Questions tagged [fake-proofs]

Seemingly flawless arguments are often presented to prove obvious fallacies (such as 0=1). This is the appropriate tag to use when asking "Where is the proof wrong?" about proofs of such obvious fallacies.

An example of a fake proof is $$1=\sqrt{-1\cdot-1}=\sqrt{-1}\sqrt{-1}=i^2=-1$$ which fails because $\sqrt{xy}=\sqrt x\sqrt y$ does not hold if $x$ or $y$ is negative. Sometimes the proof may be presented as a puzzle, the challenge being to identify the flaw.

For asking about identifying flaws in general proofs ("spot the mistake"); the tag should instead be used.

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¿Where does equality fail?

I know this: $\sqrt{x}^2 = |x|$, but $\sqrt{(-1)^2} = \sqrt{(-1)^2}$ $(-1)^\frac{2}{2} = \sqrt{-1 * -1}$ $(-1)^1 = \sqrt{1}$ $-1 = 1^2$, then $-1 = 1$ What step is wrong?
Eduardo Sebastian
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Where's the mistake in my logic for this differential equation 'solution'?

$y=ce^x$ is the only solution to $y'=y$. This is known. Let $k=\ln 2$ Let $S=\cdots+k^{-1}2^x+k^0 2^x+k^1 2^x+ k^2 2^x+\cdots$ Let $S'=\frac d{dx}S$ $S'=S$ so $S=ce^x$ $S=2^x[\cdots+k^{-1}+k^0+k^1+k^2+\cdots]$ Since…
Jacob Claassen
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Why is this -1=1 proof incorrect?

We have 1=1 1²=1² Since, (-1)²=1=1², we have 1²=(-1)² So, taking square root on both sides gives 1=-1. Obviously, this is wrong, but where is the mistake? Is it because taking the square root on both sides of 1²=(-1)² should give +or-1=+or-1,…
Stephen
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Where is my mistake ?$ (-1)^3=-1 \to \sqrt[3]{-1}=-1 $

Where is my mistake ? (In the field of real numbers) $$ (-1)^3=-1 \to \sqrt[3]{-1}=-1 $$ $$\sqrt[3]{-1}=-1=(-1)^{\frac{1}{3}}=(-1)^{\frac{2}{6}}=\sqrt[6]{(-1)^2}=\sqrt[6]{(1)^2}=1$$ $$-1=^?1 $$
Gasked71
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$4=5$. Is this possible?

we can say 4 is the number you get when you add 1 to itself 4 times. Likewise, 5 is the number we get when we add 1 to itself 5 times. Now, let's just say 4 = 5. It would have to be the case that 4-5 = 0, however, that would imply 1 = 0.…
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proof that 1/2 = -1/2

Part One: S = 1+1+1+1+1+... 1/2 = 1-1+1-1+1-1+... *if you don't understand, google grandi's series S - 1/2 = 2+2+2+... S - 1/2 = 2S S = -1/2 Part Two: S = 1+1+1+1+... 1/2 = 1-1+1-1+1... S + 1/2 = 2+2+2+... S + 1/2 = 2S S = 1/2 Conclusions: 1/2…
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If $\frac1\infty=0$ then I can prove that $0 = 1$

Given, $\dfrac{1}{\infty}=0$, then $1=0 \cdot \infty = 0$ (because $0$ times any number or values is $0$ and here that number is infinity). Which gives us $1=0$ i.e, $0=1$. Hence proved....
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What is wrong with this reasoning when calculating circle perimeter?

Looking at the following image, which was posted on the internet: Could someone tell me what is wrong? It seems true for the first 4 small images. But, when it comes to infinitesimal length, something must be wrong, i can feel it.
Hélène
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Disprove why 0 ∉ Z

Disprove this A value x is said to be an integer when floor(x) = x, where x ∈ ℝ floor(x)/x = 1 Therefore floor(x)/x ∈ Z, where x ∈ ℝ And since 0 ∈ ℝ From the definition of an integer, floor(x)/x ∈ Z, where x ∈ ℝ if 0 ∈ Z, floor(0) =…
Andrew Kor
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Trouble identifying error in fake-proof.

I was talking to a friend about the problem with this proof and I'm stomped on the illegal step: $i = (-1)^{(1/2)}= (-1)^{(2/4)} =((-1)^2)^{(1/4)} = 1^{(1/4)} = 1$
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$\iota \equiv \pm 3, \pmod{10}$

I was reading up on the properties modulo function, when I saw the property: $$-a \equiv (10-a) \space \pmod{10}$$ Which means $$-1 \equiv (10-1) \equiv 9 \space \pmod{10}$$ Now: $$\iota = \sqrt{-1}$$ Substituting $-1 \equiv 9 \pmod{10}$: $$\iota…
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Lesser-known fake proofs

I believe I have seen almost every elementary bogus proof of $0=1$ or $1=2.$ I am wondering if there are any lesser-known "proofs" where the error is harder to spot.
Display name
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Errors in a false complex proof

This proof tries to show that $ \mathbb{R} = \mathbb{C} $ $Let z \in \mathbb{C}, \exists r \in \mathbb{R}^{+}, \theta \in [0,2\pi]:z=re^{i\theta}$ $\\z=re^{i\theta\frac{2\pi}{2\pi}} = r(e^{2\pi i})^{\frac{\theta}{2\pi}}=…
Kendots
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What is wrong with the following proof that tries to prove that $2=1$?

I need to find the flaw in the following proof: $a,b\in\mathbb{R}$\ $\left\{ 0 \right\} $ such that $a=b$ 1) Multiplying both sides by $a$ yields the equality: $a^2=ab$ 2) Subtracting $b^2$ from both sides yields the equality $a^2-b^2=ab-b^2$ 3)…
Cherry_Developer
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how can be this possible? What is wrong with this.

We can see that 1^2 =1 ; 2^2 =2+2 ; 3^2=3+3+3 ; . . . x^2=x+x+x+..... (x times) differentiation on both sides gives 2x=1+1+1+....... (x times) 2x=x What's happening hear.How is this possible. Assume X be as integer and non-integer ,both cases.
user239656
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