Mathematics Stack Exchange
Mathematics Stack Exchange

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.

1281 questions
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What is wrong with this putative proof?

So I've spent about an hour trying to figure out what is wrong with this proof. Could somebody clearly explain it to me? I don't need a counterexample. For some reason I was able to figure that out. Thanks. Theorem. $\;$ Suppose $R$ is a total order…
ChemDude
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Which step is wrong in this proof

Proof: Consider the quadratic equation $x^2+x+1=0$. Then, we can see that $x^2=−x−1$. Assuming that $x$ is not zero (which it clearly isn't, from the equation) we can divide by $x$ to give $$x=−1−\frac{1}{x}$$ Substitute this back into the $x$…
Joel Birtwistle
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Fake proof that $1$ is the solution of $x^2+x+1=0$

So I have this false proof and I am honestly confused why this is happening. Consider $x^2+x+1=0$, then $x+1=-x^2$. Now by simply dividing the equation by $x$ we get $x+1+1/x=0$. Substituting $x+1=-x^2$ we get $-x^2+1/x=0$ we get $x=1$ as a…
Sorfosh
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4 answers

Where is the error in this "proof" that 3=0?

I saw this video (link at bottom), with a supposed "proof" that $3=0$. It goes as follows: Let $x$ be a solution of $$x^2+x+1=0 \tag1$$ Since $x\neq0$, we can divide both sides by $x$: $$\frac{x^2+x+1}{x}=\frac0x\implies x+1+\frac1x=0…
Just_A_User
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12
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5 answers

Prove $ 1 + 2 + 4 + 8 + \dots = -1$

Possible Duplicate: Infinity = -1 paradox I was told by a friend that $1 + 2 + 4 + 8 + \dots$ equaled negative one. When I asked for an explanation, he said: Do I have to? Okay so, Let $x = 1+2+4+8+\dots$ $2x-x=x$ $2(1+2+4+8+\dots) -…
umop aplsdn
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Having trouble debunking my friend's "proof" that "There is no real number greater than 0"

My friend and I were talking about math stuff as usual, when he brought up a fake proof for the statement: There is no real number greater than $0$. Now obviously this isn't true, because any positive number is $>0$. But I could not argue…
Aiden Chow
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Please can someone tell accurately and precisely the actual fallacies of this fake proof that 0.999... ≠ 1.000...?

Please can someone tell accurately and precisely the actual fallacies of this fake proof that 0.999... ≠ 1.000... ? Define $F: \{\text{decimals in } [0,1]\} \to \{\text{decimals in }[0,1]\}$ by $a.bcdefg... \to a.0b0c0d0e0f0g..$. Define $G:…
user953418
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4 answers

Does ((2^2)^2)^2........ = 1 or 0?

let $x = (((((2^2)^2)^2)^2)^2)^\cdots$ Therefore, $x^2 = ((((((2^2)^2)^2)^2)^2)^2)^\cdots$ But this is the same expression. Therefore $x = x^2$ Therefore, $x^2 - x = 0$ Therefore $x(x-1) = 0$ Therefore $x = 0$, or $x = 1.$ Where is the error in…
Kenshin
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3 answers

Where does the proof of $\sqrt 2$ is irrational break down when trying to prove the same for $\sqrt 4$?

I'm trying find where the common proof by contradiction that $\sqrt 2$ is irrational breaks down when trying to prove $\sqrt 4$ is irrational. Assume $\left(\frac pq\right)^2=4$ and $\gcd(p,q)=1$. I guess I could just let $p=2, q=1$ and be done, but…
Jeff
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Simple equation manipulation gives the wrong solution.

I was stumped by a false proof I came across, in which I cannot wrap my head around the exact reason why it does not work. We start from the following equation: $$x^2+x+1=0.$$ On one hand we get that $x = -1 -x^2.$ On the other we can divide the…
Matt
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Help debunk a proof that zero equals one (no division)?

Unlike the more common variant of proof that 0=1, this does not use division. So, the reasoning goes like this: \begin{align} 0 &= 0 + 0 + 0 + \ldots && \text{not too controversial} \\ &= (1-1) + (1-1) + (1-1) + \ldots && \text{by algebra}\\ &= 1…
danmcardle
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Mistake proving that $3 = 0$

I was watching this video in which they proved that $3 = 0$ and the objective is to find the mistake in the proof. They did the following: Let $x$ be a solution for the equation: $x^2 + x + 1 = 0$ $(1)$. Because $x \neq 0$ we can devide both sides…
Eduardo Magalhães
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3 answers

Is $8^x$ positive or negative?

What is wrong with this logic? $8^x = (\sqrt{8}^2)^x = (\sqrt{8}^x)^2 \Rightarrow 8^x \geq 0 \forall x$ Edit: Clarification, the answer to $8^{1/2}$ permits both $\sqrt{8}$ and $-\sqrt{8}$ (hence my confusion). That would also suggest $8^x$ is not a…
ehuang
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4
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5 answers

Is $\sqrt{x^2} = (\sqrt x)^2$?

Take $x=4$ for example: $ \sqrt{(4)^2} = \sqrt{16} = \pm4 $ However: $ (\sqrt{4})^2 = \sqrt{\pm2}$ Case 1: $ (-2)^2 = 4$ Case 2: $ (2)^2 = 4$ Solution : $+4$ How come the $ \sqrt{(4)^2} = \pm4$; but $ (\sqrt{4})^2 = 4 $ ? What is missing?
orange orange
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“Proof” that zero is equal to one by infinitely subtracting numbers

Recently, I came across a “proof” that $0=1$. Here is how it goes: Let $x = 1-1-1-1-1-1-1-\cdots$. Since $1-1=0$, $x=0-1-1-1-1-1-1-\cdots$. Now, we bracket the $1-1-1-1-1-1-\cdots$ on both sides and we get…
Pilot
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