In the example below (bottom, from http://www.purplemath.com/modules/inductn3.htm), I am fine with proving the basis, and when working, it follows as I would expect that if f(k) = 4k, then the next in the series, f(k+1) = 4k + 4. However, what I don't understand is this, which in turn lead to confusion for the rest of the example:

**Since k > 5, then 4 < 32 < 2k. Then we get**

2k + 4 < 2k + 2k = 2×2k = 21×2k = 2k+1

2k + 4 < 2k + 2k = 2×2k = 21×2k = 2k+1

Why here is it

**2k + 4 < 2k + 2k**? Why, like equalities, does it not stay as

**4k + 4 < 2k + 4**? I don't understand why/how the transformation occurred on each side of the inequality.

I'm sure the explanation is embarrassingly simple, but that's why I came here for help, right?

Thanks!

Full example from http://www.purplemath.com/modules/inductn3.htm:

* (*) For n > 5, 4n < 2n.

This one doesn't start at n = 1, and involves an inequality instead of an equation. (If you graph 4x and 2x on the same axes, you'll see why we have to start at n = 5, instead of the customary n = 1.)

Let n = 5.

Then 4n = 4×5 = 20, and 2n = 25 = 32.

Since 20 < 32, then (*) holds at n = 5.

Assume, for n = k, that (*) holds; that is, assume that 4k < 2k

Let n = k + 1.

The left-hand side of (*) gives us 4(k + 1) = 4k + 4, and, by assumption,

[4k] + 4 < [2k] + 4

Since k > 5, then 4 < 32 < 2k. Then we get

2k + 4 < 2k + 2k = 2×2k = 21×2k = 2k+1

Then 4(k+1) < 2k+1, and (*) holds for n = k + 1.

Then (*) holds for all n > 5.