Proofs on Irrational/Rational Real Numbers

Question

I'm working through a book on Abstract Algebra and I'm at the section reading into Rings. As with all textbooks they pose questions without solutions and I have solutions to two problems I was hoping to have looked over.

Problem 1:

Consider a set of real numbers of the form

$x = a + bp$

Where $a,b \in \mathbb{Q}$ and $p \in \mathbb{R} - \mathbb{Q}$

Then $\forall a,b \neq 0 \quad x \in \mathbb{R} - \mathbb{Q}$

Employ Proof by contradiction,

Let $x$ be Rational, i.e. $x = \frac{c}{d}$ with $c,d \in \mathbb{Z}$ and

$d \neq 0$ and $\gcd(c,d) = 1$

Then

$a + bp = \frac{c}{d}$

As $a,b \in \mathbb{Q}$ Let $a = \frac{e}{f}$ , $b = \frac{g}{h}$ with $e,f,g,h \in \mathbb{Z}$ and $f,h \neq 0$ and $\gcd(e,f) = \gcd(g,h) = 1$

Then

$a + bp = \frac{c}{d} $

Becomes

$\frac{e}{f} + \frac{g}{h}p = \frac{c}{d}$

$\frac{g}{h}p = \frac{c}{d} - \frac{e}{f}$

$p = \frac{h}{g} \left(\frac{c}{d} - \frac{e}{f}\right) = \frac{h\left(fc - ed\right)}{gdf} \in \mathbb{Q}$

Which is false as $p \in \mathbb{R} - \mathbb{Q}$

Thus by Proof by Contradiciton $x = a + bp \in \mathbb{R} - \mathbb{Q}$ for all $a,b \in \mathbb{Q}$ , $a,b \neq 0$

Problem 2:

If $x \in \mathbb{R} - \mathbb{Q}$

Then for $k \in \mathbb{Q}, k \neq 0$ we have $kx \in \mathbb{R} - \mathbb{Q}$

Employ Proof by Contradiction.

Let $kx$ be Rational, i.e. $kx = \frac{c}{d}$ with $c,d \in \mathbb{Z}$ and

$d \neq 0$ and $\gcd(c,d) = 1$

As $k \in \mathbb{Q}$ Let $k = \frac{e}{f}$ with $e,f \in \mathbb{Z}$ and $\gcd(e,f) = 1$

Then

$ kx = \frac{e}{f}x = \frac{c}{d}$

Thus

$x = \frac{c}{d} \frac{f}{e} = \frac{cf}{de} \in \mathbb{Q}$

Which is false as $x \in \mathbb{R} - \mathbb{Q}$

Thus by Proof by Contradiction $kx \in \mathbb{R} - \mathbb{Q}$ for $k \in \mathbb{Q}$, $k \neq 0$ and $x \in \mathbb{R} - \mathbb{Q}$

Answer 1

Your proof by contradiction seems fine.

One might present it more compactly:

For example for problem $1$, If $a+bp=x$, then $bp=x-a \in \mathbb{Q}$ since rational number is closed under subtraction and $p=\frac{x-a}{b} \in \mathbb{Q}$ is rational since division of non-zero rational number gives us rational number. This contradicts to $p$ is irrational.

Answer 2

Your arguments are correct, but unnecessarily complicated. You already know that $\mathbb Q$ is a field. If $a+bp=c$, with $c\in\mathbb Q$, then $$p=\frac{c-a} {b}\in\mathbb Q,$$ a contradiction.

Similarly, if $kp=c$, then $p=c/k$.