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?


closed as off-topic by Did, kingW3, Miha Habič, Olivier Bégassat, user99914 May 11 '15 at 0:59

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  • 3
    $\begingroup$ $\sqrt{-49}\ne -7$ $\endgroup$ – Dr. Sonnhard Graubner May 10 '15 at 15:16
  • 6
    $\begingroup$ Where to begin.... $\endgroup$ – Alec Teal May 10 '15 at 15:16
  • 3
    $\begingroup$ Do none of the many similar (answered) questions on site address your problem? $\endgroup$ – Andrew D. Hwang May 10 '15 at 15:16
  • 2
    $\begingroup$ @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!" $\endgroup$ – Alec Teal May 10 '15 at 15:18
  • 1
    $\begingroup$ @quapka I was teasing you because you said " @ AlecTeal I think i=−7, not that it matters.." <-- claiming that YOU think this :P $\endgroup$ – Alec Teal May 10 '15 at 15:30

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.


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.

  • $\begingroup$ "Therefore, what comes after therefore is wrong." Well, actually, what comes after therefore could be true. :-) $\endgroup$ – Did May 10 '15 at 15:45

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 $i^2 = -1$.


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

  • $\begingroup$ What? $\sqrt{-7}\ne7i$, last time I checked. It equals $i\sqrt7$. $\endgroup$ – Akiva Weinberger May 10 '15 at 15:24
  • $\begingroup$ @columbus8myhw Oops, thanks for catching my error. $\endgroup$ – Jimmy360 May 10 '15 at 15:28

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