I stumbled upon this problem while doing my math homework and this is the only place I know where someone can solve it. Take a look:

$\sqrt{\frac{4}{9}} = -\frac{2}{3}$


$(\sqrt{\frac{4}{9}})^2 = (-\frac{2}{3})^2$$\frac{4}{9} = \frac{(-2^2)}{(3^2)}$$\frac{4}{9} = \frac{4}{9}$???

Please help me with this, I've gone through many sources and they all just say that this doesn't check out but don't tell me why.

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    $\begingroup$ By convention, the square root is strictly non-negative, therefore rendering the initial equation false. $\endgroup$
    – abiessu
    Commented May 6 at 4:55
  • $\begingroup$ All positive number have two square roots: a positive square root and a negative square root; and they have equal absolute value. By convention/definition the one refered to be "the" square root function is the positive one. So $x^2 = w$ is not enough to say $x =\sqrt w$. We need $x^2 =w$ AND we also need $x \ge 0$ to say $x=\sqrt w$. We can say that $x^2 =w\implies |x| =\sqrt w$ and we can say that $x^2 =w \implies$ EITHER $x=\sqrt{w}$ OR $x =-\sqrt w$. $\endgroup$
    – fleablood
    Commented May 6 at 14:48
  • $\begingroup$ Also.... CONCLUSION $\implies$ Consequence $\implies$ SOMETHING TRUE, is NOT a valid prove. Consider I am the president of the united states $\implies$ The country I live in is equal to the country the president of the united states lives in $\implies$ the United States = the United States. THerefore I am the president of the United States. QED. $\endgroup$
    – fleablood
    Commented May 6 at 14:53

2 Answers 2



This statement is wrong. By definition, the square root function returns the positive square root of any real number.

Also if we have $x^2=y^2$ for some real numbers $x$ and $y$, then it is not necessary that $x$ should be equal to $y$.

In fact, if $x^2=y^2$ then we must have $|x|=|y|$ or $x=\pm y$


This result is valid only if there are no restrictions on $x$ and $y$. If we put $x$ as $\sqrt{k}$ where $k\ge0$ then we can't have $\sqrt{k}$ as $\pm y$, we'll take only that value which is positive i.e. $|y|$.

  • $\begingroup$ @peterwhy I'll do something about it $\endgroup$ Commented May 6 at 5:29
  • 1
    $\begingroup$ The square root function returns the nonnegative square root of any nonnegative real number. $\endgroup$ Commented May 6 at 10:03

Previous answers and comments are correct, but to elaborate/rephrase a little further:
The reason why the square root of 4/9 (or any other positive number) is positive and not negative is because we chose to define it that way. There's absolutely nothing wrong with your intuition that $(-\frac{2}{3})^2 = (\frac{2}{3})^2 = \frac{4}{9}$. It's just that, at some point, someone decided that the square root would not refer to all numbers that can be squared to return a given number, but rather just the positive one.
This is admittedly confusing when trying to form an intuition about square roots and inverse functions, but the advantage of it is that it allows the square root to be a function in the sense that every input (such as $\frac{4}{9}$) has one and only one output ($\frac{2}{3}$). This is sometimes helpful, even though it feels a little incomplete.
Sometimes you need to get around this limitation. The graph of the square root function ($y = \sqrt{x}$) looks like this:

half parabola

Only the top half of the parabola is there. If, as is often the case, you actually need to capture both halves of the parabola, you can simply express it as $y^2 = x$.

full parabola


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