Solutions-Class 8-Mathematics-Chapter-1-Rational & Irrational Numbers-Maharashtra Board

\(\frac{3}{2}=1\frac{1}{2}=1+\frac{1}{2}\)

\(\frac{5}{2}=2\frac{1}{2}=2+\frac{1}{2}\)

\(\frac{-3}{2}=-1\frac{1}{2}=-1-\frac{1}{2}\)

Here, the denominators of the given numbers is 2.

Therefore we divide each unit distance into 2 equal parts.

\(\frac{-3}{2}\)  is the number to the left of 0

And at the same distance, \(\frac{3}{2}\)   is on the right of 0.

Here \(\frac{7}{5}=1+\frac{2}{5}\) ,

\(\frac{7}{5}\) is at \(\frac{2}{5}\)th distance after 1 on the right side of zero

Here, we divide each unit distance into 5 equal parts.

Here \(\frac{11}{8}=1+\frac{3}{8}\) , 

\(\frac{11}{8}\) is at \(\frac{3}{8}\) th distance after 1 on the right side of zero

Here, we divide each unit distance into 8 equal parts.

\(\frac{13}{10}=1\frac{3}{10}=1+\frac{3}{10}\)

\(\frac{-17}{10}=-1\frac{7}{10}=-1-\frac{7}{10}\)

Here, we divide each unit distance into 10 equal parts.

The point B is at half distance of unit after -2 on the left side of zero. 

∴  The number -10/4 or -5/2 is indicated by the point B.

  \(1\frac{3}{4}\) means \(\frac{3}{4}\) th distance of unit after 1 on the right side of zero. This number is indicated by the point C.

\(\frac{5}{2}=2\frac{1}{2}\) i.e. Half distance of unit after 2 on the right side of zero. The point D is at this place.

∴ The statement ‘the point D denotes the number 5/2 is true.

If a and b are positive numbers such that

a >b, then -a < -b 

Here, 7 > 2  ∴ -7 < -2

Ans. -7 < -2.

On a number line, \(\frac{-9}{5}\) is to the left of zero

∴ \(\frac{-9}{5}\) < 0  i.e. 0 > \(\frac{-9}{5}\)

Ans. 0 > \(\frac{-9}{5}\)

On a number line, 8/7 is to the right of zero

∴ \(\frac{8}{7}\) > 0 i.e. 0 < \(\frac{8}{7}\)

Ans. 0 < 8/7

A negative number is always smaller than a positive number.

∴ -5/4 < 1/4

Ans. -5/4 < 1/4

The denominator of the given rational numbers is the same.

Let us compare their numerators

40 < 141 ∴ 40/29 < 141/29

 Ans. 40/29 < 141/29

Because, the denominators are the same, let us compare the numerators of the given rational numbers.

-17 < -13  ∴ -17/20 < -13/20

Ans. -17/20 < -13/20

\(\frac{15}{12}=\frac{15×16}{12×16}=\frac{240}{192}\)

\(\frac{7}{16}=\frac{15×12}{16×12}=\frac{84}{192}\)

Now \(\frac{240}{192}>\frac{84}{192}\) ∴ \(\frac{15}{12}>\frac{7}{16}\)

Ans. \(\frac{15}{12}>\frac{7}{16}\)

\(\frac{-9}{4}=\frac{-9×2}{4×2}=\frac{-18}{8}\)

Now -25<-18

∴  \(\frac{-25}{8}<\frac{-9}{4}\)

Ans. \(\frac{-25}{8}<\frac{-9}{4}\)

\(\frac{3}{5}=\frac{3×3}{5×3}=\frac{9}{15}\)

∴  \(\frac{12}{15}>\frac{3}{5}\)

Ans. \(\frac{12}{15}>\frac{3}{5}\)

\(\frac{-7}{11}=\frac{-7×4}{11×4}=\frac{-28}{44}\)

\(\frac{-3}{4}=\frac{-3×11}{4×11}=\frac{-33}{44}\)

Now -33 < -28 ∴  \(\frac{-3}{4}<\frac{-7}{11}\)

Ans.  \(\frac{-3}{4}<\frac{-7}{11}\)

(i) The point Q on the number line shows the number \(\sqrt{2}\)

(ii) A line perpendicular to the number line is drawn through the point Q. Point R is at unit distance from Q on the line l.

(iii) Right angled Δ ORQ is obtained by drawing seg OR.

l (OQ) = \(\sqrt{2}\) , l(QR) = 1

∴ by Pythagoras theorem,

[l(OR)]² = [l(OQ)]² + [l(QR)]²

= [ \(\sqrt{2}\) ]² +[1]² = [2] + [1]

= [3]

∴ l(OR) = [ \(\sqrt{3}\)]

Draw an arc with centre O and radius OR. Mark the point of intersection of the line and the arc as C. The point C shows the number \(\sqrt{3}\) .

Draw a number line l, Take OP = 2 units.

Draw a line m ⊥ line l through the point P

Take a point Q on line m such that

PQ = 1 unit.

In right angled Δ OPQ, by Pythagoras theorem,

[l (OQ)]² = [l (OP)]² + [l(PQ)]²

              = (2)²+(1)²=4+ 1 =5.

∴ l(OQ) = \(\sqrt{5}\)

With centre O and OQ as radius, draw an arc intersecting line l in the point R.

OR = \(\sqrt{5}\) on the number line l.

The point R represents the number \(\sqrt{5}\)

Draw a number line l. Show the number \(\sqrt{5}\) on this number line

Let the point R represent the number \(\sqrt{5}\)

Draw seg RS ⊥ line l. Take RS = 1 unit. With centre O and radius OS draw an arc intersecting line l at the point T.

l(OS)² =( \(\sqrt{5}\) )²+(1)² = 6

∴ l(OS) = \(\sqrt{6}\)

Point T will represent the number \(\sqrt{6}\)

Draw seg TU ⊥ line l.  Take TU = 1 unit. With centre O and radius OU draw an arc intersecting line l at the point V.

l(OU)² =( \(\sqrt{6}\) )²+(1)² = 7

∴ l(OU) = \(\sqrt{7}\)

This point V will represent the number \(\sqrt{7}\)