# Help with understanding a floor proof

## Homework Statement

Prove that for any real number x, if x - floor(x) < 1/2, then floor(2x) = 2 floor(x)

## The Attempt at a Solution

Assuming that x is a real number. Suppose that x - floor(x) < 1/2
Multiplying both sides by 2, 2x < 2 floor(x) + 1
from the definition, 2 floor(x) <= 2x
2 floor(x) is an integer, and 2 floor(x) <= 2x < 2 floor(x) + 1
Implying floor(2x) = 2 floor(x)

My question is, how does the proof imply the result? I don't see how the logic works. Thank you for any help you can provide in advance.

RUber
Homework Helper
2 floor(x) is an integer, and 2 floor(x) <= 2x < 2 floor(x) + 1
Implying floor(2x) = 2 floor(x)

My question is, how does the proof imply the result? I don't see how the logic works. Thank you for any help you can provide in advance.

The key is that floor(x) and floor(2x) are both integers and the relation you have above is 2floor(x) < 2 floor(x) + 1. You seem to be missing the part where 2floor(x) <= floor(2x).

I see... I guess I missed that part, haha, just wondering, does it come from the inequality:

floor (2x) <= 2x < floor(2x) + 1 ?

I can see how it relates... kind of, since

2 floor (x) <= 2x as well, but how does

2floor(x) <= floor (2x) ?

RUber
Homework Helper
2 floor(x) is always less than or equal to floor(2x).

If you let ## x = x_{int} + x_{frac} ## where ##x_{int}\in \mathbb{Z}, 0 \leq x_{frac}<1##
Then ##\text{floor}(x) = x_{int}=x-x_{frac}##
## 2\text{floor}(x) = 2x_{int}=2x-2x_{frac}##
But floor(2x) has 2 cases:
case 1:
##x_{frac}<.5 \implies 2x_{frac} < 1 \implies floor(2x) = 2 x_{int} = 2floor(x) ##
case 2:
##.5 \leq x_{frac}<1 \implies 2x_{frac} \geq 1 \implies floor(2x) = 2 x_{int} + 1=2floor(x) + 1 ##
In both cases, floor(2x) ##\geq ## 2floor(x) .

Greg Bernhardt