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 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 interpretation
We can also look at this geometrically and immediately see that it is false:
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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:
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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)?
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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