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## Homework Statement

My induction proof was really easy and simple, and people were having problems and doing substitutions and other complicated things. Normally I'd take pride in making a simpler proof but, there are really brilliant people in this class and I'm just starting to get the grasp of proofs, so it's not comforting.

Prove that for all real numbers:

[itex](1+\frac{1}{2})^{n} \geq 1 + \frac{n}{2}[/itex]

## Homework Equations

## The Attempt at a Solution

Let S be the set of natural numbers such that:

[itex](1+\frac{1}{2})^{n} \geq 1 + \frac{n}{2}[/itex]

Let n = 1

[itex](1+\frac{1}{2})^{1} = 1.5[/itex]

[itex]1 + \frac{1}{2} = 1.5[/itex]

Therefore, 1 E S

Assume this is true for n = k

[itex](1+\frac{1}{2})^{k} \geq 1 + \frac{k}{2}[/itex]

Check if this implies truth for n = k + 1

Left side:

[itex](1+\frac{1}{2})^{k+1}[/itex]

[itex](1+\frac{1}{2})^{k}(1+\frac{1}{2})[/itex]

Right side:

[itex]1 + \frac{k+1}{2}[/itex]

[itex]1 + \frac{k}{2} + \frac{1}{2}[/itex]

[itex](1+\frac{1}{2})^{k}(1+\frac{1}{2}) \geq 1 + \frac{k}{2} + \frac{1}{2}[/itex]

Is a true statement because our initial assumption asserts that

[itex](1+\frac{1}{2})^{k} \geq 1 + \frac{k}{2}[/itex]

Here, the greater term is multiplied by 1.5, and the lower term has .5 added to it. A number multiplied by 1.5 is always greater than the same number added to .5, except in the case of 1, in which case they are equal. Since the first term is greater than the second, and is operated on a way such that a greater result is yielded even if the two were equal,

[itex](1+\frac{1}{2})^{k}(1+\frac{1}{2}) \geq 1 + \frac{k}{2} + \frac{1}{2}[/itex]

Must be true.

Therefore S is inductive.

Therefore S = Natural Numbers.