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**Ackbeet** Right, but what are you doing here? You're *inverting* the square function. So if you let $\displaystyle f(x)=x^{2},$ and you assume that $\displaystyle f(a)=f(b),$ then I agree that it might not be the case that $\displaystyle a=b.$ However, that is not the meaning of the phrase "well-defined" as applied to functions. It's the difference in logic between the statements "If A then B" and "If B then A". One is the converse of the other, and, in general, neither implies the other. If it's true that A implies B, you cannot conclude that therefore, B implies A.

So, getting back to a well-defined function and using our $\displaystyle f(x)=x^{2}$ above, if it's true that $\displaystyle a=b$ then *it is always true that* $\displaystyle f(a)=f(b),$ or $\displaystyle a^{2}=b^{2}.$ And that's true, because the square function is well-defined. A graphical way of thinking about a well-defined function is that it passes the vertical line test: any vertical line drawn on an xy plot of the function $\displaystyle y=f(x)$ will touch the function in at most one place. So a circle is not a well-defined function. Technically, most people would say that it's not even a function.

Thinking about things from an even higher level: whether or not you want to apply a particular function to an equation depends on a number of things. Do you need to be able to reverse all the steps in a series of computations? If so, then I would advise against applying functions that are not one-to-one (the graphical way of thinking about one-to-one functions is that they pass a horizontal line test, where any horizontal line intersects the graph in at most one place). The square function is not one-to-one, and hence applying the square function is not reversible. By "reversible", I mean that you can apply the function's inverse and exactly get to where you were before. When you apply functions to an equation that are not one-to-one, you can introduce spurious solutions to the equation. So you have to be careful when you square both sides of an equation, for example, that all the roots you find satisfy the original equation (before squaring).

Does all this make sense? Any more questions?