How to explain to a 14-year-old that $\sqrt{(-3)^2}$ isn't $-3$?

Solution 1:

After having given him the square root function definition, it may be helpful to show the following: $$ \sqrt{(-3)^2}=\sqrt{9}=\sqrt{3^2}=3\neq -3. $$ Although well-definedness may be a bit of a heavy topic for a 14-year old, it should be very easy to explain why you cannot have ambiguous definitions in mathematics.

Solution 2:

This is about the difference between a function on one variable and an equation with one variable. A function should only have one result for each argument, whereas an equation can have more than one solution, or no solutions at all.

Let's consider a very simple function, $f(x) = x + 1$. For example, $f(3) = 4$, $f(4) = 5$. If $x$ doesn't change, neither should $f(x)$. You can run the function a hundred times with the same value and it should give the same result. Solve the equation $f(x) = 8$. The words "trivially simple" come to mind. This function has a one-to-one correspondence between inputs and outputs: each number $x$ is matched to a unique $f(x)$.

Now consider a slightly more complicated function, $f(x) = |x| + 1$. For example, $f(-3) = 4$. As with the first function, if $x$ doesn't change, neither should $f(x)$. But now an equation involving this function can have two solutions. If $f(x) = 8$, then $x$ can be $7$, but it can also be $-7$.

For the first function, we can define a look-up function by arithmetic means: $g(y) = y - 1$. But for the second function, if we demand a look-up function that uses arithmetic means rather than memory, we're going to be frustrated. The definition $g(y) = y - 1$ delivers the right result if $x$ was positive, or $0$. The question is whether or not this function $g(y)$ is good enough for our purposes.

That's what the principal square root function is: a function that tells us what value was input into another function if that value was positive, or $0$. The square function is defined as $x^2 = x \times x$. If $y = x^2$, the function $\sqrt{y}$ tells us what $x$ was if we know that $x$ was positive, or $0$.

Calculating $\sqrt{y}$ is not the same as solving the equation $x^2 - y = 0$. At least for this equation there are only two possible solutions: $\sqrt{y}$ and $-\sqrt{y}$. Therefore, $\sqrt{(-3)^2} = 3$, but the equation $x^2 - 9 = 0$ has two solutions, $x = 3$ and $x = -3$. We expect that if $y$ doesn't change, neither does $\sqrt{y}$, whether we run the function at noon, at midnight or any time of the day.

Canceling the square in $\sqrt{(-3)^2}$ to obtain $-3$ rather than $3$ creates a memory function rather than an arithmetic function. A memory function can be an arithmetic function only if the function being undone has a one-to-one correspondence between inputs and outputs.

Solution 3:

Give him the definition of the square root function : for $x\in\mathbf{R}_{+}$, the number $\sqrt{x}$ is by definition the unique number $y\in\mathbf{R}_{+}$ such that $y^2 = x$.