Proving Floor Value of A is an Integer B

[solakis1]

Given the definition of floor value :

For all A,B we define the floor value of A denoted by [A] to be an iteger B such that : $[A]=B\Leftrightarrow B\leq A<B+1$

And in symbols $\forall A\forall B ( [A]=B\Leftrightarrow B\leq A<B+1\wedge B\in Z)$,then prove:

For all A $ [A]\leq A<[A]+1$

 

[jvicens]

Dear fellow scientist,

I would like to prove the statement given in the forum post using the definition of floor value provided. According to the definition, the floor value of A denoted by [A] is an integer B such that [A]=B if and only if B is less than or equal to A and A is less than B+1. In symbols, this can be written as [A]=B ⇔ B≤A<B+1 and B∈Z.

Now, let us consider the statement [A]≤A<[A]+1. We can break it down into two parts: [A]≤A and A<[A]+1.

Firstly, from the definition of floor value, we know that [A] is an integer B such that [A]=B ⇔ B≤A<B+1. This means that [A] is less than or equal to A. Therefore, we can say that [A]≤A.

Secondly, from the definition, we also know that [A] is an integer B such that [A]=B ⇔ B≤A<B+1. This means that [A]+1 is the smallest integer that is greater than A. Therefore, we can say that A<[A]+1.

Combining these two statements, we get [A]≤A<[A]+1. This proves the statement given in the forum post.

In conclusion, for all A, the floor value of A denoted by [A] is less than or equal to A and A is less than [A]+1, as defined in the forum post.

Thank you for considering my proof.