Question

An operation \( \$ \) is defined as follows. For any two positive integers \(x\) and \(y\),
\[ x \$ y = \sqrt{ \sqrt{\dfrac{x}{y}} + \sqrt{\dfrac{y}{x}} } \]
Which of the following is an integer?

• A. \(4 \$ 9\)
• B. \(4 \$ 16\)
• C. \(4 \$ 4\)
• D. None of the above

Hint:

Plug the given pairs into the formula one at a time and simplify the nested square roots fully before deciding if the result is a whole number.

Answer 1

The Correct Option is D

Step 1: Understand the operation.
The symbol \(\$\) is not addition or multiplication, it is a rule made up just for this question.
For two positive integers \(x\) and \(y\), the rule says: take the ratio \(x/y\), take its square root, take the ratio \(y/x\), take its square root, add the two square roots, then take one more square root of that sum.
We just need to plug numbers in carefully and check if the final result is a whole number.

Step 2: Work out \(4 \$ 9\).
Here \(x = 4\) and \(y = 9\).
\[ \sqrt{\dfrac{4}{9}} = \dfrac{2}{3}, \qquad \sqrt{\dfrac{9}{4}} = \dfrac{3}{2} \]
Add these two fractions using a common denominator of 6:
\[ \dfrac{2}{3} + \dfrac{3}{2} = \dfrac{4}{6} + \dfrac{9}{6} = \dfrac{13}{6} \]
So \(4 \$ 9 = \sqrt{\dfrac{13}{6}}\). Since 13 and 6 share no common square factor, this square root is irrational, not a whole number.

Step 3: Work out \(4 \$ 16\).
Here \(x = 4\) and \(y = 16\).
\[ \sqrt{\dfrac{4}{16}} = \sqrt{\dfrac{1}{4}} = \dfrac{1}{2}, \qquad \sqrt{\dfrac{16}{4}} = \sqrt{4} = 2 \]
Add these:
\[ \dfrac{1}{2} + 2 = \dfrac{5}{2} \]
So \(4 \$ 16 = \sqrt{\dfrac{5}{2}}\), and this is also irrational, not an integer.

Step 4: Work out \(4 \$ 4\).
Here \(x = 4\) and \(y = 4\), so \(x/y = 1\) and \(y/x = 1\).
\[ \sqrt{1} + \sqrt{1} = 1 + 1 = 2 \]
So \(4 \$ 4 = \sqrt{2}\), which is a well known irrational number, not an integer either.

Final Answer:
None of \(4 \$ 9\), \(4 \$ 16\) or \(4 \$ 4\) comes out to a whole number, so the correct choice is none of the above. \[ \boxed{\text{None of the above}} \]