# 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 \( \color{red}{\textbf{false}}\).

Proof:Consider squaring both sides of the suggested identity.

(Equivalency should hold, since if \(x=y\), it would also be the case that \(x^2 = y^2.\))Left Hand Side: \((\sqrt{a^2 + b^2})^2 = |a^2 + b^2| = a^2 + b^2\)

(The last step is valid because \(a^2\) and \(b^2\) must both be non-negative.)Right Hand Side: \((a+b)^2 = a^2 + b^2 + 2ab\)

Comparing the two sides, \(a^2 + b^2\) and \(a^2 + b^2 + 2ab\), it is clear that the proposed identity is only true in the cases where \(2ab = 0\). This will only occur if either \(a=0\) or \(b=0\). Additionally, any non-zero variable must be positive, such that \(\sqrt{a^2 + 0} = |a+0| = a\), or similarly for \(b\). In all other situations, the equation is false. We can also look at this geometrically and immediately see that it is false:

The algebra above is a proof of why the suggested identity is false, however, the section below explains why the distributive property \((f(a+b) = f(a) + f(b)) \) cannot be applied generally to

any functions aside from simple multiplication and division.

Understanding the distributive property \((\text{"Why isn't }\sqrt{a+b} = \sqrt{a} + \sqrt{b} ~?\text{"})\):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.

This is why the distributive property

only applies to multiplication(and division, as it is multiplication by a proper fraction). Exponential functions including \(x^2\), other exponents, square roots, logarithms, etc., do not obey the distributive property.For multiplication, \(f(x) = mx\), and as the region between two points on that line, \((a,f(a))\) and \((a+b,f(a+b))\), shifts to the right, the difference between \(f(a)\) and \(f(a+b)\) remains fixed. And, as might be expected, this difference equals \(f(b)\):

\[f(a+b) - f(a) = m(a+b) - ma = ma + mb - ma = mb = f(b).\]

And so, the distributive property is true for multiplication in that for any function \(f(x) = mx,\)

\[f(a+b) = f(a) + f(b).\]

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

Reply: Nope, 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/