[leflyCode]

Is there a way to prove that:

Code:

Code

floor(x) + 1 = ceiling(x)

?

I think it is true, but I cant think how to prove the damn thing! Cheers.

[stastony]

Is there a way to prove that:

Code:

Code

floor(x) + 1 = ceiling(x)

?

I think it is true, but I cant think how to prove the damn thing! Cheers.

If x is a real integer, then the equation no longer holds.

[leflyCode]

If x is a real integer, then the equation no longer holds.

How so?

edit - Forgot to mention the restriction on x is x is a positive integer

[stastony]

How so?

edit - Forgot to mention the restriction on x is x is a positive integer

Maybe I'm not understanding the problem. Here's how I look at it.

Say x = 2

Floor(x) = 2
Ceiling(x) = 2

Floor(x) + 1 = 3

3 != 2

[leflyCode]

LMAO! Oh my, how could I miss that? Cheers [thumbup]

[jesyflowers]

it could be true if floor and ceiling are variables, not necessarily the same.

[leflyCode]

Oh my, I read the question entirely wrong

Code:

Code

floor(log2(x)) +1 = ceiling(log2(x))  x is positive integer

That is log to base 2. I read the wrong question for the last one. I think this one is true...

edit - Ah, doesnt work for 1

[stastony]

Oh my, I read the question entirely wrong

Code:

Code

floor(log2(x)) +1 = ceiling(log2(x))

Code:

Code

  x is positive integer

That is log to base 2. I read the wrong question for the last one. I think this one is true...

I think it's still false. My reasoning:

Let y = log2(x)

Therefore, x = 2^y

Since x = positive integer, y = positive integer too.

Therefore, floor(y) + 1 != ceiling(y)

[leflyCode]

Cheers, thats a good proof to go with my counter example [thumbup] Many thanks.