Show that for any real number x ≥ 0 the set (x, x + 1]∩n is a singleton set

First, a singleton set is the set with only one element, for example, the set {∅} or set "null" is a singleton.

now, X ≥ 0, and X ∈ R.

(x,x + 1]∩N is the intersection between the set x and x+1 with the set of natural numbers, so the result of this operation is the set of all the natural numbers between x and x+1.

if x is a rational (or irrational) number, such as 1.5, then x +1 = 2.5

in (1.5, 2.5] there is only one natural element, the 2, then (x,x + 1]∩N is the singleton {2}

if x is a positive integer number, such as 1, then x + 1 = 2

the set is (1,2] is important to notice that 1 ∉ (1,2] because the set is open in the lower limit. so this set only contains one natural number, the 2; so again; the intersection with the natural set is a singleton

So this is proven for x rational, irrational and integers.

This happens because each element in the natural set is separated by the next one by 1, and in the set (x,x + 1]  although the distance between the lower and upper limits is 1, the set is open in the lower limit, so the distance is < 1. then you never can have two natural numbers on this set.

4

step-by-step explanation:

6 (donuts) divided by 3 is 2.

so 2/3 means 2 times 2, which is 4/6.

you get two and your friend gets four.