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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!

Edit:

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.

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Let $x = n + r$ where $n$ is an integer and r is such $1/2 \le r < 1$. So $x - \lfloor x \rfloor = r$

Now this should be trivially easy.

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