Just a disclaimer before I proceed with my question and the proof I wrote up: I know that this question has been asked before, for example here, but I am more interested in being critiqued on how I wrote the proof and its completeness.
In addition, I am having trouble seeing how it proves what I first set out to prove.
Theorem: if $r$ is irrational, then ${ r }^{ \frac { 1 }{ 5 } }$ is irrational
Proof: We prove the contrapositive: if ${ r }^{ \frac { 1 }{ 5 } }$ is rational, then $r$ is rational
1) Assume that ${ r }^{ \frac { 1 }{ 5 } }$ is rational, then there exists $a,b\in\mathbb{Z}$ such that:
${ r }^{ \frac { 1 }{ 5 } }=\frac { a }{ b }$ where $a,b$ are coprime and $b\neq 0$
2) Therefore, ${ r }=\frac { a^{ 5 } }{ b^{ 5 } } $
3) If $a,b\in\mathbb{Z}$, then $a^5,b^5\in\mathbb{Z}$ as well.
4) Therefore $r\in\mathbb{Q}$
Q.E.D.
When it comes to proving that the square root of $2$ is irrational, I can quickly see and understand why $\sqrt { 2 } $ is indeed irrational. However, with this proof, I don't understand how this actually proves the theorem I set out to prove. Does proving it by using the contrapositive just make it that much simpler? Or did I miss some vital steps in my proof?
Please feel free to give me constructive criticism about my proof writing techniques as well.
Best Answer
The contrapositive of a statement is logically equivalent, so if it is convenient for us we may prove the contrapositive instead.
In this particular problem, the statement involves assuming $r$ is irrational. The problem is that we can't really say much about what $r$ looks like.
However, when we look at the contrapositive, we are assuming a number is rational and so we have an idea of what it looks like. So yes using the contrapositive really does make this problem much easier.
On the subject of your exposition, it looks great! The comment about noting that $b^5 \neq 0$ is really the only thing to improve on.