# Help with understanding a floor proof

1. Feb 4, 2015

### Temp0

1. The problem statement, all variables and given/known data
Prove that for any real number x, if x - floor(x) < 1/2, then floor(2x) = 2 floor(x)

2. Relevant equations

3. 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.

2. Feb 4, 2015

### RUber

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).

3. Feb 5, 2015

### Temp0

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) ?

4. Feb 6, 2015

### RUber

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) .