# Compare A ,B

• June 24th 2009, 09:40 AM
dhiab
Compare A ,B
• June 24th 2009, 10:44 AM
Bruno J.
• June 24th 2009, 10:45 AM
Soroban
Hello, dhiab!

Quote:

Compare $A\text{ and }B\!:\quad\begin{array}{ccc}
A &=& \dfrac{1.0000004}{(1.0000006)^2} \\ \\[-3mm] B &=& \dfrac{(0.9999995)^2}{0.9999998} \end{array}$

$\text{Let }d \:=\:\frac{a}{10^7},\:\text{where }a\text{ is a digit } \leq 6$

Then: . $\begin{array}{ccc}A &\approx & \dfrac{1+d}{(1+d)^2} \\ \\ [-3mm]B & \approx& \dfrac{(1-d)^2}{1-d} \end{array}$

We have: . $\overbrace{\frac{1+d}{(1+d)^2}}^{A} \quad^{>}_{<}\quad \overbrace{\frac{(1-d)^2}{1-d}}^{B}$

Then: . $(1+d)(1-d)\quad ^{>}_{<}\quad (1-d)^2(1+d)^2$

. . . . . . . . . . $1 - d^2 \quad ^{>}_{<}\quad (1-d^2)^2$

. . . . . . . . . . $1 - d^2 \quad ^{>}_{<}\quad 1 - 2d^2 + d^4$

. . . . . . . . . . . . $d^2\quad ^{>}_{<}\quad d^4$

Divide by positive $d^2\!:\quad 1 \quad ^{>}_{<}\quad d^2$ .[1]

Since $d < 1$, then $d^2 < 1$

Hence, [1] becomes: . $1 \;\;{\color{red}>}\;\;d^2$

Therefore: . $A \;{\color{red}>}\;B$

• June 24th 2009, 12:34 PM
dhiab
Quote:

Originally Posted by Bruno J.