What is the fallacy in this proof?

Question

I came across this funny proof-

$$4$$

$$=4+\frac 92-\frac 92$$

$$=\sqrt{(4-\frac 92)^2}+\frac 92$$

$$=\sqrt{16+\frac{81}{4}-36}+\frac 92$$

$$=\sqrt{25+\frac {81}{4}-45}+\frac 92$$

$$=\sqrt{(5-\frac 92)^2}+\frac 92$$

$$=5-\frac 92+\frac 92$$

$$=5$$

I suspect,that the error is in the second line where operation on negative is done before positive violating $BODMAS$ rule.But,$\sqrt{(4-\frac 92)^2}=4-\frac 92$ and the similar solving continues.So, where is the real error?

Thanks for any help!!

Answer 1

The error is going from the second to third line. $$\sqrt{x^2}=|x|$$ so $$\sqrt{\left(4-\frac{9}{2}\right)^2}=\left|4-\frac{9}{2}\right|\neq 4-\frac{9}{2}$$

Answer 2

The error comes in when $4 - \frac92$ is squared, which causes $-\frac12$ to become $+\frac14$, and the square root taken from this is erroneously taken as $+\frac12$.

Answer 3

The effective problem seems to be in switching between solutions, but I'd like to point out the relevance of the numbers involved in the game (using other numbers it wouldn't have the same effect): $$ (4- \frac{9}{2})^2=(16+\frac{81}{4}-36) $$ $$^{(*)}(16+9-9+\frac{81}{4}-36)= (25+\frac{81}{4}-45) $$ $$(25+\frac{81}{4}-45)=(5-\frac{9}{2})^2 $$

The game is based on the fact that the numbers involved are $4$, $5$ and $\frac{9}{2}$ indeed is $$4-\frac{9}{2}=-0.5$$ $$-4+\frac{9}{2}=0.5$$ while $$5-\frac{9}{2}=0.5$$ $$-5+\frac{9}{2}=-0.5$$ So what's happening is that adding and subtracting 9 in (*) is sufficient to confuse and switch between the two solution 0.5 and -0.5 when ricomposing the binomial square

Answer 4

The equality $$\sqrt{\left(4-\frac92\right)^2} = \sqrt{\left(5-\frac92\right)^2}$$ is equivalent to $$\sqrt{\left(-\frac12\right)^2} = \sqrt{\left(\frac12\right)^2}$$ which of course is true, but that does not imply $$-\frac12 = \frac12$$

Adding the first square and square root is invalid, this equality:

$$4+\frac 92-\frac 92$$ $$=\sqrt{\left(4-\frac 92\right)^2}+\frac 92$$

does not hold.