 Welcome back to another example of proof-by-mathematical induction and this time we're going to look at an example that involves an inequality for all positive integers n and n is less than 2 to the nth power. So just to kind of establish the anatomy of what we're working with here, this expression right here, that's our p of n. That's our predicate that is going to be true or false depending on what we plug in for n. And what we're claiming here is that this predicate is true for all positive integers. So this is a perfect time to use mathematical induction and we just want to keep in mind what that p of n is because that becomes pretty important. So as with all proofs by induction we need to do two things. We need to establish the base case. I need to prove that p of 1 is true and then I need to prove the inductive step. I want to assume p of k is true and then prove that p of k plus 1 is true to get from one step to the next. Base case is very very easy this time but I want to do is show that p of 1 is true and that would just be showing that 1 is less than 2 to the first power. So there's not much to say here. We're just going to notice that 1 is less than 2. I think everybody knows that. So we could just turn this around and say 1 is less than 2 to the first and that very firmly establishes the base case. That was really really simple this time. Almost nothing to say. You just almost say that the base case is trivial. Step here where it's slightly less trivial and we need to pull out some tricks here. So we're going to assume that for all for some positive integer k that the p of k statement is true. So we're going to assume that for some positive integer k and you know a positive integer for us is just the same thing as a natural numbers. I'm just going to write for some k in the natural numbers. p of k is true. That means k is less than 2 to the k. What we're going to prove is that k plus 1 is less than 2 to the k plus 1 power. So we're trying to prove an inequality and the way we're going to work this is very similarly to how we work with equations. We're going to work with one side only. Just go to the left side and start doing some valid mathematical steps to try to arrive at the inequality that we want. Go to a new slide here and remember I'm just going to write down we are assuming just to remind myself we were assuming that k is less than 2 to the k for some positive integer k. So now we want to show that k plus 1 is less than 2 to the k plus 1. So like I said let's start on the left hand side of my inequality and just write k plus 1. Now I'm going to do a little trick here that's kind of standard working with inequalities. I'm trying to establish an inequality that goes this direction for k plus 1. So what I'm going to do is replace something in k plus 1 here some term in k plus 1 here with something that's larger than yet. Like for example if I take this 1 or replace it with a 2 the new expression would be larger right. So I'm going to take the 1 and replace it with a k another copy of k. So this is going to k plus 1 is definitely less than k plus, watch this, k plus k. I just make this little replacement right there and I pick up a new inequality. Now why did I do that? Well again I'm looking ahead to proving this inequality right here and so I need to get that inequality in the picture somehow. And also it allows me to use my inductive hypothesis. Remember every time I see a k I can replace it with a 2 to the k and make it an inequality. So I'm going to use the inductive hypothesis right now and maybe I'll put this in green to indicate that that by the inductive hypothesis this k right here is less than 2 to the k and this k right here is also less than 2 to the k. So again I'm taking each term of the sum and swapping it out for something larger and that makes the inequality flow in the right direction. Now let's just see what happens from here. What I have here are two copies of 2 to the k and so this is equal to 2 times 2 to the k. Well 2 times 2 to the k that's the same thing as 2 to the first times 2 to the k and this is equal to 2 to the k plus 1. Now that actually ends the proof. Do you see it? Because if you read the stack of inequalities and inequalities all the way from the top to bottom I have shown that k plus 1 is less than 2 to the k plus 1 and that is what I wanted to show and I have shown it and I've established the base case and so that ends the proof and so now we know that for all positive integers k or sorry positive integers n, n is less than 2 to the nth power.