# 2.7

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**n = 1**
The single element array is already its max. Loop is not entered. Max is returned

Let for **n=k**, the algorithm is true

For **n = k+1** ,two cases arise :

**1)** a[k+1] is max

**2)** a[k+1] is not max.

If **1)** holds then at the last iteration when i = k+1, m = a[k+1] is assigned. Hence max is returned

Else if **2)** holds then we are left with the task of finding the max among **n = k** elements, which we already assumed that the algorithm does correctly.

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