Proof Errors: 4=5 Using sin

Question

Many fake proofs are somewhat obvious involving division by 0 for example. I've come across this one which I have not seen discussed anywhere and I'm not too sure where the error lies, if there even is one.

I'm not so good with MathJax but the opening line given is:

"4 = 5 iff 4/5 = 1" : Not sure if use of double arrow is correct.

The 'proof' is as follows:

$$ 4=5 \iff \frac{4}{5} = 1$$ $$ \Rightarrow \frac{4}{5} = \sin\left(\frac{\pi}{2}\right)$$ $$Let \space \theta = \frac{\pi}{2} $$ $$\Rightarrow \sin \left( \theta\right) = \frac{4}{5}$$ $$\Rightarrow \sin\left(\pi - \theta\right) = \frac{4}{5}$$ $$\Rightarrow \pi - \theta = \arcsin\left(\frac{4}{5}\right) $$ $$\Rightarrow\pi - \arcsin\left(\frac{4}{5}\right) = \arcsin\left(\frac{4}{5}\right) $$ $$\Rightarrow\arcsin\left(\frac{4}{5}\right) = \frac{\pi}{2} $$ $$\Rightarrow\frac{4}{5} = \sin\left(\frac{\pi}{2}\right) = 1$$ $$\therefore 4 = 5$$

Is there an error with restricted domain of $\arcsin?$ Is it that there are a family of solutions when $\sin$ is introduced? Or is it just a fancy manipulation which ultimately does nothing and leaves the original statement? My gut feeling is that it's just a string of fancy ways to re-write $1$, but I'm really not sure and am therefore asking here.

What's the way to argue against this?

Answer 1

As I pointed out in the comments, the main flaw of the solution is poor exposition, which makes it difficult to find the finer, intended error. A better version of this puzzle would be:

Let $\theta = \arcsin \frac{4}{5}$ so that $\sin \theta = \frac{4}{5}$. Then also $\sin( \pi - \theta ) = \frac{4}{5}$, thus $\pi - \theta = \arcsin \frac{4}{5}$. Matching it together, we get $\theta = \pi - \theta$ so $\theta = \frac{\pi}{2}$, therefore $\frac{4}{5} = \sin \frac{\pi}{2} = 1$.

The obvious fallacy is drawing the conclusion that $\pi - \theta = \arcsin \frac{4}{5}$ from the equation $\sin( \pi - \theta ) = \frac{4}{5}$, since $\arcsin$ is only a partial inverse of $\sin$.

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One could even argue I just made up the puzzle above myself, because there is so little similarity to the original puzzle in the structure of the proof. Well, I wouldn't deny.

But maybe the mess in the puzzle was intended to make it more obscure...

Answer 2

Many times in mathematics your "assumption" can be false, but it can lead to a satisfying and believing proof.

All the fake proofs start off with something false, and so whatever the result is from then on, its all gibberish, and can be true, or it can be false. We don't know.

The proper proofs assume something that is known to be true.

This is just another case where it just "works", but since your assumption that $4=5$ is already false from the start, the proof from then on has no meaning.

Answer 3

The 'proof' ultimately does nothing at all. It just walks around in a circle then returns to the blatantly false* claim that $4=5$. Since $\theta$ is set to $\frac{\pi}{2}$ from the outset, all the intermediate steps are ultimately the same** expression: $\sin{\theta}=\frac{4}{5}$. You could swap $4$ and $5$ for any two numbers you desired and the 'proof' would work exactly the same; but unless your two numbers were equal and non-zero, the 'proof' will be invalid because it would assert that $\sin{\frac{\pi}{2}}\neq1$ and that $\sin{\frac{\pi}{2}}=1$ simultaneously, which is obviously fallacious.

* Sorry Big Brother.

** If we consider $\arcsin$ to have codomain $(-\pi,\pi]$, or some other codomain which contains $\frac{\pi}{2}$.