We have, (-8) + ______ = 0
(-8) + 8 = 0 [ \(\because\) additive inverse of -8 is 8 ]

LHS = (-25) - (-42) = -25 + (additive inverse of -42)
= -25 + 42 = 17
RHS = (-42) - (-25) = -42 + (additive inverse of -25)
= -42 + 25 = -17
Here, 17 is a positive integer and -17 is a negative integer.
\(\therefore\)    17 > -17

LHS = 45 - (-11) = 45 + 11 [ \(\because\) additive inverse of -11 is 11 ]
= 56
RHS = 57 + (-4) = 53 + 4 + (-4) = 53 + 0 = 53
Here, both are posiive integers, but 56 is to the right of 53.
\(\therefore\)      56 > 53

LHS = (-21) - (-10) = -21 + (Additive inverse of -10)
= -21 + 10 = -11
RHS = (-31) + (-11) = -42
Here, both are negative inegers, but -42 is to the left of -11
\(\therefore\)    -11 > -42

LHS = (-3) + (-6) = -9
RHS = (-3) + (-6) = (-3) + (Additive inverse of -6)
= -3 + 6 = 3
Since, -9 is a negative integer and 3 is a positive integer.
\(\therefore\)  -9 < 3.

The integers lying between -7 and 6 are -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5. Whole number are 0, 1, 2, 3, 4 and 5.
\(\therefore\)  The number of whole numbers = 6

The integers lying between -2 and 2 are -1, 0 and 1. So, there are 3 integers.

The predecessor of -1 = -1 - 1 = -2
\(\therefore\)  The required predecessor is -2.

5 units to the right of 0 is +5.

We know that, all the negative integers are less than zero and they have '-' sign.