# A Collection of Bogus Proofs

Hello M.S.E. people, This question is just for fun, don't take it seriously :). We have all encountered Bogus Proofs, which seem logical and reasonable, but they prove some claims which are completely wrong and unreasonable. Lets collect a list of very convincing, yet Bogus Proofs. Let me show one down below;

I understand this isn't really a 'question', but its just a fun post to make M.S.E. more interesting and fun.

Here's a classic one; \begin{align*} \frac{\text{d}}{\text{d}x}\left(x^2\right) &=\frac{\text{d}}{\text{d}x}\left(\underbrace{x+x+\cdots+x}_{x\text{ times}}\right)\\ &=\underbrace{\frac{\text{d}}{\text{d}x}\left(x\right)+\frac{\text{d}}{\text{d}x}\left(x\right)+\cdots+\frac{\text{d}}{\text{d}x}\left(x\right)}_{x\text{ times}}\\ &=\underbrace{1+1+\cdots+1}_{x\text{ times}}\\ &=x. \end{align*}

But since $$\frac{\text{d}}{\text{d}x}\left(x^2\right)=2x,$$ we have; \begin{align*} 2x &=x\\ \implies 2 &=1. \end{align*}

• This one is oddly satisfying until you see $2=1$ :). Commented May 11, 2022 at 15:46
• Where's the mistake for positive $x$? Commented May 11, 2022 at 15:58
• Here in the derivative, $x$ is an unfixed variable, while when we say "$x$ times", we are fixing it and taking it as a constant. That too, it would have to be an integral constant for "$x$ times". Commented May 11, 2022 at 16:08
• It would be fine if we applied the multivariate chain rule :) The following technically works out to the right answer of $2x$: $$\underbrace{\frac{\text{d}}{\text{d}x}\left(x\right)+\frac{\text{d}}{\text{d}x}\left(x\right)+\cdots+\frac{\text{d}}{\text{d}x}\left(x\right)}_{x\text{ times}} + \underbrace{x + x + \dots + x}_{\frac{\text{d}}{\text{d}x}(x) \text{ times}}$$ Commented Apr 30, 2023 at 4:00

Theorem. Every horse has the same color.

Proof. Let's show by induction on $$n$$ that every horse in a set with $$n$$ horses have the same color.

• If $$n=1$$, the statement is trivially true.
• If the statement is true for $$n$$, consider a set of $$n+1$$ horses. If we remove one of the horses from the set, the remaining ones should all have the same color.

Now we can add the removed horse back again in the set and remove some other random horse.

Again we have a set of $$n$$ horses, so they all should have the same color. In particular, the first removed horse have the same color as the previous, concluding our proof. $$\square$$

$$A=B=1$$

$$A=B$$

$$A^2=AB$$

$$A^2-B^2=AB-B^2$$

$$(A+B)(A-B)=B(A-B)$$

$$A+B=B$$

$$1+1=1$$

$$2=1$$

(The fallacy of dividing by zero in moving from the fifth line to the sixth line)

$$-20=-20$$

$$25-45=16-36$$

$$5^2-(5×9)=4^2-(4×9)$$

$$5^2-(5×9)+\frac{81}{4}=4^2-(4×9)+\frac{81}{4}$$

$$(5-\frac{9}{2})^2=(4-\frac{9}{2})^2$$

$$5-\frac{9}{2}=4-\frac{9}{2}$$

$$5=4$$

(The fallacy in moving from the fifth line to the sixth line, since when writing the square root of the square of a negative number, the sign under the root must be reversed)

$$\pi=4$$

$$\pi=\infty$$

(Squiggly lines can be used to approximate area only, not to approximate perimeter)

$$\sqrt{-1}=(-1)^\frac{1}{2}$$

$$=(-1)^\frac{2}{4}$$

$$=((-1)^2)^\frac{1}{4}$$

$$=1^\frac{1}{4}$$

$$=1$$

The error occurs at the end, as we neglected the quadruple roots of $$1$$ , which are $$1$$ , $$-1$$ , $$i$$ , $$-i$$ , and contented ourselves with With only one root.

• Even with your resolution there's still a fallacy: you get that $\sqrt{-1}=1^{1/4}=\pm 1,\pm i$, but $\sqrt{-1}$ is only $\pm i$. Commented Mar 13 at 13:51