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This question already has an answer here:

Can anybody please describe how and why it is even possible? If there is anything wrong what it is,?

Prove: $2+2=5$

$$-20=-20$$

$$16-36=25-45$$

$$16-36+\frac{81}{4}=25-45+\frac{81}{4}$$

$$4^2-2\cdot4\cdot\frac92+\left(\frac92\right)^2=5^2-2\cdot5\cdot\frac92+\left(\frac92\right)^2$$

$$\left(4-\frac92\right)^2=\left(5-\frac92\right)^2$$

$$4=5-\frac92+\frac92$$

$$4=5$$

$$2+2=5$$

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marked as duplicate by Rob Arthan, Dietrich Burde, mathematics2x2life, JonMark Perry, qwr Sep 29 '17 at 5:04

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    $\begingroup$ That silly "proof" is too dumb to die, it was old already in my youth. It's a waste of space, too, the economy version is $1^2=(-1)^2\to1=-1$. $\endgroup$ – Professor Vector Sep 28 '17 at 20:01
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    $\begingroup$ @Sandeep I've edited your question's formatting and written out the algebra from the picture. Let me know if you think anything's awry. $\endgroup$ – Jam Sep 28 '17 at 20:05
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    $\begingroup$ This is dreary beyond belief. If you want to find where it goes wrong just evaluate the formulas (and tara-tara, you'll find that $x^2 = y^2$ does not imply $x = y$). How can you possibly need help with that? My close vote is because you have made no attempt to solve the "problem" yourself of to make it of any interest to anybody else. $\endgroup$ – Rob Arthan Sep 28 '17 at 20:12
  • $\begingroup$ @ProfessorVector : How is the age of this argument relevant? $\endgroup$ – Michael Hardy Sep 28 '17 at 20:44
  • $\begingroup$ @Michael Hardy "This is getting old" is about age? Sorry, I didn't know that. $\endgroup$ – Professor Vector Sep 29 '17 at 3:38
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As the other answers have pointed out, the flaw lies in the step between $\left(4-\frac92\right)^2=\left(5-\frac92\right)^2$ and $\left(4-\frac92\right)=\left(5-\frac92\right)$. There are a number of ways to explain why you can't do this:

One reason you can't do it is because $\sqrt{x}$ takes $2$ different values. Namely, $\sqrt{x}$ and $-\sqrt{x}$. This is because $x^2=(-x)^2$. When you do the incorrect step in your proof, you're taking the square root of both sides but really, you should be taking either $\sqrt{x}$ or $-\sqrt{x}$. So, in fact, it would be correct to say that $\left(4-\frac92\right)=-\sqrt{\left(5-\frac92\right)^2}=-\left(5-\frac92\right)$.

Another way of explaining why that step is wrong is that for any number, $x$, you get the same number from $x^2$ as $(-x)^2$. Case in point, $3^2=(-3)^2$ and, as isyoung points out, $0.5^2=(-0.5)^2$. So, this fake-proof is designed to have one side equal to $0.5^2$ but the other side equal to $(-0.5)^2$. Specifically, the fake-proof says that $(4-4.5)^2=(5-4.5)^2$ since $4-4.5=-0.5$ and $5-4.5=0.5$. This trick lets the fake-proof have you thinking that $0.5=-0.5$. All other parts of the fake-proof are just puffery disguising that step.

In summary, $x^2=y$ doesn't just mean $x=\sqrt{y}$, it can also mean $x=-\sqrt{y}$.

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Cancellation of the square is wrong in the 4th line from the bottom. $4-\frac{9}{2}=-\frac{1}{2}$ is negative, and $5-\frac{9}{2}=\frac{1}{2}$ is positive. They cannot be equal, but their squares are equal.

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  • $\begingroup$ Clear and concise. $\endgroup$ – Mark Viola Sep 28 '17 at 20:09
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On the 4th from bottom line, you "cancel" the ${}^2$ sign. When doing this you should take the absolute value - $$\left|4-\dfrac92\right|=\left|5-\dfrac92\right|$$ which would be correct.

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When they remove the powers is where they go wrong.

$$\bigg(4-\frac{9}{2}\bigg)^2 = \bigg(-\frac{1}{2}\bigg)^2 = \bigg(\frac{1}{2}\bigg)^2 = \bigg(5-\frac{9}{2}\bigg)^2$$

$$\bigg(-\frac{1}{2}\bigg) \neq\bigg(\frac{1}{2}\bigg)$$

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\begin{align} & \left( 4 - \frac 9 2 \right)^2 = \left( 5 - \frac 9 2 \right)^2 \\[15pt] \text{Therefore } & 4 - \frac 9 2 = \pm\left( 5 - \frac 9 2 \right) \\[10pt] \text{i.e. } \quad & 4 - \frac 9 2 = \left\{\text{either } \quad +\left(5 - \frac 9 2\right) \quad\text{ or }\quad -\left( 5 - \frac 9 2 \right)\right\}. \end{align}

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  • $\begingroup$ @MarkViola : That someone would insist that $\text{“} \pm \text{''}$ must have the interpretation that you gave it is new to me. $\qquad$ $\endgroup$ – Michael Hardy Sep 28 '17 at 20:14
  • $\begingroup$ I didn't "insist" Michael. Rather, I said that one "might" interpret that $\pm$ means that both solutions are viable. I just gave an up vote. I suggest that you take a different tact. $\endgroup$ – Mark Viola Sep 28 '17 at 20:27
  • $\begingroup$ A typesetting oddity: Notice the difference between these two lines: $$\begin{align} 4 - \frac 9 2 = \left\{\text{either } \quad +\left(5 - \frac 9 2\right) \quad\text{ or }\quad -\left( 5 - \frac 9 2 \right)\right\} \\ \\ 4 - \frac 9 2 = \left\{\text{either } \quad +{\left(5 - \frac 9 2\right)} \quad\text{ or }\quad -{\left( 5 - \frac 9 2 \right)} \right\} \end{align}$$ In the second format, the plus sign in the second line is barely, but visibly, closer to the left parenthesis than it was in the first line (and I expected that) but the minus sign is no closer. $\endgroup$ – Michael Hardy Sep 28 '17 at 20:27
  • $\begingroup$ @MarkViola : You said that $\cdots= \pm\left( 4 - \frac 9 2\right)$ means that both "solutions" (as you called them) are possible. I wouldn't have taken it to mean that. $\endgroup$ – Michael Hardy Sep 28 '17 at 20:40
  • $\begingroup$ Michael, I'm curious. When you solve the quadratic equation $ax^2+bx+c=0$ and write $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, how do you interpret the $\pm$ sign? Are both solutions viable? $\endgroup$ – Mark Viola Sep 28 '17 at 20:44
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Unless a function $f$ is 1-1, it needn't follow from $f(a)=f(b)$ that $a=b$.

You're using the function $f(x)=x^2$ to make this (erroneous) conclusion.

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  • $\begingroup$ I agree that this is the most general explanation why OP's reasoning fails but if someone posts a fake-proof this basic, they're likely to not know what one-to-one functions are. $\endgroup$ – Jam Sep 28 '17 at 20:23

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