Let x and y be any numbers at all with x ≤ y. show that the number of integers between x and y is [y] - [x] +1.that is show that the number of integers between x and y is = (the floor of y) - (the ceiling of x) +1

the explanation is given below.

Step-by-step explanation:

Here what is applied is assumption of range of values of number from say 1 - 100In total, i stopped at 100 on the dot.

from this, the lowest number is 1 and the highest number is 100

hence the range of the numbers = Difference between Highest and Lowestrange = 100 - 1 = 99, the 99 gotten as the range is indicative that a number has been missing.In order to make up the 100, an integer is added to the difference = 99, i.e 99 is added to 1 to make up the 100.Furthermore, if 0 is exclusively out when numbers are counted up 100 with 0 inclusive, in such case, the first and last number are excluded from the counting. as such the integers will be {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20, 99} since both 0 and 100 are not included.Here, if we try to get the range = highest - lowest = 99 - 1 = 98, it implies that to make up the 99, an integer is added to the result of the difference = 98+1 = 99As such, the number of integers between two numbers is the difference between the highest and the lowest number plus 1 i.e highest - lowest + 1 = y - x +1 = (the floor of y) - ( the ceiling of x) + 1

)because it's going in order of the math sign