Q. 1174.5( 4 Votes )

Match the following


Answer :

For (a) a × 1 = a

So, we can match (a) with (vi).


For (b) 1 = Multiplicative identity


1 is multiplicative identity because any integer multiplied with 1 gives the same integer.


So, we can match (b) with (iii).


For (c) (-a) ÷ (-b) = a ÷ b


As


…(i)


And …(ii)


Comparing equations (i) and (ii), we get


(-a) ÷ (-b) = a ÷ b


So, we can match (c) with (v).


For (d) a × (-1) = -a


As a × (-1) = -a × 1 = -a


So, we can match (d) with (vii).


For (e) a × 0 = 0


0 multiplied with any integer equals to 0.


So, we can match (e) with (viii).


For (f) (-a) ÷ b = a ÷ (-b)


As …(i)


And


…(ii)


Comparing equations (i) and (ii), we get


(-a) ÷ b = a ÷ (-b)


So, we can match (f) with (iv).


For (g) 0 = additive identity


0 is additive identity because any integer added to 0 gives the same integer.


So, we can match (g) with (ii).


For (h) a ÷ (-a) = -1


As



a ÷ (-a) = -1


So, we can match (h) with (ix).


For (I) -a = Additive inverse of a


(-a) is additive inverse because integer a added with –a gives 0.


So, we can match (I) with (i).


Arranging it into table:



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