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I found out this on Google+ yesterday and I was thinking about what's the trick. Can you tell?

How can you prove $3=2$?

This seems to be an anomaly or whatever you call in mathematics. Or maybe I'm just plain dense.

See this illustration:

$$ -6 = -6 $$

$$ 9-15 = 4-10 $$

Adding $\frac{25}{4}$ to both sides:

$$ 9-15+ \frac{25}{4} = 4-10+ \frac{25}{4} $$

Changing the order

$$ 9+\frac{25}{4}-15 = 4+\frac{25}{4}-10 $$

This is just like $a^2 + b^2 - 2a b = (a-b)^2$. Here $a_1 = 3, b_1=\frac{5}{2}$ for L.H.S, and $a_2 =2, b_2=\frac{5}{2}$ for R.H.S. So it can be expressed as follows:

$$ \left(3-\frac{5}{2} \right) \left(3-\frac{5}{2} \right) = \left(2-\frac{5}{2} \right) \left( 2-\frac{5}{2} \right) $$

Taking positive square root on both sides:

$$ 3 - \frac{5}{2} = 2 - \frac{5}{2} $$

$$ 3 = 2 .$$

I think it's something near the root.

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4=4 => (2) x (2) = (-2) x (-2) => ("taking positive roots on both sides") => 2 = -2 – leonbloy Oct 1 '11 at 1:27
Why the -1 point? – user9209 Oct 1 '11 at 1:32
When you take square roots of both sides, you should write "$\pm$" in front of one of them. Then no paradox arises. – Michael Hardy Oct 1 '11 at 2:56
+1. This adds another example to our collections of fallacious proofs with which to challenge students to identify the error. (Although it's not really all that different from some standard ones.) – Michael Hardy Oct 1 '11 at 2:57
up vote 27 down vote accepted

$2-(5/2)$ is not a positive square root.

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But [2- (5/2)][2- (5/2)] is positive. – user9209 Oct 1 '11 at 1:21
@CHarlie: Yes, $(-0.5)\times(-0.5)$ is indeed positive, but $-0.5$ (one of the square roots) is not positive. – anon Oct 1 '11 at 1:30
Fair enough. I guess. – user9209 Oct 1 '11 at 1:32

Back when I was in academia, I taught the "how to prove stuff" course, and one of the first problems that I'd give (which, I admit, I borrowed from my graduate adviser) was along the same vein, namely: criticize the "proof" of the following "theorem" or rethink your life!

"Theorem": You have all the money you need.

"Proof:" Let $M$ denote the amount of money you have and $N$ denote the amount of money you need. Let $A=\frac{M+N}{2}$ be the average of $M$ and $N$. Then, we have:


$2A(M-N)=(M+N)(M-N)=M^2 - N^2$

$M^2-2AM = N^2-2AN$

$M^2-2AM + A^2 = N^2-2AN + A^2$

$(M-A)^2 = (N-A)^2$

And taking the square root of both sides, we have $M-A=N-A$, and hence $M=N$. $\blacksquare$

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Also, please don't show this answer to politicians! – user5137 Oct 26 '11 at 20:31

HINT $\ $ You erroneously inferred $\rm\ x^2 =\: (-x)^2\ \Rightarrow\ x\: =\: -x\:,\ $ for $\rm\ x\:=\:1/2\:.$

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On the right side, when you say to take the positive square root of $(2-5/2)(2-5/2)$, you're taking a $-.5 [(2-5/2)]$ instead of $.5$

It's easy to see if you multiply out all the numbers in each step.

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This one is too easy. You cannot divide for (a-b) since a = b. – user9209 Oct 1 '11 at 1:31
(I deleted it since it's not part of the answer) – xpda Oct 1 '11 at 1:35

It's simply not true in the reals that if $ x^2=y^2 $ then x=y. For example, if $ (-2)^2=2^2 $, but 2 does not equal -2. The scheme of inference used in the last step in general isn't valid.

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