Proof Question using Proof By Contradiction, irrationality of $a + \sqrt[b]{5}$ I answered this question but am not quite sure if i did what was correct. If anything is wrong please point it out, thanks.
Question:Prove that any number of the form $a + \sqrt[b]5$ is irrational, where $a,b$ are integers and $b \ne 0$
My Answer:
Proof by Contradiction:
Assume $a + \sqrt[b]5$ is rational:
Assume $p$/$q$ is in lowest terms where GCD($p$,$q$) = $1$ where $p$, $q$ are integers and $q \ne $0
$$\begin{align}
a + \sqrt[b]{5} &= {p \over q} & \\\
    \sqrt[b]{5} &= \frac pq - a \\
              5 &= \left(\frac pq - a\right)^b \\
              5 &= \frac{(p - qa)^b}{q^b}   & \text{(Multiplying q to be the denominator of the whole equation)} \\
           5q^b &= (p - qa)^b         &\text{(Isolating q and this shows }(p-qa)^b\text{ is a multiple of 5)}
\end{align}$$
Assuming $5c = p - qa$:
$$5q^b = 5^bc^b$$
How do i show q^b is a multiple of 5? Because if the question had a integer other than b then I could prove for odd and even numbers using (5p+1)^b and (5p+2)^b where b is an integer such as 1, 2, ... b then I can expand it and show there is a common divisor of 5 but I cannot do that in this case because I dont know what b is. The assumption to show this equation is rational if and only if b$\sqrt5$ is rational will work but I already knew that my problem is actually showing b$\sqrt5$ is rational.
Then,
$$5q^b = 5\left(5^{b-1}c^b\right) ~~~~\text{(which implies q is a multiple of 5)}$$
Since $q$ is a multiple of $5$:
$$\begin{align}
q &= 5x &\text 
p - 5xa &= 5c \\
p &= 5(xa + c)
\end{align}$$
This implies that both $p$ and $q$ are divisible by $5$. Which contradicts GCD($p$,$q$) = $1$ since $5$ is a greater common divisor than $1$. Thus the assumption of $a + \sqrt[b]5$ is rational is false.
 A: Your last paragraph is wrong:

This implies that both p and q are multiples of 5 so there is a contradiction because the equation is irrational so there cannot be any common factors for it to be irrational.

It's like you've seen a proof of contradiction before, borrowed some words from it, and had no idea what the actual logic was.  Memorizing won't get you far in mathematics.  It looks like random incoherent rambling.
To correct your proof, first, your statement

For some integer p,q where q is not zero

is incomplete.  You need to specify that $\frac pq$ is in lowest common terms, in other words that $\operatorname{GCD}(p,q) = 1$.
Second, your conclusion should be:


*

*Therefore $p$ and $q$ are both divisible by $5$

*This contradicts the assumption that $\operatorname{GCD}(p,q) = 1$

*Therefore the assumption that $a + \sqrt[b]5$ is rational must be false

A: *

*$b \in \mathbb{Z^+}\\ \sqrt[b]5=\dfrac{p}{q}, \operatorname{gcd}(p,q)=1\\\implies 5=\dfrac{p^b}{q^b} \\ \implies 5q^b=p^b\\ \implies p \mid 5 \\\implies p=5 \\\implies q^b=5^{b-1}$


$b-1>0 \implies 5\mid q \implies q=5k \implies 5^b \cdot k^b=5^{b-1} \implies 5k^b=1$ 
$\therefore \not \exists p,q\mid \operatorname{gcd}(p,q)=1 \land \sqrt[b]5=\dfrac{p}{q}$ ,$\forall b \in \mathbb{N}$ $\implies$ $\sqrt[b]5 \in \mathbb {R} \setminus \mathbb{Q} \implies a+\sqrt[b]5 \in \mathbb {R} \setminus \mathbb{Q}$ , $\forall a \in \mathbb {Z}$
$b \in \mathbb{Z^-} \implies \exists t\in \mathbb {N} \mid b=-t \implies \sqrt[b]5=\sqrt[-t]5=\dfrac{1}{5^{\dfrac{1}{t}}}$
$\exists r,s \mid \operatorname{gcd}(r,s)=1, t \in\mathbb {N} \land \sqrt[-t]5=\dfrac{r}{s} \implies \sqrt[t]5 = \dfrac{s}{r}$ !!
