Question

let y>0 and n a positive integer. Prove that (1+y)^n >= 1+ ny

Answer 1

Hey...@mhannahw are you here...
Like I seen this post like 4 times now and I have responded to all three + this one now.

Answer 2

Oh yeah sorry I'll look at this one I got confused

Answer 3

So the first positive integer is 1.

Answer 4

But the way the method we are going to prove this by is induction.

Answer 5

Like that is the method I'm familiar if anyways.

Answer 6

The base case is the first case (the case we first need to prove it for)
Since we are trying to prove this for all positive integers
That means we will need to start with n=1
Does the inequality hold for n=1?

Answer 7

\[(1+y)^1 \ge 1+(1)y \]

Answer 8

If this inequality is true then we move on to the next step of the proof..
The part that is actually kinda tough to really tough depending on what kinda algebra/some other matematical magic is involved

Answer 9

You suppose \[ 1+ky \le (1+y)^k \text{ for some integer } k>0 \\ \text{ And you want to show } 1+(k+1)y \le (1+y)^{k+1}\]

Answer 10

So starting with
the 1+(k+1)y thing ...we want to show this is less than or equal to that other thing on the other side of the inequality and we want to some how do it by using what we have which is
\[1+ky \le (1+y)^k\]

Answer 11

So given that you are starting with 1+(k+1)y ...can you somehow using the thing that we assumed was true...

Answer 12

wait how did you get 1+(k+1)y≤(1+y)^k+1
(the second line when you were supposing some positive integer k)

Answer 13

Mathematical Induction:
1) Show for base case
2) Assume the statement holds for n=k, then show the statement holds for n=k+1

Answer 14

Yeah but where did the 1+... and on the other side, the power k+1 come from in that line?

Answer 15

You are trying to prove...
\[1+ny \le (1+y)^n \text{ right? } \]

Answer 16

I replaced the n's with (k+1)'s
This inequality with (k+1)'s is the inequality we are trying to prove
using the inequality with the k's.

Answer 17

Okay, I see that. Can you explain why you replaced the n's with (k+1)'s?

Answer 18

Well {1,2,3,4,5,6,...,k,k+1}
The way I always thought of it is...
if we can see if the first case holds ...
and then assume it holds for that k and show it holds for also k+1 then it holds for all the positive integers
like this will give us all the positive integers
like k could be 6 and k+1 could be 7
k could be 100 and k+1 could be 101
So what you are doing is showing it always holds for the next number in the set

Answer 19

Would you like me to go through the process with another inequality/equality...

Answer 20

Like I haven't proved your inequality here...I was trying to give you what you need to do it. But it seems like mathematical induction might actually be new to you.

Answer 21

Precisely that! It is new to me. How would you show with a few cases that it works for every case?

Answer 22

the k (k>1) can be any number in the set with k+1 always following afterwards
We are technically showing it for all positive integers

Answer 23

We aren't just pluggin in 6 and 7
or 91 and 100

Answer 24

but we are choosing k and k+1

Answer 25

which can represent any consecutive positive integer

Answer 26

integers*

Answer 27

\[1+2+3+\cdots +n=\frac{n(n+1)}{2} \text{ for all integer } n >0 \\ \text{ BASE CASE } \\ \text{ For } n=1: 1=\frac{1(1+1)}{2} \text{ holds ! } \\ \text{ ALL OTHER CASES (the induction case) } \\ \text{ Assume for some integer } k>0 \text{ we have } 1+2+3+\cdots +k=\frac{k(k+1)}{2} \\ \text{ Now we want to show } 1+2+3+\cdots +(k+1)=\frac{(k+1)(k+2)}{2} \\ \text{ So that task begins here } 1+2+3+\cdots+k+(k+1) \\ \\ = \frac{k(k+1)}{2}+k+1 \\ =\frac{k(k+1)}{2} +\frac{2(k+1)}{2} \\ =\frac{k(k+1)+2(k+1)}{2} \\ =\frac{(k+1)(k+2)}{2} \\ \text{ Therefore for all integer } n >0 \text{ we have }1+2+3+\cdots +n=\frac{n(n+1)}{2} \]\\

Answer 28

This above was an example of the induction proof.

Answer 29

Right! okay I can follow that

Answer 30

The line after 1+2+3+...+k+(k+1) is where I used my induction hypothesis

Answer 31

aka the part we assumed was true

Answer 32

So in yours we had
\[\text{ Induction hypothesis was: } 1+ky \le (1+y)^k \\ \text{ and we want to show } 1+(k+1)y \le (1+y)^{k+1}\]

Answer 33

We want to somehow reword the 1+(k+1)y so we can use what 1+ky is less than

Answer 34

Are you doing okay? Do you need a hint to start off?

Answer 35

I sleep now. Good luck.

Answer 36

I think so, I'm finishing up! You helped much more than I could do on my own!

Answer 37

Use the binomial Theorem to show that\[
(1+y)^n= 1+ ny +\text {something positive}\\
(y+1)^2=1+2y + y^2\\
(1+y)^3=1 + 3 y + 3 y^2 + y^3\\
(1+y)^4=1 + 4 y + 6 y^2 + 4 y^3 + y^4

\]
and you are done