For what values of \(x\) and \(y\) do these statements make sense?

What are the possible values for the expressions?

- \((xy)^\frac{1}{2}=x^\frac{1}{2}y^\frac{1}{2}\)

Is the left-hand side the same as the right-hand side?

\((xy)^\frac{1}{2}= \sqrt{xy} = \sqrt{x}\sqrt{y} = x^\frac{1}{2}y^\frac{1}{2}\)

It would seem so from the reasoning above, however what happens if \(x = -2\) and \(y = -8\)? \(\sqrt{xy}\) will give us an answer of \(4\) but \(\sqrt{x}\sqrt{y}\) will not be defined.

We can only get a real value from a square root when the number is not negative, so \(x≥0\) and \(y≥0\).

The overall value of the expression \((xy)^\frac{1}{2}\) (and therefore \(x^\frac{1}{2}y^\frac{1}{2}\)) will also be \(≥ 0\).

We have said that the value of the expressions \((xy)^\frac{1}{2}\) and \(x^\frac{1}{2}y^\frac{1}{2}\) will be \(≥ 0\). But will they always be positive? We may think not, as a square root has two solutions: a positive and a negative one.

However, the square root symbol \(\sqrt{}\), and writing something as raised to the power of \(\frac{1}{2}\), denotes the *principal* root, which is always the positive root. If we take a square root, as part of rearranging an equation then we may need to write \(\pm\sqrt{}\) or \(\pm(\quad)^\frac{1}{2}\) depending on the situation.

Can you think of an equation which involves taking a square root where you will always want the positive root only?

- \((xy)^{\frac{5}{3}}=x^{\frac{5}{3}}y^{\frac{5}{3}}\)

\((xy)^{\frac{5}{3}} = \left(\sqrt[3]{xy}\right)^5 = \left(\sqrt[3]{x}\right)^5 \left(\sqrt[3]{y}\right)^5 = x^{\frac{5}{3}}y^{\frac{5}{3}}\)

Is there anywhere this chain of reasoning breaks down? Does it make a difference if we write \(\sqrt[3]{x^5}\) instead of \(\left(\sqrt[3]{x}\right)^5\)?

We can cube root any real value, positive or negative, so \(x, y \in \mathbb{R}\). What values can \((xy)^{\frac{5}{3}}\) take?

- \((xy)^{\frac{2}{3}}= x^{\frac{2}{3}}y^{\frac{2}{3}}\)

\((xy)^{\frac{2}{3}}= \left(\sqrt[3]{xy}\right)^2 = \left(\sqrt[3]{x}\right)^2 \left(\sqrt[3]{y}\right)^2 = x^{\frac{2}{3}}y^{\frac{2}{3}}\)

We know we can cube root any number, so \(x, y \in \mathbb{R}\). The value of \((xy)^{\frac{2}{3}}\) however, is restricted. How is it different from the value of \((xy)^{\frac{5}{3}}\) and why is that?

- \((xy)^{-\frac{1}{2}}= x^{-\frac{1}{2}}y^{-\frac{1}{2}}\)

This is similar to the first question. It only involves square roots, so we know \(x\) and \(y\) cannot be negative. What happens to either side of the equation if \(x\) or \(y = 0\)?

- \((xy)^\frac{1}{2} = x^\frac{1}{3}y^\frac{2}{3}\)

On first glance, we might easily dismiss this as being incorrect. The question that was asked though, is ‘for what values of \(x\) and \(y\) do these statements make sense?’ Are there any values of \(x\) and \(y\) when this equation would be true?

If \(x=0\) or \(y = 0\), then the equation is satisfied, but to see if there are any other possibilities we should try to solve this equation.

\[(xy)^\frac{1}{2} = x^\frac{1}{3}y^\frac{2}{3}\]

For the left hand side to be defined, \(xy≥0\). In addition, \(x\) and \(y\) cannot be negative. Can you explain why?

We can write \[x^{\frac{1}{2}}y^{\frac{1}{2}}=x^{\frac{1}{3}}y^{\frac{2}{3}},\]

then divide by \(x^{\frac{1}{3}}y^{\frac{1}{2}}\), assuming \(x,y \neq 0\),to get

\[x^{\frac{1}{6}} = y^{\frac{1}{6}}.\]

Therefore

\[x = y\]

is a solution for \(x, y >0\), along with \(x=0\) or \(y=0\).

This is perhaps more solutions than we might have first thought. Can you see why this happens? Try some numbers if you are not sure.

- \((xy)^{-2}= x^{2}y^{2}\)

Once again, this is not true for all values, but may still be true for certain values of \(x\) and \(y\).

It can also be written

\[\dfrac{1}{(xy)^2} = x^{2}y^{2},\]

so we can see that if \(x\) or \(y\) was zero the left-hand side would not exist, therefore \(x,y \neq0\).

We can multiply by \((xy)^{2}\)

\[1= x^4y^4,\]

then take the fourth root. This gives us a positive and negative value so we get

\[1 = xy \text{ or} -1 = xy,\]

giving us the solutions

\[x = \frac{1}{y} \text{ and } x = -\frac{1}{y}.\]

- \(\left(\frac{x}{y}\right)^{-\frac{1}{3}} = x^{-\frac{1}{3}}y^{\frac{1}{3}}\)

From the \(\left(\frac{x}{y}\right)\) on the left hand side we can see that \(y \neq0\), even though it could be on the right hand side.

On the right hand side we have \(x^{-\frac{1}{3}}\) which is \(\frac{1}{\sqrt[3]{x}}\) and so \(x \neq0\).

Otherwise we can cube root any value so \(x,y \in \mathbb{R}\) except for \(x, y \neq0\) and the value of the expressions can also take any value except \(0\).

- \(\left(\frac{x}{y}-\frac{1}{y}\right)^{\frac{1}{2}}= y^{-\frac{1}{2}}(x-1)^\frac{1}{2}\)

On the left hand side we require \(y \neq0\) and \(\frac{x}{y}-\frac{1}{y} ≥ 0\), which implies that \(x ≥ 1\)or \(x≤1\) depending on whether \(y\) is positive or negative.

On the right hand side we have \(y^{-\frac{1}{2}}\) which requires \(y >0\), and \((x-1)^\frac{1}{2}\) which also requires \(x ≥ 1\).

Putting these together gives us \(y >0\) and \(x≥1\) and the value of either side of the equation must be \(≥0\).

Remember that although square rooting something gives us a positive and negative solution, the notation of ‘to the power of \(\frac{1}{2}\)’ denotes the *principal root* which is positive.