Why is this proof of $i = -1$ wrong?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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

Answer 4

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 $i^2 = -1$.