Example: Let's presume one was attempting to isolate m below:
A common mistake would be: $k^2 = m^2 + n^2 \to k = m +n$
Even though: $k^2 = m^2 + n^2 \to k \neq m +n$
If you apply a square root to both sides of the equation, you will have an inequality.
Why is this true?
If you put the question in a broader context you may understand it better.
Suppose $f$ is some function you can apply to numbers. It might be squaring, or square-rooting, or inverting, or raising to some other power, or taking logarithms or $\sin$ or $\cos$ or just adding 15. In none of these cases is $f(x+y)$ the same as $f(x) + f(y)$ (you should check). In general, you would not expect that coincidence.
The special case in which it is true is the function "multiply by a fixed quantity". That's the distributive law:
$$ c \times (x + y ) = c \times x + c \times y . $$
Many of the most common errors students make in algebra or precalculus come from thinking that those other functions behave this way too.
Edit: Just in case you missed @Rahul 's comment: What we need here is a cure for the “law of universal linearity”: Pedagogy: How to cure students of the "law of universal linearity"?
This was a bit unintuitive for me. Edit: It is possible to prove that these functions exist, but probably not possible to write one down explicitly ... very strange!
– Zubin Mukerjee Mar 29 '16 at 01:56"Why is this true?" (that $f(x+y) \ne f(x)+f(y)$).
Because the functions that satisfy $f(x+y)=f(x)+f(y)$ and are not badly behaved (continuous will work) are all linear, so that $f(x) = cx$ for some real $c$.
So if you try $f(x) = \sqrt{x}$, you can not have $f(x+y) = f(x)+f(y)$.
There are (at least) two ways to prove this.
First, if $f(x+y) = f(x)+f(y)$, then $\sqrt{x+y} = \sqrt{x}+\sqrt{y}$, then, squaring, we get $x+y = x+y+2\sqrt{xy}$, so that $2\sqrt{xy} = 0$, so that $xy = 0$, or at least one of $x$ and $y$ is zero.
Second, if we use the result that $f(x) = cx$ for some real $c$, then $\sqrt{x} = cx$. Squaring and dividing by $x$, we get $1=c^2x$, which can not hold for any constant $c$.
Well this can be shown by the triangle inequality. Another way to look at it is for $k^2 = m^2 + n^2$ and $k = m + n$ you have that $(m+n)^2 = m^2 + n^2$ which is only true when either m or n are 0. In other words, the inequality only holds if you assume $m, n$ are nonzero.
Consider $k,m,n$ as the sides of a triangle. Then:
$k^2 = m^2 + n^2$ means that the triangle is a right triangle.
$k = m +n$ means that the triangle is a degenerate triangle.
Clearly, no right triangle can be degenerate!
(Unless one of the sides is zero, in which case, yes, $k^2 = m^2$ implies $k=m$.)
