Why is the square root of a sum not equal to the square root of each its addends?

Question

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?

Answer 1

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"?

Answer 2

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.

Answer 3

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 $k=m$.)

Answer 4

"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 $c$.

Answer 5

Otherwise $\sqrt n = n$ for all natural number $n$.

More generally $\sqrt q = q$ for all positive rational number $q$.

Homework:

Is it true that: $\sqrt x = x$ for all positive real number $x$?