Page 106 - New Medley 4 Semester 1

Z Z We can obtain the successor of a number by adding 1 to it.

Z successor of 23245 is 23245 + 1 = 23246.

successor of 318320 is 318320 + 1 = 318321.

Z Z We can add 1 to any number, so every number has a Successor.

Z Z The number that comes just before a number is called its Predecessor.

Z 54072 comes just before 54073. So 54072 is predecessor of 54073.

749099 comes just before 749100. So 749099 is predecessor of 749100.

Z Z We can obtain predecessor of a number by subtracting 1 from it.

Z predecessor of 54073 is 54073 – 1 = 54072.

predecessor of 749100 is 749100 – 1 = 749099.

Note: We cannot subtract 1 from 0, so 0 has no predecessor.

Example 18: (a) Give the successor of 35195.

(b) Give the predecessor of 314707.

(c) Give two numbers just after 83499.

(d) Give two numbers just before 458178.

Solution: (a) Successor of 35195 = 35195 + 1 = 35196.

(b) Predecessor of 314707 = 314707 – 1 = 314706.

(c) Successor of 83499 = 83499 + 1 = 83500.

Successor of 83500 = 83500 + 1 = 83501.

Required numbers are 83500 and 83501.

(d) Predecessor of 458178 = 458178 – 1 = 458177.

Predecessor of 458177 = 458177 – 1 = 458176.

Required numbers are 458177 and 458176.

Skip Counting

In skip counting we start with a given number and write more numbers
with a fixed gap between the successive numbers.

104