Question

Find an integer Nsuch that \({2^n} > {n^4}\) whenever nis greater

than N. Prove that your result is correct using mathematical

induction.

Short answer

The integer\(N = 16\), such that\({2^n} > {n^4}\)whenever\(n\)is an integer greater than 16.

Step-by-step solution

Principle of Mathematical Induction

Consider the propositional function \(P\left( n \right)\). Consider two actions to prove that\(P\left( n \right)\)evaluates to accurate for all set of positive integers \(n\).

Consider the first basic step is to confirm that\(P\left( 1 \right)\) true.

Consider the inductive step is to demonstrate that for any positive integer k the conditional statement\(P\left( k \right) \to P\left( {k + 1} \right)\)is true.

Prove the basis step

Given statement is

\({2^n} > {n^4}\)

If\(N = 16\)

\(\begin{array}{l}{2^{16}} = {16^4}\\65,536 = 65,536\end{array}\)

Therefore, if\(n > 16\),\({2^n} > {n^4}\)is true

In the basis step, we need to prove that\(P\left( 1 \right)\)is true

Since integer\(n\)is greater than 16.

Therefore, in the basis step, we need to prove that\(P\left( {17} \right)\)is true.

For finding statement\(P\left( {17} \right)\)substituting\(17\)for\(n\)in the statement.

Therefore, the statement\(P\left( {17} \right)\)is

\(\begin{array}{c}{2^{17}} > {17^4}\\131,072 > 83,521\end{array}\)

The statement \(P\left( {17} \right)\) is true this is also known as the basis step of the proof.

Prove the Inductive step

In the inductive step, we need to prove that, if\(P\left( k \right)\)is true, then\(P\left( {k + 1} \right)\)is also true.

That is,

\(P\left( k \right) \to P\left( {k + 1} \right)\)is true for all positive integers k.

In the inductive hypothesis, we assume that\(P\left( k \right)\)is true for any arbitrary positive integer\(k\).

\({2^k} > {k^4}\) …… (i)

Now we must have to show that\(P\left( {k + 1} \right)\)is also true.

Therefore, replacing\(k\)with\(k + 1\)in the left-hand side of the statement

\(\begin{array}{l}{2^{k + 1}} = 2 \cdot {2^k}\\{2^{k + 1}} > 2 \cdot {k^4}\end{array}\)

Now, replacing\(k\)with\(k + 1\)in the right-hand side of the statement.

\(\begin{array}{c}{\left( {k + 1} \right)^4} = {k^4} + 4{k^3} + 6{k^2} + 4k + 1\\ < {k^4} + 4{k^3} + 6{k^3} + 4{k^3} + 1\\ = {k^4} + 14{k^3}\end{array}\)

Since we can see,\(14{k^3} < {k^4}\), for\(k > 16\)

Therefore

\({2^{k + 1}} > {\left( {k + 1} \right)^4}\)

From the above, it is clear that\(P\left( {k + 1} \right)\)is also true.

Hence,\(P\left( {k + 1} \right)\)is true under the assumption that\(P\left( k \right)\)is true. This

completes the inductive step.

Hence, It is shown that,\(N = 16\)and\({2^n} > {n^4}\)whenever\(n\)is an integer greater than 16.