This problem is from Epp, S. Discrete Mathematics with Applications (2010, 4th ed.). More specifically, it's problem 13 from p. 212.
"If an integer greater than 1 is a perfect square, then its cube root is irrational." No solution for this problem is given in the book, but there's also an additional hint: "Can you think of any integers x and y that are greater than 1 and have the property that x2 = y3?" However, the hint doesn't seem very helpful to me (it might be for everyone else, though).
The statement in the problem expressed in more definite terms appears to be: [MATH]\forall{n}\in\mathbb{Z}[(n > 1 \land \sqrt{n} \in \mathbb{Z}) \rightarrow \sqrt[3]{n} \notin \mathbb{Q}] [/MATH].
I can get a proof started, but only started. I'm fairly certain the statement is true, so I'm approaching it from the angle of proving it correct.
This is what I've come up with:
To be proven: If an integer greater than 1 is a perfect square, then its cube root is irrational ie. [MATH]\forall{n}\in\mathbb{Z}[(n > 1 \land \sqrt{n} \in \mathbb{Z}) \rightarrow \sqrt[3]{n} \notin \mathbb{Q}] [/MATH].
Proof:
Suppose [MATH]n\in\mathbb{Z}[/MATH] such that [MATH]n > 1 \land \sqrt{n} \in \mathbb{Z}[/MATH]. Then [MATH]\exists{s} \in \mathbb{R}(s^3 = n)[/MATH].
So, [MATH]\sqrt{n} = \sqrt{s^3} = (s^3)^{1/2} = s^{3/2}[/MATH].
What's left to be shown is that [MATH]s^{3/2} \notin \mathbb{Q}[/MATH], but I don't see how to proceed.
This should be one of the easy problems in the book, and I've managed to crank out correct answers to almost all problems in this section. Except this one ...
"If an integer greater than 1 is a perfect square, then its cube root is irrational." No solution for this problem is given in the book, but there's also an additional hint: "Can you think of any integers x and y that are greater than 1 and have the property that x2 = y3?" However, the hint doesn't seem very helpful to me (it might be for everyone else, though).
The statement in the problem expressed in more definite terms appears to be: [MATH]\forall{n}\in\mathbb{Z}[(n > 1 \land \sqrt{n} \in \mathbb{Z}) \rightarrow \sqrt[3]{n} \notin \mathbb{Q}] [/MATH].
I can get a proof started, but only started. I'm fairly certain the statement is true, so I'm approaching it from the angle of proving it correct.
This is what I've come up with:
To be proven: If an integer greater than 1 is a perfect square, then its cube root is irrational ie. [MATH]\forall{n}\in\mathbb{Z}[(n > 1 \land \sqrt{n} \in \mathbb{Z}) \rightarrow \sqrt[3]{n} \notin \mathbb{Q}] [/MATH].
Proof:
Suppose [MATH]n\in\mathbb{Z}[/MATH] such that [MATH]n > 1 \land \sqrt{n} \in \mathbb{Z}[/MATH]. Then [MATH]\exists{s} \in \mathbb{R}(s^3 = n)[/MATH].
So, [MATH]\sqrt{n} = \sqrt{s^3} = (s^3)^{1/2} = s^{3/2}[/MATH].
What's left to be shown is that [MATH]s^{3/2} \notin \mathbb{Q}[/MATH], but I don't see how to proceed.
This should be one of the easy problems in the book, and I've managed to crank out correct answers to almost all problems in this section. Except this one ...
