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I know that $-7^2 = -49$

Therefore $\sqrt{-49} = -7$

Because $\sqrt{-1} = i$ we can then expand it to $\sqrt{-49} = -7 = 7i$

And therefore $-7 = 7i$, divide both sides by 7 and you get

$-1 = i$

And I know that is not true, because $i = \sqrt{-1}$. What is wrong with my proof?

user239425
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    $\sqrt{-49}\ne -7$ – Dr. Sonnhard Graubner May 10 '15 at 15:16
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    Where to begin.... – Alec Teal May 10 '15 at 15:16
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    The error is when you take the square root of -49. There is a difference between $(-7)^2$ and $-7^2$. – felani May 10 '15 at 15:16
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    Do none of the many similar (answered) questions on site address your problem? – Andrew D. Hwang May 10 '15 at 15:16
  • There was pretty much the same question yesterday, can't find it. – quapka May 10 '15 at 15:16
  • Suppose $\sqrt{-49}=-7$ then you are saying $(-7)*(-7)$ is $-49$ - can you see how stupid that is? – Alec Teal May 10 '15 at 15:17
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    @user86418 I call it "upvote whoring" - if you ask a question accessible to people on the other SE sites they'll come here and be like "THIS - this is a fine question!" – Alec Teal May 10 '15 at 15:18
  • @quapka that was $i=-i$ or some such nonsense. – Alec Teal May 10 '15 at 15:19
  • @AlecTeal I think $i = -7$, not that it matters.. – quapka May 10 '15 at 15:20
  • @quapka pat yourself on the head, from me. You need it. – Alec Teal May 10 '15 at 15:21
  • @AlecTeal Not sure, whether you mean that in a good way (as in "pat on the back") or not :), anyway, hopefully we won't receive e.g. "Why is this proof for i = -13" tomorrow (not that I want to offend those, who have troubles understanding the concept of imaginary unit). – quapka May 10 '15 at 15:28
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    @quapka I was teasing you because you said " @ AlecTeal I think i=−7, not that it matters.." <-- claiming that YOU think this :P – Alec Teal May 10 '15 at 15:30
  • @user86418 Which similar questions are these, and how do you propose the OP should have gone about finding them? Look up "false proof" and sift through the hundreds of questions in case one of them happens to be the exact same one as in this question? – Jack M May 10 '15 at 19:53
  • @JackM: My comment was intended to be good-natured; perhaps it came across otherwise? That said, the second result in a site search for "fake proof" is an argument that $-1 = 1$ via square roots, and that's linked to other similar questions. To my knowledge this question is not a duplicate, but it's conceivable that perusing a typical question of the linked type would have suggested an answer. (And I hope it's clear I answered because the question seemed worthwhile.... :) – Andrew D. Hwang May 10 '15 at 20:44
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    @user86418 Fair enough, I was sensitive because of the very negative response this question is getting outside of your comment. – Jack M May 10 '15 at 21:13
  • possible duplicate of Why $\sqrt{-1 \times {-1}} \neq \sqrt{-1}^2$? –  May 11 '15 at 00:59

4 Answers4

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The meaning of the first line is $-(7^2)=-49$, not $(-7)^2=-49$, the second one is wrong.

Therefore, what comes after therefore is wrong.

N. S.
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  • "Therefore, what comes after therefore is wrong." Well, actually, what comes after therefore could be true. :-) – Did May 10 '15 at 15:45
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At risk of sounding tart, it's easier to say what's right with the proof:

  • $-7^{2} = -49$ is true (because with normal precedence of operations "$-7^{2}$" means "$-(7^{2})$".

  • $\sqrt{-1} = i$ is fine as notation, but will reliably start an argument in some quarters on Math.SE. (What vexes people is that $\sqrt{\ }$ is a function in the sense understood by The General Public only if the radicand is a non-negative real number, and $-1$ is not a non-negative real number.)

What's wrong:

  • $-(7^{2}) \neq (-7)^{2}$, so "Therefore $\sqrt{-49} = -7$" doesn't follow.

  • On using $\sqrt{-1} = i$ to expand $\sqrt{-49}$, two falsehoods get daisy-chained, $\sqrt{-49} = -7$ and $-7 = 7i$. Ironically, the combined effect, "$\sqrt{-49} = 7i$" is correct in the sense that $(7i)^{2} = -49$. This shows that in mathematics, two wrongs (sometimes) do make a right.

  • If we include errors of style, the introduction of $7$ into the proceedings does nothing to clarify. The entire proof could have been reduced to one erroneous step:

I know that $-1^{2} = -1$.

Therefore $-1 = i$.

What is wrong with my proof?

And then the sign error would have been apparent: $-1 = -1^{2} = -(1^{2}) \neq (-1)^{2} = 1$.

The real moral is the same as with debugging code: Always strive to reduce a contradiction to the smallest and/or simplest problematic situation possible.

Andrew D. Hwang
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There are several mistakes that I notice. First off, $\sqrt{-49}\neq -7$, since $(-7)^2 = 49,$ not $-49$ (also note that $-7^2 \neq (-7)^2$).

Also, it is very dangerous to say that $i = \sqrt{-1},$ without full understanding that the square root is multi-valued. That is, since $(-i)^2 = i^2 = -1$, we also have that $\sqrt{-1} = -i$. What you do know, for sure, is that $i^2 = -1$.

Eff
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You confused $-x^2$ with $(-x)^2$.

Jimmy360
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