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Recently, I found out someone has tried to prove that $x^2 + 1 = 0$ in $\mathbb{R}$, Here is the detail:

\begin{align*} x^2 + 1 = 0 \\ (x^2 + 1)(x^2 - 1) = 0(x^2 - 1)\\ x^4 - 1 = 0 \\ x^4 = 1\\ \Longrightarrow x \in \{-1,1\} \end{align*}

This is a repaired version of the original he threw at me, after I point out that it is wrong. The original one use the notation "$\Longrightarrow$" to assume his multiplication. Here is how it looks like:

\begin{equation*} x^2 + 1 = 0 \Longrightarrow (x^2 + 1)(x^2 - 1) = 0(x^2 - 1) \Longrightarrow x^4 - 1 = 0 \Longrightarrow x^4 = 1 \end{equation*}

From my knowledge of logic theory to the very least, by the way of using "$\Longrightarrow$", you are "implying" something. Suppose that we are given two statements, $A$ and $B$. $A ⇒ B$ means that you are implying that "If $A$ then $B$", or from $A$ you implies that $B$ is something like this or that. The statement $A⇒B$ doesn't make any claim about whether $B$ also implies $A$. Therefore, even if for a "wrong" transformation, the statement $B$ does not implies that it will be applied inversely to $A$, thus the prove is likely to be nothing. This is already discussed in this post on MathStackExchange:

The statement $A⇒B$ doesn't make any claim about whether $B$ also implies $A$. Defining it this way makes certain things a lot easier, since we can now say (e.g.) $x>7⇒x>5$. This is a true mathematical statement $(A⇒B)$ for real x. In other words, $A⇒B$ is true. But what if we let $x=6$? Now $B$ is true but $A$ is false; yet you'd still agree that $A⇒B$ is true. It just so happens that $A$ is false in this one case; that doesn't impact the overall truth or falsehood of $A⇒B$.

Now that's is for the "original" proof he gave me. After that, he typically "fix" it, by reading it as purely as a algebraic manipulation task. This is my least trial of proving he is wrong.

  • First, the statement is already false, geometrically. To "prove" that $x^2 + 1 = 0$ basically means you are solving to find where the function $f: x \longmapsto x^2 + 1$ cross the x-axis. $x^2 + 1$ from the beginning has the $+1$ as the translation up, that is, the vertex will always be $(1,0)$, and considering that this is a even quadratic function, it is unlikely that there will be any intersection even if you tried to search for it along all $\mathbb{R}$

  • Second, multiplying anything with $0$ will result in $0$. By "factor" out $x^2 -1$ from both side is wrong, since you are altering the solution sets of the total equation.

  • Last, don't use what you are wanting to prove to prove itself. If you use itself to prove it, then it is similar as you are assuming it, therefore, you don't have to prove it anymore.

The question here is: Am I right, how can I fix my reasoning if I was wrong, and what is your attempt to de-prove his proof?

区なしま
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    You have multiplied by $x^2-1$ in the second line, so by zero for $x=\pm 1$. This is not allowed. – Dietrich Burde Aug 31 '22 at 16:18
  • $x-1=0$ so $(x-1)(x-2)=0$ so $x=1$ or $x=2.$ What you've shown is that if $x^2+1=0$ and $x$ is real, then $x=\pm1.$ But that's because there is no real solution to $x^2+1=0.$ – Thomas Andrews Aug 31 '22 at 16:19
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    In other words, imagine taking $5=3$ (which is false), multiplying by $0$ on both sides, arriving at $0=0$ (which is true), and thinking that you've proven something deep. – Randall Aug 31 '22 at 16:20
  • You should have $\longleftrightarrow$ for each step instead of $\rightarrow$. Can you see at which step it fails if you move backward? – Etemon Aug 31 '22 at 16:24
  • If $x \in \mathbb R$, then the equation $x^2=-a \ \ \forall \ a\geq 0$ has no solution. A squared number is always positive, as an area of a square is positive. – callculus42 Aug 31 '22 at 16:27
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    This argument correctly shows: if $x$ is a real solution of $x^2+1=0$, then $x \in {-1,1}$. That is correct. But of course an "if/then" statement $P \rightarrow Q$ is true when $P$ is false. This is what happens for any real number $x$. – GEdgar Aug 31 '22 at 16:32
  • @Randall this is not my proof, it's only a thing that I met but kind of confused of how he could come up with the proof. Still, thanks for clarify me. – 区なしま Aug 31 '22 at 16:36
  • @Etemon I just copy the original proof, and it is exactly what it is up there. – 区なしま Aug 31 '22 at 16:37
  • The final step picks out only two of the four solutions to the quartic - they happen to be the two that do not solve the original quadratic. [in this context] If $p(x)q(x)=0$ then $p(x)=0$ or $q(x)=0$, but it does not follow that if $p(x)=0$ then $q(x)=0$. – Mark Bennet Aug 31 '22 at 16:39
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    Is it clear to you that if you set $x=\pm1$ that $x^2+1=2?$ – Chickenmancer Aug 31 '22 at 16:41
  • @DietrichBurde Multiplying by $0$ is allowed. The resulting statement is not equivalent to the original one, but one implication is valid and that can be useful. – jjagmath Aug 31 '22 at 17:22
  • You should us MathJax also in the title. – jjagmath Aug 31 '22 at 17:23
  • @jjagmath Sorry, I wanted to say "Multiplying by 0 is not allowed for equivalence". I was assuming that we speak about equivalent equations. – Dietrich Burde Aug 31 '22 at 18:12
  • @jjagmath Thanks, this is my first time so I am very thankful for your advice. – 区なしま Sep 01 '22 at 04:05
  • I came across this question when I was doing the close vote. But I really like this question and the comments. Let’s keep it open :) – JacobsonRadical Sep 03 '22 at 00:26

1 Answers1

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Consider this only equation/condition : $x^4 = 1$

The solutions of this equation are : $±1$ and $±i$ .

Now, consider these equations/conditions: $x$ must be $1$ and $x$ satisfy $x^2 = 1$ .

Now if you solve only condition 2 i.e. $x^2 = 1$ , you'll get that $x = ±1$; however you cannot claim that "I have disproved condition first that $x$ must be $1$" because $x$ is not $1$

Similarily , you have assumed previously that $x = ±1$ is not a solution of $x^2 = 1$ ( because these values don't satisfy it ) . And hence , if you later get these values as solutions of a modified/another equation , you cannot claim that it is the solution of previous problem .

Get_ Maths
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