14. Show that [2x] + [2y] >= [x] + [y] + [x + y] whenever x and y are real numbers.
Solution: We have to show that
[2x]+[2y]>=[x]+[y]+[x+y]
Where x,y are real numbers
We write x,y as
x=n+\epsilon
y=m+\delta
Where n,m are integers and \epsilon,\delta are non-negative real numbers less than 1.
There are two possibilities for the right hand side.
n+m+(n+m)=2n+2m
Or
n+m+(n+m+1)=2n+2m+1
If
\epsilon+\delta>=1
There are two possibilities for the left hand side.
\lceil2x\rceil=\begin{cases} 2n & \epsilon<1/2\\ 2n+1 & otherwise \end{cases}
And
\lceil2y\rceil=\begin{cases} 2m &\delta<1/2\\ 2m+1 & otherwise \end{cases}
We are trying to prove that the right hand side is always less than or equal to the left hand side. We construct a proof by contradiction and assume that right hand side is greater than left hand side. So the right hand side must be 2n+2m+1 and the left hand side must be 2n+2m
But then,
\epsilon+\delta>=1
So at least one of them must be equal to or greater than 1/2
So the left hand side cannot equal 2n+2m
Hence we arrive at a contradiction and end the proof.
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