(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Prove that the following statement is either true or false:

For all real numbers y, there is a real number x such that y = x + ceiling[x].

Note: ceiling[x] = n such that n-1 < x ≤ n. For example, if x = 3.2, then ceiling[x] = 4.

2. Relevant equations

3. The attempt at a solution

I think the statement is false. So I am going to prove its negation is true.

Negation: There exists a real number y such that for all real numbers x, y ≠ x + ceiling[x].

Proof:

Suppose x is a real number.

Suppose y = 2x + 1. Then y is a real number because x is a real number. (I think I can do this because I just have to show that my statement holds for at least one y.)

Now suppose that y = x + ceiling[x].

Define x_{1}= ceiling[x] - x. Then 0 ≤ x_{1}< 1.

So ceiling[x] = x_{1}+ x. Plug this into y = x + ceiling[x].

So y = x + x_{1}+ x = 2x + x_{1}

But by one of my assumptions, y= 2x + 1. Therefore, x_{1}= 1.

However, x_{1}< 1. This is a contradiction.

So y ≠ x + ceiling[x]. Negation of the statement is proved, so the original statement is true.

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Is this correct? I showed it to a friend who told me this doesn't prove the right thing. Apparently this only shows that y ≠ x + ceiling[x], and doesn't actually prove my original statement. I'm confused because I think it works. Help!

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# Homework Help: Ceiling Proof

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