Discussion
Consider the following functions for non-zero positive integers, \(p\) and \(q\).
\[ \begin{aligned} f(p, q) &= \underbrace{p \times p \times p \times \cdots \times p}_{q\ \text{terms}} \\ &= p^{q}; \quad f(p, 1) = p \end{aligned} \]
\[ g(p, q) = p^{p^{p^{p^{\cdot^{\cdot^{\cdot}}}}}} \]
(up to \(q\) terms); \(g(p, 1) = p\)
Which one of the following options is correct based on the above?
Answer: A
The correct option is \(f(2,2) = g(2,2)\).
\(f(p, q)\) multiplies \(q\) copies of \(p\), so \(f(p, q) = p^{q}\). \(g(p, q)\) stacks \(q\) copies of \(p\) as a power tower. With two terms both give the same thing: \(f(2,2) = 2^{2} = 4\) and \(g(2,2) = 2^{2} = 4\).
The other options are false:
- \(g(2,2) = 4\), so \(f(g(2,2), 2) = f(4, 2) = 4^{2} = 16\) and \(f(2, g(2,2)) = f(2, 4) = 2^{4} = 16\). They are equal, so 'less than' fails.
- \(g(2,1) = 2\) and \(f(2,1) = 2\), so 'not equal' fails.
- \(f(3,2) = 3^{2} = 9\) while \(g(3,2) = 3^{3} = 27\), so \(f(3,2)\) is smaller, not greater.
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