Show that,with the array representation for sorting an n-element heap, the leaves are the nodes indexed by n⌊n/2⌋+1,⌊n/2⌋+2,…,n The Question of the CLRS $6.1-7$ exercise reads as:


  
    Show that, with the array representation for sorting an n-element heap, the leaves are the nodes indexed by $\lfloor n / 2 \rfloor + 1, \lfloor n / 2 \rfloor + 2, \ldots, n⌊n/2⌋+1,⌊n/2⌋+2,…,n$.
  


I looked for the solution here:
https://walkccc.github.io/CLRS/Chap06/6.1/

The solution was provided like this:

Let's take the left child of the node indexed by $\lfloor n / 2 \rfloor + 1.$

\begin{aligned} \text{LEFT}(\lfloor n / 2 \rfloor + 1) &amp; = 2(\lfloor n / 2 \rfloor + 1) \\ &amp; &gt; 2(n / 2 - 1) + 2 \\ &amp; = n - 2 + 2 \\ &amp; = n. \end{aligned}

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I can't understand this statement:
$LEFT(⌊/2⌋+1) &gt; 2(/2−1)+2$

Please help me out.
Thank you.

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So, basically in heap representation, $LEFT(i)$ refers to the index of $i's$ left child. What we want to show is that index $⌊/2⌋+1$ is a leaf and is not a middleware node which can be proved if we could show the index of the left child is larger than the number of elements in the heap. 

On the other hand, $LEFT(⌊/2⌋+1) = 2(⌊/2⌋+1) = 2⌊/2⌋+2 $ and with removing those brackets around the $n/2$ we can show that it is larger than $2(n/2-1)+2 = n$.