- #1

iamsmooth

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## Homework Statement

Prove that:

[tex]2^n \geq n + 5[/tex] for all integer n > 5

## Homework Equations

n/a

## The Attempt at a Solution

This is what I did, just want to see if it's right.

Using mathematical induction:

Basis step:

When n = 3,

[tex]2^3 \geq 3 + 5[/tex]

Since 8 = 8, base case is proven

Inductive step:

Assume [tex]2^k \geq k + 5[/tex] for all integers k greater or equal to n.

We want to prove that [tex]2^{k+1} \geq k + 1 + 5[/tex]

Using our assumption, [tex]2^k \geq k + 5[/tex], I know that:

[tex]2(2^k) \geq 2(k + 5)[/tex]

=[tex]2^{k+1} \geq 2k + 10[/tex]

From here, if I can prove 2k+10 is greater or equal to k+6, then it will prove

[tex]2^{k+1} \geq k + 6[/tex]

So:

[tex]2k+10 \geq k + 6[/tex]

[tex] 2k \geq k - 4[/tex]

[tex] k \geq -4[/tex]

Thus is proven because k was said to be greater or equal to n which is at least 3. So the inequality is true.

Since inductive step is proven, [tex]2^3 \geq 3 + 5[/tex] must be true.

QED

Does this proof stand? I haven't had much practice with inequalities... so I'm very unsure about this proof.

Thanks.

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