# Does $\sqrt{x^2+y^2}=x+y?$

This is part of a series on common misconceptions.

Is this conception true or false?

$\large \sqrt{a^2 + b^2} = a + b$

**Why some people say it's true:** It's an example of the distributive property which works since exponents are just repeated multiplication. So, just like $5(a+b) = 5a + 5b, \sqrt{a^2 + b^2} = \sqrt{a^2} + \sqrt{b^2} = a + b.$

**Why some people say it's false:** The distributive property doesn't work for the square root functions, so you can't say that $\sqrt{a^2 + b^2} = \sqrt{a^2} + \sqrt{b^2}$.

The statement is $\color{#D61F06}{\textbf{false}}$.

Proof 1:Provide a counter example.Let $a = 3 , b = 4$.

Then, $LHS = \sqrt{ 3^2 + 4^2 } = \sqrt{ 25 } = 5$.

However, $RHS = 3 + 4 = 7$.

Hence, we do not have an identity.

Proof 2:By contradiction. Let's assume that the identity is true.If we square both sides, we obtain

$a^2 + b^2 = (a+b)^2 = a^2 + 2ab + b^2.$

This is clearly not an identity since $2ab$ is not always 0. Thus, we can conclude that this is an identity if and only if $2ab = 0$.

Proof 3:Geometric interpretationWe can also look at this geometrically and immediately see that it is false:

Understanding why the distributive property does not hold for square roots:The distributive property is an identity that relates addition and multiplication: $m(a+b) = ma + mb$. However, it does not apply to the square root function. To understand why, consider the shape of the square-root function:

Consider any two points on this curve, $(a, \sqrt{a})$ and $(a+b, \sqrt{a+b})$, for $b > 0$. What is the effect on the difference between $\sqrt{a}$ and $\sqrt{a+b}$ as the two points shift together, along the function and to the right (in other words, as $a$ increases, but $b$ remains fixed)?

As the region shifts to the right, the $\sqrt{x}$ curve flattens out, which, algebraically, means that the difference between $\sqrt{a}$ and $\sqrt{a+b}$ drops to 0 as $a$ increases and $b$ stays constant. On the other hand, the difference between $\sqrt{a}$ and $\sqrt{a} + \sqrt{b}$ is always $\sqrt{b}$, which doesn't change of course, as the region shifts right, since $b$ doesn't change.

Rebuttal: For the most part, I agree, but $\sqrt{x^2} = \pm x,$ not $|x|$.

Reply: That's another misconception. See the "$\text{Is }\sqrt{x^2} = \pm x?$" misconception page for details.

Rebuttal: Consider a linear function, $f(x) = mx +b$. As a region defined by two points a fixed width apart moves right, it stays the same height; does that mean that all linear functions obey the distributive property? (That is, if $f(x) = mx + c$ for any real number $c,$ then $f(a+b) = f(a) + f(b).$)

Reply: No. There's actually a second condition in order for the distributive property to be applicable to a function: $f(0)$ must equal 0. Otherwise, $f(a) = f(a+0) = f(a) + f(0)$ could not possibly be true. Therefore, functions of the form $f(x) = mx + b$ only obey the distributive property on the domain of real numbers if $b=0$.

Sherry is excited to learn about the distributive law, and thinks that it applies to every possible operation. As such, she claims that

$\sqrt{a^2+b^2} = \sqrt{a^2} + \sqrt{b^2} .$

How many ordered pairs of integers $(a, b)$ are there, such that $-10 \leq a \leq 10, -10 \leq b \leq 10$ and

$\sqrt{ a^2 + b^2} = \sqrt{a^2} + \sqrt{b^2} ?$

**See Also**

**Cite as:**Does $\sqrt{x^2+y^2}=x+y?$.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/is-square-root-of-a2b2-equal-a-b/