# Thread: Proving inequality through induction

1. ## Proving inequality through induction

$\displaystyle 2^k > 4k$

$\displaystyle 2^k > 4k = 2^k - 4k < 0$

$\displaystyle 2^{k+1}-4(k+1) = 2 x 2^k -4(k+1)$
$\displaystyle > 2 x 4k - 4(k+1)$, using $\displaystyle 2^k > 4^k,$
$\displaystyle =8k -4k - 4 = 4k -4$

I don't understand what has happened in the second to last step here.

2. ## Re: Proving inequality through induction

It's not true. Try k = 2, we have \displaystyle \displaystyle \begin{align*} 2^2 = 4 \end{align*} and \displaystyle \displaystyle \begin{align*} 4 \cdot 2 = 8 \end{align*}, so the inequality does not hold...

3. ## Re: Proving inequality through induction

The text says "This expression is greater than 0 for K > 1. So Proof by mathematical induction works, except we cannon start at n = 1."

I just don't follow the working.

4. ## Re: Proving inequality through induction

I have just proved to you that the inequality is not true. Arguing that it is will not change that. Either state that \displaystyle \displaystyle \begin{align*} k \geq 5 \end{align*} or stop trying.

5. ## Re: Proving inequality through induction

Yes sorry I should have stated that k has to be equal to or larger than 5. I didn't mean to try to argue with you. It's just that I'm trying to understand what is happening in the second to last step in the working. I can't see how the expression on the left is converted into an equality and then what it means to have an equality there.

6. ## Re: Proving inequality through induction

Originally Posted by alyosha2
Yes sorry I should have stated that k has to be equal to or larger than 5.
Because $\displaystyle 2^5>4\cdot 5$, the base case it true.

Suppose that $\displaystyle N>5$ and $\displaystyle 2^N>4\cdot N$.
\displaystyle \begin{align*}2^{N+1}=2\cdot 2^N&\ge 2(4\cdot N)\\&= (4\cdot N)+ (4\cdot N)\\&\ge 4(N+1)\end{align*}

7. ## Re: Proving inequality through induction

Originally Posted by Plato
Because $\displaystyle 2^5>4\cdot 5$, the base case it true.

Suppose that $\displaystyle N>5$ and $\displaystyle 2^N>4\cdot N$.
\displaystyle \begin{align*}2^{N+1}=2\cdot 2^N&\ge 2(4\cdot N)\\&= (4\cdot N)+ (4\cdot N)\\&\ge 4(N+1)\end{align*}
I'm sorry I don't understand

how does $\displaystyle 2^{N+1}=2\cdot 2^N&\ge 2(4\cdot N)$ How does an integer equal an equality?

how do you get from this line to the last?

How does this relate to the original working I posted?

I'm trying to understand how we go from an integer expression to an equality and then back again and how I am supposed to understand the meaning of this equality. To put it simply one equals a number and the other is true or false so how can they be equal?

Thanks.

8. ## Re: Proving inequality through induction

Originally Posted by alyosha2
I'm sorry I don't understand
how does $\displaystyle 2^{N+1}=2\cdot 2^N&\ge 2(4\cdot N)$ How does an integer equal an equality?
One step at a time.
1) Do you understand $\displaystyle 2^{N+1}=2\cdot 2^N&~?$

9. ## Re: Proving inequality through induction

Originally Posted by Plato
One step at a time.
1) Do you understand $\displaystyle 2^{N+1}=2\cdot 2^N&~?$
yes.

10. ## Re: Proving inequality through induction

Originally Posted by alyosha2
yes.
OK then:
2) the inductive step says if $\displaystyle N>5$ then $\displaystyle 2^N>4\cdot N$.
Therefore $\displaystyle 2\cdot 2^N>2(4\cdot N)$. Do you see that?

yes.

12. ## Re: Proving inequality through induction

Originally Posted by alyosha2
yes.
Good. 3)
$\displaystyle 2(4\cdot N)=4\cdot N+4\cdot N$

But $\displaystyle 4\cdot N>4$

So $\displaystyle 4\cdot N+4\cdot N>4\cdot N+4=4(N+1)$

13. ## Re: Proving inequality through induction

I'm sorry I don't see how we go from an equality on the left to an number on the right.

Also, what is the significance of 4N being larger than 4?

14. ## Re: Proving inequality through induction

Originally Posted by alyosha2
I'm sorry I don't see how we go from an equality on the left to an number on the right.
Also, what is the significance of 4N being larger than 4?