Rational and Irrational Numbers — ICSE Class 9 Maths Important Questions

Answer: (c) $\sqrt7$7.

$\sqrt{16}=4$√16=4 and $\dfrac{22}{7}$22/7 are rational, $0.\overline{3}=\dfrac13$0.3=13 is a recurring (rational) decimal, but $\sqrt7$7 is non-terminating and non-recurring, hence irrational.

Answer: (b) $\sqrt5$5.

Multiplying $\sqrt5\times\sqrt5=5$5×5=5, a rational number, so $\sqrt5$5 is the rationalising factor.

Answer: (b) $\sqrt3+\sqrt2$3+2.

$\dfrac{1}{\sqrt3-\sqrt2}\times\dfrac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}=\dfrac{\sqrt3+\sqrt2}{3-2}=\sqrt3+\sqrt2.$1/3-2×3+2/3+2=3+2/3-2=3+2.

Answer: (a) $2\sqrt5$25.

$\dfrac{1}{x}=\dfrac{1}{\sqrt5+2}=\dfrac{\sqrt5-2}{5-4}=\sqrt5-2$1/x=1/5+2=5-2/5-4=5-2. So $x+\dfrac1x=(\sqrt5+2)+(\sqrt5-2)=2\sqrt5.$x+1x=(5+2)+(5-2)=25.

Answer: (c) $\sqrt2+\sqrt3$2+3 is indeed irrational, so A is true. But R is false: the sum of two irrationals need not be irrational, e.g. $(2+\sqrt3)+(2-\sqrt3)=4$(2+3)+(2-3)=4, which is rational.

As decimals, $\dfrac13=0.333\ldots$13=0.333 and $\dfrac12=0.5$12=0.5.

Rational number: $0.4=\dfrac{2}{5}$0.4=2/5 lies between them.

Irrational number: $0.4040040004\ldots$0.4040040004 (a non-terminating, non-recurring decimal) lies between $0.333\ldots$0.333 and $0.5$0.5, so it is a valid irrational number in the interval.

Write each surd in terms of $\sqrt5$5:

$\sqrt{45}=\sqrt{9\times5}=3\sqrt5$√45=√9×5=35 and $\sqrt{20}=\sqrt{4\times5}=2\sqrt5$√20=√4×5=25.

$\therefore\ \sqrt{45}-3\sqrt{20}+4\sqrt5=3\sqrt5-3(2\sqrt5)+4\sqrt5=3\sqrt5-6\sqrt5+4\sqrt5=\sqrt5.$√45-3√20+45=35-3(25)+45=35-65+45=5.

Multiply numerator and denominator by the conjugate $3+\sqrt2$3+2:

$\dfrac{3+\sqrt2}{3-\sqrt2}\times\dfrac{3+\sqrt2}{3+\sqrt2}=\dfrac{(3+\sqrt2)^2}{(3)^2-(\sqrt2)^2}.$3+2/3-2×3+2/3+2=(3+2)^2/(3)^2-(2)^2.

Numerator: $(3+\sqrt2)^2=9+6\sqrt2+2=11+6\sqrt2$(3+2)^2=9+62+2=11+62.

Denominator: $9-2=7$9-2=7.

$\therefore\ \dfrac{3+\sqrt2}{3-\sqrt2}=\dfrac{11+6\sqrt2}{7}.$3+2/3-2=11+62/7.

Multiply by the conjugate $\sqrt7+\sqrt5$7+5:

$\dfrac{1}{\sqrt7-\sqrt5}\times\dfrac{\sqrt7+\sqrt5}{\sqrt7+\sqrt5}=\dfrac{\sqrt7+\sqrt5}{(\sqrt7)^2-(\sqrt5)^2}=\dfrac{\sqrt7+\sqrt5}{7-5}=\dfrac{\sqrt7+\sqrt5}{2}.$1/7-5×7+5/7+5=7+5/(7)^2-(5)^2=7+5/7-5=7+5/2.

So the rationalised form is $\dfrac{\sqrt7+\sqrt5}{2}$7+5/2, i.e. $\dfrac12\sqrt7+\dfrac12\sqrt5$127+125.

Assume, to the contrary, that $\sqrt3$3 is rational. Then $\sqrt3=\dfrac{p}{q}$3=p/q where $p,q$p,q are integers with $q\neq0$q≠0 and $\gcd(p,q)=1$(p,q)=1 (in lowest terms).

Squaring: $3=\dfrac{p^2}{q^2}\Rightarrow p^2=3q^2$3=p^2/q^2 p^2=3q^2. So $3\mid p^2$3 p^2, hence $3\mid p$3 p. Let $p=3k$p=3k.

Then $(3k)^2=3q^2\Rightarrow 9k^2=3q^2\Rightarrow q^2=3k^2$(3k)^2=3q^2 9k^2=3q^2 q^2=3k^2, so $3\mid q^2$3 q^2, hence $3\mid q$3 q.

Thus $3$3 divides both $p$p and $q$q, contradicting $\gcd(p,q)=1$(p,q)=1. Therefore our assumption is wrong and $\sqrt3$3 is irrational.

First rationalise $x$x:

$x=\dfrac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}\times\dfrac{\sqrt5+\sqrt3}{\sqrt5+\sqrt3}=\dfrac{(\sqrt5+\sqrt3)^2}{5-3}=\dfrac{5+2\sqrt{15}+3}{2}=\dfrac{8+2\sqrt{15}}{2}=4+\sqrt{15}.$x=5+3/5-3×5+3/5+3=(5+3)^2/5-3=5+2√15+3/2=8+2√15/2=4+√15.

Similarly $y=4-\sqrt{15}$y=4-√15.

Now $x+y=(4+\sqrt{15})+(4-\sqrt{15})=8$x+y=(4+√15)+(4-√15)=8 and $xy=(4+\sqrt{15})(4-\sqrt{15})=16-15=1$xy=(4+√15)(4-√15)=16-15=1.

Use $x^2+y^2+xy=(x+y)^2-xy$x^2+y^2+xy=(x+y)^2-xy:

$x^2+y^2+xy=(8)^2-1=64-1=63.$x^2+y^2+xy=(8)^2-1=64-1=63.

Construction steps:

1. Draw a number line and mark $O$O at $0$0 and $A$A at $1$1, so $OA=1$OA=1 unit.

2. At $A$A, draw $AB\perp OA$AB OA with $AB=1$AB=1 unit. Then $OB=\sqrt{1^2+1^2}=\sqrt2$OB=√1^2+1^2=2 (by Pythagoras).

3. At $B$B, draw $BC\perp OB$BC OB with $BC=1$BC=1 unit. Then $OC=\sqrt{(\sqrt2)^2+1^2}=\sqrt{2+1}=\sqrt3$OC=√(2)^2+1^2=√2+1=3.

4. With centre $O$O and radius $OC$OC, draw an arc cutting the number line at point $P$P. Then $OP=\sqrt3$OP=3, so $P$P represents $\sqrt3$3.

(i) $\sqrt{50}=\sqrt{25\times2}=5\sqrt2$√50=√25×2=52 cm.

(ii) $\sqrt{72}=\sqrt{36\times2}=6\sqrt2$√72=√36×2=62 cm.

(iii) $\dfrac{1}{\sqrt2}=\dfrac{1}{\sqrt2}\times\dfrac{\sqrt2}{\sqrt2}=\dfrac{\sqrt2}{2}$1/2=1/2×2/2=2/2 cm.

(iv) $\sqrt{50}+\sqrt{72}=5\sqrt2+6\sqrt2=11\sqrt2$√50+√72=52+62=112 cm.