# Mathematical Induction Question

• Feb 23rd 2008, 08:34 PM
shawn
Mathematical Induction Question
I was hoping someone could help me with a proof. I have attached an image.
I am going to prove it using induction (rather than well-ordering), and the base cases where n=1,2 and 3 are straightforward.

Thank you in advance.
• Feb 23rd 2008, 10:14 PM
Jen
Inductive Hypothesis:

$\left(\frac{5}{4}\right)^k-\left(\frac{3}{4}\right)^k>\frac{k}{2}$

Show:

$\left(\frac{5}{4}\right)^{k+1}-\left(\frac{3}{4}\right)^{k+1}>\frac{k+1}{2}$

Working with the left side...

$\left(\frac{5}{4}\right)\left(\frac{5}{4}\right)^k-\left(\frac{3}{4}\right)\left(\frac{3}{4}\right)^k >\left(\frac{5}{4}\right)\left(\frac{5}{4}\right)^ k-\left(\frac{5}{4}\right)\left(\frac{3}{4}\right)^k$

$\left(\frac{5}{4}\right)\left(\frac{5}{4}\right)^k-\left(\frac{3}{4}\right)\left(\frac{3}{4}\right)^k >\frac{5}{4}\left[\left(\frac{5}{4}\right)^k-\left(\frac{3}{4}\right)^k\right]$

On the right hand side we have 5/4 times our inductive hypothesis, substituting we get...

$\left(\frac{5}{4}\right)\left(\frac{5}{4}\right)^k-\left(\frac{3}{4}\right)\left(\frac{3}{4}\right)^k >\frac{5}{4}\left(\frac{k}{2}\right)=\left(1+\frac {1}{4}\right)\left(\frac{k}{2}\right)=\frac{k}{2}+ \frac{k}{8}>\frac{k}{2}+\frac{1}{2}$

$\frac{k}{8}>\frac{1}{2}$ Because k>4

So, by transitivity, we get...

$\left(\frac{5}{4}\right)\left(\frac{5}{4}\right)^k-\left(\frac{3}{4}\right)\left(\frac{3}{4}\right)^k >\frac{k}{2}+\frac{1}{2}$

$W^5$ "Which Was What We Wanted"

Sorry I didn't know how to do the greater than or equal to symbol. (Crying)

Hope that was clear enough.
• Feb 23rd 2008, 10:27 PM
Jhevon
Quote:

Originally Posted by Jen
Sorry I didn't know how to do the greater than or equal to symbol. (Crying)

just type \ge

type \le for less than or equal to

type \ne for not equal to
• Feb 23rd 2008, 10:27 PM
Jen
Oooops, lets try this again.
• Feb 23rd 2008, 10:29 PM
shawn
Thank you very much Jen. Now I just need to do c using part b as my solution.
• Feb 23rd 2008, 10:40 PM
Jen
Given that:

$\left(\frac{5}{4}\right)-\left(\frac{3}{4}\right)^x>\left(\frac{x}{2}\right )$

Show:

$3^x+4^x<5^x$

Using the given inequality... (multiply both sides by $4^x$, note that this is legal because exponential functions are non-negative)

$4^x\left[\left(\frac{5}{4}\right)-\left(\frac{3}{4}\right)^x>\left(\frac{x}{2}\right )\right]$

$5^x-3^x>4^x\frac{x}{2}$

Adding $3^x$ to both sides we get...

$5^x>4^x\frac{x}{2}+3^x>4^x+3^x$

The inequality on the right is possible because x is always greater than or equal to 2 so x/2 is always greater than or equal to 1.

Once again by transitivity we get

$5^x>4^x+3^x$

(Clapping)

Once again, hope that made sense...