# Quod Erat Demonstrandum

## 2008/05/03

### Wrong method, but correct answer

Filed under: Additional / Applied Mathematics,HKCEE,Junior Form Mathematics — johnmayhk @ 7:11 上午

Obviously, the methods throughout the following sets are completely wrong in general; however, correct answers will be obtained. Try to verify them and explain why it happens. You may read the explanation at the end for confirmation.

(Set 1) Erase the indices

$\frac{5^3 + 2^3}{5^3 + 3^3} = \frac{5 + 2}{5 + 3}$
$\frac{7^3 + 3^3}{7^3 + 4^3} = \frac{7 + 3}{7 + 4}$

(Set 2) Erase the logarithm

$\frac{\log 2}{\log 4} = \frac{2}{4}$
$\log \frac{9}{4}\div \log \frac{27}{8} = \frac{9}{4} \div \frac{27}{8}$

(Set 3) Product to sum

$\frac{8}{7} \times 8 = \frac{8}{7} + 8$
$\frac{11}{10} \times 11 = \frac{11}{10} + 11$

(Set 4) Out of radical sign

$\sqrt{5\frac{5}{24}} = 5\sqrt{\frac{5}{24}}$
$\sqrt{7\frac{7}{48}} = 7\sqrt{\frac{7}{48}}$

(Set 5) Exchange indices

$(\frac{5}{7})^2 + \frac{2}{7} = \frac{5}{7} + (\frac{2}{7})^2$
$(\frac{\pi}{4})^2 + \frac{4 - \pi}{4} = \frac{\pi}{4} + (\frac{4 - \pi}{4})^2$

(Set 6) Sum to product

$\sec^238^{o} + \csc^238^{o} = \sec^238^{o}\csc^238^{o}$
$\sec^254^{o} + \csc^254^{o} = \sec^254^{o}\csc^254^{o}$

Explanation
Try to prove the following identities to see the reasons why we have equations from set 1 to set 6.

(Set 1) $\frac{a^3 + b^3}{a^3 + (a - b)^3} = \frac{a + b}{a + (a - b)}$
(Set 2) $\log(\frac{m + 1}{m})^m \div \log(\frac{m + 1}{m})^{m + 1} = (\frac{m + 1}{m})^m \div (\frac{m + 1}{m})^{m + 1}$
(Set 3) $\frac{n + 1}{n}\times (n + 1) = \frac{n + 1}{n} + (n + 1)$
(Set 4) $\sqrt{a + \frac{a}{a^2 - 1}} = a\sqrt{\frac{a}{a^2 - 1}}$
(Set 5) $(\frac{1}{n})^2 + \frac{n - 1}{n} = \frac{1}{n} + (\frac{n - 1}{n})^2$
(Set 6) $\sec^2x + \csc^2x = \sec^2x\csc^2x$