Question

Prove that \[{{2}^{n}}>n\] for all positive integers n.

Answer 1

Hint: To solve this question, we will use mathematical induction. In that, we will assume P(n) to be the statement that is to be proved and then by using the Principle of Mathematical Induction, we will get the result.

Complete step by step answer:
We are given n as an integer. The process of solving of proving using Mathematical Induction is given by
1. First we will assume P(n) to be a statement that is to be proved.
2. We will show for n = 1, P(n) = P(1) holds true.
3. Then we will assume for k > 1, P(k) is true.
4. Finally, using the assumed hypothesis that P(k) is true, we will prove that P(k + 1) to be true.
This is the process that is to be followed in this question as well.
Let P(n) be the statement showing that \[P\left( n \right){{2}^{n}}>n\forall n=\text{positive integers}\text{.}\]
Let n = 1. Consider, P(n) = P(1).
\[\Rightarrow {{2}^{1}}>1\]
\[\Rightarrow 2>1\]
Hence, 2 > 1. Therefore, P(1) is true.
P(1) is a true statement……(i)
Now for the second step, let us assume k > 1. P(k) is true.
P(k) holds true that \[{{2}^{k}}>k\] is correct or holds true……(ii)
Finally, we have to show that P(k + 1) is true. Consider that \[P\left( k+1 \right)={{2}^{k+1}}>k+1.\] To show this is true, we will do as follows. By equation (ii), we have, \[{{2}^{k}}>k\] which is true.
Multiplying 2 on both the sides of the above equation, we have,
\[\Rightarrow {{2.2}^{k}}>2k\]
\[\Rightarrow {{2}^{k+1}}>2k\]
Now, as k > 1, adding both sides with k, we have,
\[\Rightarrow k+k>k+1\]
\[\Rightarrow 2k>k+1\]
Hence, we have,
\[\Rightarrow {{2}^{k+1}}>2k>k+1\]
\[\Rightarrow {{2}^{k+1}}>k+1\] holds true.
Therefore, P(k + 1) holds true…..(iii)

From equations (i), (ii) and (iii), we have, \[P\left( n \right)={{2}^{n}}>n\] for all n positive integers is a true statement.

Note: The point where we have used \[{{2}^{k}}>k.\] \[\Rightarrow {{2.2}^{k}}>2k\] is true if and only if 2 > 0. Now, since 2 > 0 then this implies \[{{2}^{k}}>k.\] \[\Rightarrow {{2.2}^{k}}>{{2}^{k}}.\] In short if a > 0, then b > c. Therefore, ab > ac. Also, at the step where we used k > 1, k + k > k + 1. This was true as k was also a positive integer. If k was not a positive integer, here then k > 1, k + k > k + 1 would be true as well as false.