'Prove' 2 = 0 Using Square Roots. Can You Find The Mistake?

I received an email with a challenge from a viewer of MindYourDecisions. Wolfgang came up with a false proof that 2 = 0. No one in his class, not even his teacher, could spot the mistake. Can you figure it out?

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[imissmyoldchannel]

i seriously doubt that the teacher couldn't find the mistake

[LedZeta]

When you work with real numbers, a square root is always positive, so square root 1 is 1, but when you work with complex numbers, for a root with index n you have exactly n solutions, so for a square root you have two solutions. In this case, square root -1 = +\- i, now you can find two solutions: 2=2 that is correct, or2=0 that is false. Breaking a square root in that way is not an error if you are working with complex numbers so that is not an error, the error is not considering the solution -i

[AgentMidnight]

I thought for sure the solution was going to be that the one step of the equation was dividing by a fancy form of zero, but then it turned out to be an entirely different fake-proof.

[suryanshsinh]

2=0
Can you spot the mistake!??
Hmm
Me:
Yes I can!!
The mistake was to write = instead of ≠!!
2≠0

[franchello]

As soon as you went to the wrong step, I was like "I don't think I like that".

[sidhantrastogi]

Honestly his teacher wasn't very good if he couldn't spot this obvious of a mistake..

[physicsinanutshell]

I think the student must get his school changed.

[MandhanAcademy]

The rule that i learned in class 11 is applicable : You cannot split ROOT while both numbers have -ve Sign.
Reply if something else is applicable.

[arpitpandey]

sqrt((-1)(-1)) is not equal to sqrt(-1)*sqrt(-1)

[BlinkLed]

Whenever you take a square root, you always have to add "+ or -." I've always really loved these false proofs and I think they would be great problems for school kids learning math, because they really require you to understand the rules of what you're doing, and why the rules exist.

[lythd]

It's because of the +- of the square root. If one of the sqrt(-1) terms is negative and the other is positive it works. When there is a negative in the product it can mess up whether you need a plus or minus to retain the identity. If you started with +-sqrt(1) then nothing would be wrong but since you set it to the positive sqrt(1) you need to be careful when you split it with negative terms because you will get problems like in this example.

[jamesdurtka]

My favorite version of this proof is one where there is a division by zero which is then obscured by using multiple variables which are equal to each other, and things like that. You can form an incredibly convoluted "proof" to "prove" any two numbers are equal to each other, and make it VERY difficult to spot the mistake.

[Littlefighter]

2:11 I've spotted it, but I didn't know that such a restriction existed.
Good to know.

[BALAGE]

sqrt(a*b) by definition only equals sqrt(a)*sqrt(b), if a and b are positive or zero.

[seanwilkinson]

You can avoid this mistake and still get 2=0 as follows:
You can also arrive at 2=-2 by applying the same math to both ones. Also, 2=0 when shifting to one-bit binary (00=10, as the 2^0 bit in both representations is 0), and 2=-2 when absolute value (magnitude) is applied to both sides.

[unterbawr]

when you said "so we take the square root of each term" I was like "....wait, why??"

[kishorbhushan]

Brilliant presh telwakar I love your videos

[shikhanshu]

i never paid attention to these caveats like "if a>0, b>0" etc. in my high school... they are important kids!

[Mycroft]

I went a different road:


The square root of -1 isn't just i. It's also -i. If we turn one i into -i, the product becomes +1, summing with 1 to equal 2. This also proves that we could in fact take the separate square roots to get the correct answer. Thus, the problem with the proof is the all-too-common error of overlooking the fact that square roots have both positive and negative answers.

[zlotnleo]

If you're working with complex numbers, the square roots aren't that straightforward. The splitting the square root is only legal if either both (-1) and (-1) are >=0 (which isn't true) OR you rewrite sqrt as a number to power of 1/2. But (-1)^0.5 can be either i or -i.