Question

Question 117

Match the following :
$\begin{array}{cc}& \\ \text{Column I}& \text{Column II}\\ \text{(a) a}\times 1& \text{(i) Additive inverse of a}\\ \text{(b) 1}& \text{(ii) Additive identity}\\ \text{(c) (-a)}÷\text{(-b)}& \text{(iii) Multiplicative identity}\\ \text{(d) a}\times (-1)& \text{(iv) a}÷\text{(-b)}\\ \text{(e) a}\times 0& \text{(v) a}÷\text{b}\\ \text{(f) (-a)}÷\text{b}& \text{(vi) a}\\ \text{(g) 0}& \text{(vii) -a)}\\ \text{(h) a}÷\text{(-a)}& \text{(viii) 0}\\ \text{(i) -a}& \text{(ix) -1}\end{array}$

Answer 1

$\text{(a)}\to \left(vi\right)a\times 1=a$
$\text{(b)}\to \left(iii\right)$
1 is multiplicative identity
$\text{(c)}\to \left(v\right)-a÷(-b)=a÷b$[both signs are cancelled with each other]
$\text{(d)}\to \left(vii\right)a\times (-1)=-a$
$\text{(e)}\to \left(viii\right)a\times 0=0$[any value, when multiplies with 0 becomes zero]
$\text{(f)}\to \left(iv\right)(-a)÷b=a÷(-b)$
$\text{(g)}\to \left(ii\right)$
0 is an additive identity.
$\text{(h)}\to \left(ix\right)a÷(-a)=\frac{a}{-a}=-1$
$\text{(i)}\to \left(i\right)$
​​​​​​​-a is additive inverse of a.​