Help: Proof that 1 = -1?

I came across something the other day involving complex numbers.

It is obvious that: $\frac {1}{ -1} = -1 = \frac {-1}{1}$ .

If we take square roots on both sides, we get: $\frac {1}{i} = \frac {i} {1}$ .

Multiplying gives $1 = 1 \times 1 = i \times i = -1$ .

So, $1 = -1$ ?

I don't understand what is wrong. Can someone help?

#HelpMe! #Proofs #Math

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

I believe the problem here is that $\sqrt{-1}$ is not an ordinary square root, as it is not real and thus it does not follow the same square-rooting rules as real square roots..

$\sqrt{\frac{x}{y}}$ is not necessarily equal to $\frac{\sqrt{x}}{\sqrt{y}}$

Which goes back to the basic precalc premise that $\sqrt{x}\sqrt{y} \neq \sqrt{xy}$

Tim Ye - 8 years, 5 months ago

Why not ? $\sqrt{\dfrac{x}{y}}$ can be written as $(\dfrac{x}{y})^{\dfrac{1}{2}}$ which is $\dfrac{x^{\dfrac{1}{2}}}{y^{\dfrac{1}{2}}}$ which is ${\dfrac{\sqrt{x}}{\sqrt{y}}.}$

Sai Ram - 5 years, 10 months ago

Yes, Tim is right. Notice that $\sqrt{-4}\sqrt{-9}=(2i)(3i)=-6$ , while $\sqrt{(-4)(-9)}=\sqrt{36}=6$ . This is why the premise below works for non-negative real numbers only.

Raoul Danniel Manuel - 8 years, 5 months ago

That is a great observation.

Do you know why we have to make such an exception when it comes to dealing with complex numbers, but not with the non-negative reals?

Hint: $\sqrt{x} : \mathbb{C} \rightarrow \mathbb{C}$ is interpreted as the inverse of the square function $x: \mathbb{C} \rightarrow \mathbb{C}, x \rightarrow x^2$ . Same is true of $\sqrt{x} : \mathbb{R} \rightarrow \mathbb{R}$

Calvin Lin Staff - 8 years, 5 months ago

Yes, as Krishna pointed out, the inverse of the square function is NOT a function, since it can take on 2 values. As such, when we talk about $\sqrt{x}$ , we have to be very certain about which value we are actually referring to. We cannot immediately conclude that $\sqrt{1} = 1$ , since there is the possibility that it is actually $-1$ . It can be impossible to tell from the context, and we have to be aware that only one of the values might make sense.

For exponents, if the exponent is not an integer, there can be ambiguity about what the value actually refers to. However, the laws of indices will still hold. For example, regardless of what value we choose for $(-1)^{0.5}$ , then $(-1)^ {0.5} \times (-1)^{0.5} = -1$ holds.

Calvin Lin Staff - 8 years, 5 months ago

Also, the square roots of 1 are 1, and -1, so it's ambiguous

Krishna Kinnal - 8 years, 5 months ago

when the laws of indices were first proved they were proved for postive radicands so as tim says one cannot directly apply them for the negative radicands.also in the books that i have read it is mostly given that a+ib is a complex number where i is the square root of -1 and i square is -1 .so its mostly treated as an axiom

pranav chakravarthy - 8 years, 5 months ago

complex numbers are imaginary but natural numbers are real we cannot deal the imaginary with real no

keshav kumar - 7 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$