One Equals Minus One
From ProofWiki
Contents |
Paradox
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle -1\) | \(=\) | \(\displaystyle \sqrt {-1} \sqrt {-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \sqrt {-1 \times -1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \sqrt 1\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle 1\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
Resolution
This is a falsidical paradox arising from incorrect reasoning about the nature of square roots.
Explanation
The property:
- $\sqrt {a} \times \sqrt {b} = \sqrt {ab}$
can only be used when:
- $a \ge 0$ and $b \ge 0$
The above is subtle enough to merit some proper analysis rather than just to rely on the bald statement. This would best be documented on a separate page.
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Since $-1 < 0$, this property of square roots cannot be applied to this statement.
This is how the analysis is to go:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle -1\) | \(=\) | \(\displaystyle \sqrt {-1} \sqrt {-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle i \sqrt 1 \cdot i \sqrt 1\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | as $\sqrt {-a}$ is defined as being $i \sqrt a$ for $a > 0$ | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left({i \cdot i}\right) \sqrt 1 \sqrt 1\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle -1 \cdot 1\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | as $i^2 = -1$ | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle -1\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
... and all is well with the world.
$\blacksquare$
Also see
Sources
- For a video presentation of the contents of this page, visit the Khan Academy.
