Use mathematical induction to prove the inequality for the specified integer values of n. (1)/(√

Use mathematical induction to prove the inequality for the...

Key Concepts

Inequality Proofs

Inequality proofs involve techniques for showing that one expression exceeds or is less than another. These proofs often require careful manipulation of algebraic expressions and a deep understanding of the properties of inequalities. When combined with mathematical induction, they provide a systematic method for proving inequalities that depend on natural number parameters.

Inductive Hypothesis and Inductive Step

In the inductive step, one assumes that the statement holds for a particular natural number (the inductive hypothesis) and then uses this assumption to prove that the statement holds for the next natural number. This process creates a chain reaction that extends the validity of the statement to all appropriate natural numbers.

Mathematical Induction

Mathematical induction is a proof technique used to demonstrate that a statement holds for all natural numbers. It involves proving a base case and then showing that if the statement is true for an arbitrary integer, it must also be true for the next integer, thereby establishing the statement for all integers in the domain.

Base Case

The base case is the initial step in an inductive proof where the statement is verified for the starting value, typically the smallest value in the domain. Establishing the base case is essential as it serves as the foundation upon which the inductive argument is built.

Transcript

00:02 All right, this proof that we're going to do here is different from all the ones that we had summations for.

00:09 Although we could write this left -hand side as a summation, that's not the different part about it.

00:17 The different part is that we have a greater than right here in the sentence that we're trying to prove.

00:25 Okay, everything else had an equal sign or most of them had equal signs.

00:29 And so you just do them the same way.

00:31 Okay, this one's going to have some sort of trick in the middle where we're going to have to figure out how something is bigger than something else.

00:39 Okay, so just prepare yourself.

00:42 All right, so here we go, proof by induction.

00:49 Notice that this one says it's true for n greater than or equal to two.

00:53 So that means instead of showing that it works for n equals one, we have to first show it works for n equals two.

01:01 So show it is true for n equals 2.

01:09 So for n equals 2, we get 1 over the square to 1 plus 1 over the square root of 2.

01:16 This is 1 plus, okay, 1 with a square root of 2, i recognize as the sign and cosine of 45 degrees, which i know is 0 .707 approximately.

01:30 So this is approximately 1 .707.

01:35 And then the right hand side will be the square root of 2, which i know is 1 .414 approximately.

01:47 So now here's what i just figured out since 1 over the square root of 1 plus 1 over the square root of 2 is approximately equal to 1 .707, which is greater than 1 .414, which is approximately equal to the square root of 2.

02:03 This is true for n equals 2.

02:17 Okay next i'm going to assume that it's true for n equals k.

02:32 That is i'm assuming that when i add 1 over the square root of 1 plus 1 over the square to 2 plus 1 over the square to 3 plus all the way up to where i get to the square root of k i know that that's bigger than the square root of the k.

02:55 So i'm assuming that that's true.

02:58 So if k was 13, then i'm assuming that when i add all of these up to the square to 13, i get a number that's bigger than the square root of 13.

03:08 Okay? now i want to show it is true for n equals k plus 1.

03:20 That is, i want to show, i want to show, 1 over the square of 1 plus 1 over the square of 2 plus 1 over the square to 3 plus dot dot dot plus 1 over the square of k plus the next term is greater than the square root of k plus 1 so that's what i'm going to try and show okay now it's going to work just like before i'm going to start over here on the left hand side i'm going to break this into two pieces the first k parts of it plus the extra part so that i can make this substitution right here.

04:12 When i see this, i know that it is bigger than the square root of k.

04:18 Okay, so you gotta be careful with these greater thans and less than and equals in here.

04:24 All right, here we go.

04:24 One plus one square to one, one over the square to two, one over the square to three, plus mm -mm.

04:36 Okay, remember, in the line above right here, i just wrote down what i was, is trying to show.

04:42 Now i'm getting ready to show it.

04:45 Okay, so i'm going to use the distributive law, which says i can, sorry, the associate law, which says i can group this however i want.

05:02 All right.

05:03 I know that this stuff in the parentheses right here is greater than the square root of k.

05:12 So that's this.

05:15 And then here's this plus sign.

05:17 And then here's this.

05:23 Let me straighten that up because that looks horrible messy.

05:40 Hang on just a minute.

05:53 Okay, one over, whoa.

05:59 So that is bigger than the square root of k plus one over the square root of k plus one.

06:06 All right now, remember what we're trying to get is that this thing here is greater than square root of k plus one.

06:19 And this is what i have here.

06:21 So somehow i've got to figure out how to combine these or do something with them to show that they are, this thing is greater than the square of k plus one.

06:35 Okay, so i get out another piece of paper and i'm thinking about it.

06:43 Okay, this is not in the paper that i'm writing my homework on...

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