Sum of a rational number and an irrational number ....

[Math Amateur]

I am trying without success to provide a rigorous proof for the following exercise:

Show that the sum of a rational number and an irrational number is irrational.Can someone please help me with a rigorous solution ...I am working from the following books:

Ethan D. Bloch: The Real Numbers and Real Analysis

and

Derek Goldrei: Classic Set TheoryBoth use a Dedekind Cut approach to the construction of the real numbers (but Goldrei also uses Cauchy Sequences ... )
I am taking the definition of an irrational number as equivalent to an irrational cut as defined by Bloch as follows:https://www.physicsforums.com/attachments/7014To assist those members reading this post I am providing Bloch's definition of a Dedekind Cut plus a Lemma indicating the that there are at least as many of them as there are rational numbers ...https://www.physicsforums.com/attachments/7015Help will be much appreciated ...

Peter

 

[Math Amateur]

I am trying without success to provide a rigorous proof for the following exercise:

Show that the sum of a rational number and an irrational number is irrational.Can someone please help me with a rigorous solution ...I am working from the following books:

Ethan D. Bloch: The Real Numbers and Real Analysis

and

Derek Goldrei: Classic Set TheoryBoth use a Dedekind Cut approach to the construction of the real numbers (but Goldrei also uses Cauchy Sequences ... )
I am taking the definition of an irrational number as equivalent to an irrational cut as defined by Bloch as follows:To assist those members reading this post I am providing Bloch's definition of a Dedekind Cut plus a Lemma indicating the that there are at least as many of them as there are rational numbers ...Help will be much appreciated ...

Peter

After reflecting on this problem ... I now think that the answer to the exercise above may be disappointingly trivial ...

Consider the following:

Let \(\displaystyle a \in \mathbb{Q}\) and let \(\displaystyle b \in \mathbb{R}\) \ \(\displaystyle \mathbb{Q}\)

Then suppose a + b = r

Now ... assume r is rational

Then b = r - a ...

But since r and a are rational ... we have r - a is rational ..

Then ... we have that an irrational number b is equal to a rational number ...

Contradiction!

So ... r is irrational ..
Is that correct?

Peter

 

[HallsofIvy]

No need for "Dedekind cuts". All that is needed is that the rational numbers are closed under subtraction: if $$x= \frac{a}{b}$$ and $$y= \frac{c}{d}$$ are rational numbers, then $$x- y= \frac{a}{b}- \frac{c}{d}= \frac{ad- bc}{bd}$$ is rational number.

Now, in contradiction, suppose a is rational and b is irrational and that there sum, c, is rational. From a+ b= c, we have a= c- b, contradicting the fact that the difference of two rational numbers is rational.

 

[Math Amateur]

No need for "Dedekind cuts". All that is needed is that the rational numbers are closed under subtraction: if $$x= \frac{a}{b}$$ and $$y= \frac{c}{d}$$ are rational numbers, then $$x- y= \frac{a}{b}- \frac{c}{d}= \frac{ad- bc}{bd}$$ is rational number.

Now, in contradiction, suppose a is rational and b is irrational and that there sum, c, is rational. From a+ b= c, we have a= c- b, contradicting the fact that the difference of two rational numbers is rational.

Thanks HallsofIvy ... appreciate the help ...

Peter