Problem with operations on exponents

Okay, this is a very basic doubt, and I would like to know where exactly the mistake is.

Let's take the equality $1^2=1$

Now I do it this way Step 1: $1^2 = 1^\frac{4}{2}$

Step2: $\implies (1^4)^\frac{1}{2}$

Step3: $\implies 1^\frac{1}{2} = \pm 1$

The answer should be only one but I am also getting minus one. I know it obviously is a mistake, but I would like to know which exact step and in which particular operation the mistake is....

Given a positive number $a$, $a^\frac12$ is defined to be the positive number $b$ such that $b^2=a$. In particular, $1^\frac12=1,$ even though $(-1)^2=1,$ too.

• Okay but how about for a complex number? For example: $i^2 = i^\frac{4}{2} = (i^4)^\frac{1}{2} = 1^\frac{1}{2}$ which is equal to 1 if your definition is true for complex numbers....... – Quark May 30 '13 at 4:32
• Ah! The definition I gave is for what is called the principal branch of the square root function. For a given non-zero complex number and a positive integer $n$, there are $n$ distinct ways that we can define $z^\frac1n$. (We usually like to define this function so it is continuous "almost everywhere," and so that it agrees with our usual real definition of $n$th root.) Unfortunately, the usual exponent rules needn't all apply in general if we're dealing with complex numbers, as your example shows. – Cameron Buie May 30 '13 at 4:37
• One way to get around that is by declaring the $n$th roots to be "multi-valued functions." That has its upsides and downsides, though. – Cameron Buie May 30 '13 at 4:40
• Alright but can I know which exponent rules apply and which don't? (while solving problems in variables things are not very obvious, and doing certain operations lead to wrong answers) – Quark May 30 '13 at 5:55
• You can find more details here and in particular here. – Cameron Buie May 30 '13 at 6:22

Cameron's answer is probably the best one to give, but there's another aspect which I think is worth pointing out.

Look at the following argument: $$x=0\implies x(x-1)=0\implies x=0,1$$

This is completely true! There's nothing actually wrong with the conclusion that "IF $x=0$ THEN $x=0,1$". You're just being less specific. What you can't do is reverse the logic.

• exactly, but look at your second step, you multiplied both sides of the equation by (x-1). We can do this only if x-1 is not equal to zero, implying x not equal to one, and hence 1 is not a valid solution. Sk the mistake is precisely in the second step. Now I would like to know what the mistake in my calculation is. – Quark May 30 '13 at 4:12
• No! You don't need $x-1\neq 0$ in order to multiply by it, only to divide by it. Hence the logic works from left to right as written, just you can't reverse it. My point is that your conclusion in the question is valid when interpreted as $1=\pm 1$ meaning "either $1=+1$ or $1=-1$". In general, $x^2=a^2\implies x=\pm a$ is true even though we might have more information which rules out one possibility. – Sharkos May 30 '13 at 6:22
• Just to make clear, $x=\pm 1$ does not usually mean "$x=-1$ and $x=1$ are both solutions to the current problem". It means "any solution must be either 1 or -1" which is very different. – Sharkos May 30 '13 at 6:26
• However, as Cameron says, your specific statement does indeed have more information. Because we know $1^{1/2}$ is the correct expression for the number sought, one can conclude it must be $+1$ specifically. In general, one adopts a convention for writing $z=re^{i\theta}$ like $-\pi<\theta\le\pi$ and $r\ge 0$, then defines $z^{1/2}=\sqrt r e^{i\theta/2}$. This is an unambiguous definition of the principal branch of the square root. – Sharkos May 30 '13 at 6:54