So clearly I did something very wrong! Can anyone correct my error?
Okay, let's address this somewhat more broadly.
You have an equation that you attempt to "solve". (
Strictly speaking, that isn't the case, because there is nothing to solve about that equation, and what you actually want to do is evaluate the limit of a certain sequence. But since you resorted to doing it with algebraic means, let's call it solving the equation.)
So you perform subsequent operations on the equation, and you get new and different equations, and you're hoping to get to one you would like. (
Do you English-speaking guys have a better term for this than "simplifying"? A more general one that would equally well describe the operations in either direction?)
Now, the crucial thing here is: when performing those operations, you want to get
equivalent equations. Two equations are equivalent if they have the same set of solutions. When you stick to a certain set of algebraic operations, you will always get an equivalent equation in the next step, so when you finally find the solution in the last step, you will have solved the original equation as well.
The error you did was that you used some operations that didn't satisfy this fundamental requirement. In some steps, you got from an equation A to an equation B, such that any solution to A would also satisfy B. That's good, you're half-way there - but the opposite implication must also hold: any solution to B must satisfy A. If that isn't true, you don't get the equivalence.
Let's review your steps:
1. x = c
(I will use c to denote the numeric expression that you're trying to evaluate)
2. x
2 = c
2
Here's your first error. Equation 2 isn't equivalent to equation 1. Anything that satisfies eq.1 will satisfy eq.2, but not everything that satisfies eq.2 will satisfy eq.1. Namely, x = -c is a solution to eq.2, but not to eq.1.
Squaring is not an operation that you can rely on to get equivalent equations.
3. x
2 = 7 + c
This is a silent implicit step you made, and it is a valid one, because 7+c=c
2, so eq.3 is equivalent to eq.2.
4. x
2 = 7 + x
Here's your second error. You assumed that x=c, so you can make that substitution, but that would only work in the forward direction. In the opposite direction it fails, because eq.4 can have some solution other than x=c, and then the backward step is not valid (solution to eq.4 will not necessarily satisfy eq.3).
Here's a more obvious rendering of the error:
x=2
x=x (since we know that x=2)
But clearly, the second equation has solutions (such as x=3) that don't satisfy the first.
The rest of the steps seem fine.
One more thing: you actually
can use steps that are only valid in the forward direction (i.e. that any solution to A will satisfy B, but not necessarily the other way around). If you do that, at least all solutions of the original equation will be among the solutions of the final equation - but you must keep in mind that the final equation may also have some other solutions that may not be solutions to the original equation. What you can do then is verify the solutions you get against the original equation, to see which ones are valid.
If we allow negative square roots, then there are an infinite number of solutions,
To what?
