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Explanation:-
Given that
a = -20
b = -10
c = 5
⟨i⟩ LHS = a÷b
⇛ -20÷-10
⇛ -20/-10
⇛ 2
and
RHS = b÷a
⇛ -10÷-20
⇛ -10/-20
⇛ 1/2
LHS ≠ RHS
a÷b ≠ b÷a
⟨ii⟩
LHS = (a÷b)÷c
⇛ (-20÷-10)÷5
⇛ 2÷5
⇛ 2/5
RHS = a÷(b÷c)
⇛ -20÷(-10÷5)
⇛ -20÷-2
⇛ -20/-2
⇛ 10
LHS ≠ RHS
(a÷b)÷c ≠ a÷(b÷c)
⟨iii⟩ LHS = a÷(b+c)
⇛ -20÷(-10+5)
⇛ -20÷(-5)
⇛ -20/-5
⇛ 4
RHS = (a÷b)+(a÷c)
⇛ (-20÷-10)+(-20÷5)
⇛ 2+(-4)
⇛ 2-4
⇛ -2
LHS ≠ RHS
a÷(b+c) ≠ (a÷b)+(a÷c)
Given that
a = -40
b = 10
c = -5
⟨i⟩ LHS = a÷b
⇛ -40÷1
⇛ -40/1
⇛-40
and
RHS = b÷a
⇛ 10÷-40
⇛ 10/-40
⇛ -1/4
LHS ≠ RHS
a÷b ≠ b÷a
⟨ii⟩.
LHS = (a÷b)÷c
⇛ (-40÷10)÷-5
⇛ -4÷-5
⇛ 4/5
RHS = a÷(b÷c)
⇛ -40÷(10÷-5)
⇛ -40÷-2
⇛ -40/-2
⇛ 20
LHS ≠ RHS
(a÷b)÷c ≠ a÷(b÷c)
⟨iii⟩ LHS = a÷(b+c)
⇛ -40÷(10-5)
⇛ -40÷(5)
⇛ -40/5
⇛ -8
RHS = (a÷b)+(a÷c)
⇛ (-40÷10)+(-40÷-5)
⇛ (-4)+8
⇛ 8-4
⇛ 4
LHS ≠ RHS
a÷(b+c) ≠ (a÷b)+(a÷c)
Conclusion :-