I had this problem as a homework assignment and had to write a proof for it. I've tried some approaches but keep getting stuck.
This is what I have so far:
Suppose that x is any particular but arbitrarily chosen real number such that x-⌊x⌋ ≥ 1/2.
Multiplying both sides by 2 gives
2x - 2⌊x⌋ ≥ 1 or 2x ≥ 2⌊x⌋ + 1.
Now by the definition of floor, ⌊x⌋ ≤ x. Hence, 2⌊x⌋ ≤ 2x. Since an integer plus 1 is always greater than the integer itself, 2⌊x⌋ ≤ 2x + 1. Putting the two inequalities together gives
2⌊x⌋ < 2⌊x⌋ + 1 ≤ 2x
From here I am not sure where to go in order to prove that ⌊2x⌋= 2⌊x⌋+ 1. Any help would be appreciated!
Would it be correct to from here say that since the definition of floor is n ≤ x < n + 1, assume that ⌊2x⌋ + 1 is equal to n so, therefore, n + 1, would be ⌊2x⌋ + 2.
From there, the new equality would be,
2⌊x⌋ + 1 ≤ 2x < ⌊2x⌋ + 2.
And by the definition of floor, therefore, ⌊2x⌋ = 2⌊x⌋ + 1.