Ten Problems Puzzles and Brain Teasers » Problem Six: an eighth is greater than a quarter?

Problem Six: an eighth is greater than a quarter?

Oct 27th, 2006 by tony

Since Percy posted the 1=2 thing a few days ago, I’m tempted to make up something of my own. So here it is…

To proof:

$\dfrac{1}{8} > \dfrac{1}{4}$

proof:

3 > 2

multiply both sides by $\lg\dfrac{1}{2}$ , we have

$3\lg\dfrac{1}{2} > 2\lg\dfrac{1}{2}$

From property of logarithms, we have

$\lg\left(\dfrac{1}{2}\right)^3 > \lg\left(\dfrac{1}{2}\right)^2$

Therefore, $\left(\dfrac{1}{2}\right)^3 > \left(\dfrac{1}{2}\right)^2$ , and

$\dfrac{1}{8} > \dfrac{1}{4}$

What went wrong?

Update: Solutions are posted!

Posted in Mathematical, Paradox | 1 Comment

on 31 Oct 2006 at 1:07 am Antony Rheneus

Log(1/2) is a negative value, hence the greater then sign (>) should be changed to lesser than sign (