Question: The Question of the CLRS $6.1-7$ exercise reads as:

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> > Show that, with the array representation for sorting an n-element heap, the leaves are the nodes indexed by $\lfloor n / 2 \rfloor + 1, \lfloor n / 2 \rfloor + 2, \ldots, n⌊n/2⌋+1,⌊n/2⌋+2,…,n$.
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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) & = 2(\lfloor n / 2 \rfloor + 1) \\ & > 2(n / 2 - 1) + 2 \\ & = n - 2 + 2 \\ & = n. \end{aligned}

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

Please help me out.
Thank you.

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Answer: 
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 $2(n/2-1)+2 = n$.
It's mathematical ture that

```
floor(n/2) > n/2 - 1

```

Maybe rearrange the equation can help understanding the question.

```
Left[floor(n/2) + 1]
=   2 * (floor(n/2) + 1)
>   2 * (n/2 - 1 + 1)
=   2 * n/2
=   n

```