Question

Fill in the blanks with the correct symbol out of $>,=$ and $<:$
(i) $-\frac{3}{7}\dots \dots -\frac{6}{13}$

(ii) $\frac{5}{-13}\dots \dots -\frac{35}{91}$

(iii) $-2\dots \cdots -\frac{13}{5}$

(iv) $-\frac{2}{3}\dots \dots \frac{-5}{8}$

(v) $0\dots \dots \frac{-3}{-5}$

(vi) $-\frac{8}{9}\dots \dots \frac{-9}{10}$

Answer 1

(i) Given $-\frac{3}{7}\dots \cdots -\frac{6}{13}$

L.C.M. of $7$ and $13$ is $91$.

$-\frac{3\times 13}{7\times 13}=-\frac{39}{91}$ and $-\frac{6\times 7}{13\times 7}=-\frac{42}{91}$

Thus, $-\frac{39}{91}>-\frac{42}{91}$

Therefore, $-\frac{3}{7}>-\frac{6}{13}$

(ii) Given $\frac{5}{-13}\dots \cdots -\frac{35}{91}$
L.C.M. of $13$ and $91$ is $91.$

$\begin{array}{cc}13& 13,19\\ 7& 1,7\\ & 1,1\end{array}$

$\frac{5\times (-7)}{-13\times (-7)}=-\frac{35}{91}$

Thus, $-\frac{35}{91}=-\frac{35}{91}$

Therefore, $\frac{5}{-13}=-\frac{35}{91}$
(iii)$-\mathrm{2.......}-\frac{13}{5}$

L.C.M of $1$ and $5$ is $5$.

$\frac{-2\times 5}{1\times 5}=-\frac{10}{5}$

Thus, $-\frac{10}{5}>-\frac{13}{5}$

Therefore, $-2>-\frac{13}{5}$

(iv) $-\frac{2}{3}\dots \cdots -\frac{5}{8}$

L.C.M. of $8$ and $3$ is $24$.

$\begin{array}{cc}2& 8,3\\ 2& 4,3\\ 2& 2,3\\ 3& 1,3\\ & 1,1\end{array}$

$-\frac{2\times 8}{3\times 8}=-\frac{16}{24}$ and $-\frac{5\times 3}{8\times 3}=-\frac{15}{24}$

$-\frac{16}{24}<-\frac{15}{24}$

$-\frac{2}{3}<-\frac{5}{8}$

(v) Given $0\dots \dots (\frac{-3}{-5}=\frac{3}{5})$

L.C.M. of $1$ and $5$ is $5.$

$\frac{0\times 5}{1\times 5}=\frac{0}{5}$

Thus, $\frac{0}{5}<\frac{3}{5}$

Therefore, $0<\frac{-3}{-5}$

(vi) Given $-\frac{8}{9}\dots \cdots -\frac{9}{10}$

L.C.M. of $9$ and $10$ is $90$.

$\begin{array}{cc}3& 9,10\\ 3& 3,10\\ 5& 1,10\\ 2& 1,2\\ & 1,1\end{array}$

$-\frac{8\times 10}{9\times 10}=-\frac{80}{90}$ and $-\frac{9\times 9}{10\times 9}=-\frac{81}{90}$

Thus, $-\frac{80}{90}>-\frac{81}{90}$

Therefore,$-\frac{8}{9}>-\frac{9}{10}$