# Do Square Roots Always Multiply?

This is part of a series on common misconceptions.

True or False?For all real numbers \(a\) and \(b,\)

\[ \sqrt{a} \times \sqrt{b} = \sqrt{ab}. \]

**Why some people say it's true:**

By the rules of exponents, \(\sqrt{a}\times \sqrt{b} = a^{\frac{1}{2}} \times b^{\frac{1}{2}} = \left(ab\right)^{\frac{1}{2}} = \sqrt{ab}.\) This seems to work out for positive real numbers; e.g., if \(a=4\) and \(b=9,\) then \(\sqrt{4}\times \sqrt{9} = 2\times 3 = 6,\) just as \(\sqrt{4\times 9} = \sqrt{36} =6.\)

**Why some people say it's false:**

There are other cases to consider besides positive real numbers. In those other cases, this "identity" may fail.

This statement is \( \color{red} {\textbf{false}}\). In particular, \(\sqrt{a} \times \sqrt{b} = \sqrt{ab}\) is true

except when \(a\) and \(b\) are both negative.

Proof:

When \(x \ge 0,\) square roots are fairly straightforward, but keep in mind the \(\sqrt{\cdot}\) operator is a function that gives exactly one value. For example, \(\sqrt 4 = 2.\) However, when \(x<0,\) there is no real number whose square is \(x.\) Thus, when \(x < 0,\) we define \(\sqrt{x} = i\sqrt{-x},\) where \(i=\sqrt{-1}\) (the imaginary unit). For example, \(\sqrt {-4} = 2i.\)First, let's check what happens when

exactly oneof \(a\) or \(b\) is negative. For example, if \(\color{red}{a}<0\) and \(\color{green}{b}> 0,\) then \(\sqrt{\color{red}{a}} \times \sqrt{\color{green}{b}} = i\sqrt{-\color{red}{a}}\times \sqrt{\color{green}{b}}.\) Since \(-\color{red}{a}>0\) and \(\color{green}{b} > 0\), we can multiply these square roots to obtain\[i\sqrt{-\color{red}{a}}\times \sqrt{\color{green}{b}} = i \sqrt{-\color{red}{a} \times \color{green}{b}}.\]

Then, since \(\color{red}{a} \times \color{green}{b} < 0,\) we have \(\sqrt{\color{red}{a}\times \color{green}{b}} = i \sqrt{-\color{red}{a} \times \color{green}{b}},\) so the two sides are equal!

However, we also need to check what happens when

both\(a\) and \(b\) are negative. In this case,\[\sqrt{a} \times \sqrt{b} = i \sqrt{-a} \times i \sqrt{-b} = i^2 \sqrt{ab} = -\sqrt{ab}.\]

On the other hand, since \(a\) and \(b\) are both negative, \(ab>0.\) Thus,

\[\sqrt{a} \times \sqrt{b} = -\sqrt{ab}\ \color{red}{\ne} \sqrt{ab},\]

so the statement

does nothold true. \(_\square\)

Rebuttal:Do we really need to take the square rootbeforemultiplying the radicands?

Reply:Yes! The square root sign is essentially a fractional exponent, and we have to take care of exponents first according to order of operations.

Rebuttal:Why doesn't the "squaring both sides" of the equation argument work?

Reply:Keep in mind that \(x^2 = y^2\)does notimply that \(x=y.\) While \(\left(\sqrt{a}\times \sqrt{b}\right)^2 = \left(\sqrt{ab}\right)^2,\) this does not mean that \(\sqrt{a} \times \sqrt{b}\) must be equal to \(\sqrt{ab}.\)

Rebuttal:My teacher told me that \(\sqrt{a} \times \sqrt{b} = \sqrt{ab}.\)

Reply:Well, hopefully your teacher mentioned that this only holds true under certain conditions! In particular, it is true when \(a\) and \(b\) are both positive, which is where you usually encounter such equations. It is also true if exactly one of \(a\) or \(b\) is negative, but it is false when both are negative, as shown above.

Rebuttal:Why must we say that \(\sqrt{-x} = i \sqrt{x}\) for \(x<0?\) Couldn't we set \(\sqrt{-x} = -i\sqrt{x}\) as well? For example, \(\left(-2i\right)^2 = -4.\)

Reply:No! Remember that for \(x\ge 0,\) \(\sqrt{x}\) is defined as thenon-negativenumber whose square is \(x.\) Similarly, the convention is that, for \(x<0,\) \(\sqrt{x}\) is defined as \(\sqrt{-1}\) times thenon-negativenumber whose square is \(-x.\)

Note that \(\sqrt{\cdot}\) is a function, so it's important that it is defined to have a single output for each input. Many complicated issues (which are tackled in the subject of complex analysis) would arise if we did not use these conventions!

Want to make sure you've got this concept down? Try these problems:

**See Also**

**Cite as:**Do Square Roots Always Multiply?.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/does-sqrtatimes-sqrtb-sqrtab/