irrational.

Take $c=\dfrac{a+b}{2}$c=(a+b)/2. The sum of two rationals is rational and half of a rational is rational, so $c$c is rational.

Since $a<b$a<b: $a+a<a+b<b+b$a+a<a+b<b+b, i.e. $2a<a+b<2b$2a<a+b<2b. Dividing by $2$2, $a<c<b$a<c<b. So $c$c lies between $a$a and $b$b.

Repeating the step between $a$a and $c$c, then between $a$a and that number, and so on, always gives a new rational number. The process never ends, so there are infinitely many rationals between $a$a and $b$b. This property is called the density of rational numbers.

Steps:

1. Mark a point $O$O and draw $OP_1=1$OP₁=1 unit.

2. At $P_1$P₁ draw $P_1P_2$P₁P₂ of $1$1 unit perpendicular to $OP_1$OP₁. Join $OP_2$OP₂.

3. At $P_2$P₂ draw $P_2P_3$P₂P₃ of $1$1 unit perpendicular to $OP_2$OP₂. Join $OP_3$OP₃.

4. Continue in the same way to get $P_4$P₄, $P_5$P₅ and $P_6$P₆.

Why $OP_k=\sqrt{k}$OP_k=√k: each triangle $OP_kP_{k+1}$OP_kP_k+1 is right-angled at $P_k$P_k. By Pythagoras, $OP_{k+1}^2=OP_k^2+1^2$OP_k+1²=OP_k²+1². Starting from $OP_1^2=1$OP₁²=1, the squares are $2,3,4,5,6,\dots$2,3,4,5,6,…, so $OP_2=\sqrt2$OP₂=√2, $OP_3=\sqrt3$OP₃=√3, $OP_4=\sqrt4=2$OP₄=√4=2, $OP_5=\sqrt5$OP₅=√5, $OP_6=\sqrt6$OP₆=√6.

(i) Suppose $\sqrt3$√3 is rational: $\sqrt3=\dfrac{p}{q}$√3=p/q with integers $p,q$p,q having no common factor other than $1$1, $q\neq0$q≠0.

Then $p^2=3q^2$p²=3q², so $3$3 divides $p^2$p².

If $3$3 did not divide $p$p, then $p=3m+1$p=3m+1 or $p=3m+2$p=3m+2, and $p^2=9m^2+6m+1$p²=9m²+6m+1 or $9m^2+12m+4$9m²+12m+4, which leave remainder $1$1 when divided by $3$3. So $3$3 must divide $p$p. Write $p=3m$p=3m.

Then $9m^2=3q^2$9m²=3q², so $q^2=3m^2$q²=3m², and by the same reasoning $3$3 divides $q$q.

So $3$3 is a common factor of $p$p and $q$q, a contradiction. Hence $\sqrt3$√3 is irrational.

(ii) If $2\sqrt3=r$2√3=r were rational, then $\sqrt3=\dfrac{r}{2}$√3=r/2 would also be rational, contradicting (i). Hence $2\sqrt3$2√3 is irrational.

(i) Side $=\sqrt{50}$=√50 cm. Since $50$50 is not a perfect square, $\sqrt{50}$√50 is irrational.

(ii) $7^2=49<50<64=8^2$7²=49<50<64=8², so $7<\sqrt{50}<8$7<√50<8: the side lies between $7$7 cm and $8$8 cm.

(iii) $7.07$7.07 is a terminating decimal, equal to $\dfrac{707}{100}$707/100, so it is rational. It only approximates the irrational exact value $\sqrt{50}$√50.

(iv) For example $7.075=\dfrac{7075}{1000}=\dfrac{283}{40}$7.075=7075/1000=283/40, which lies between $7.07$7.07 and $7.08$7.08.