Proving the sum of a rational and irrational number is irrational

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Homework Statement



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.

Homework Equations



I am working from the following books:

Ethan D. Bloch: The Real Numbers and Real Analysis

and

Derek Goldrei: Classic Set Theory

Both 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:

?temp_hash=cee404bddcc16a3ee55c431470ff5b05.png


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 are relevant ... and read as follows:

?temp_hash=cee404bddcc16a3ee55c431470ff5b05.png

The Attempt at a Solution



I have been unable to make a meaningful start on this problem ...Peter*** EDIT ***

Reflecting on this problem ... it has become apparent to me that Bloch's definition of the addition of real numbers (in terms of Dedekind Cuts is relevant ... so I am providing the relevant definition ... as follows:

?temp_hash=5e153c4d3963047e0f6ee5c2aa26e760.png
The note preceding the above definition mentions Lemma 1.6.8 which reads as follows:

?temp_hash=5e153c4d3963047e0f6ee5c2aa26e760.png
 
Attachments
  • Bloch - Irrational Cut ....png
    Bloch - Irrational Cut ....png
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  • Bloch - Dedekind Cut ... plus Lemma ....png
    Bloch - Dedekind Cut ... plus Lemma ....png
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  • Bloch - Defn 1.7.3 ... ....png
    Bloch - Defn 1.7.3 ... ....png
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  • Bloch - Lemma 1.6.8 ... ....png
    Bloch - Lemma 1.6.8 ... ....png
    18.7 KB · Views: 741
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After reflecting on this problem ... I now think that the answer to the exercise above may be disappointingly trivial ...

Consider the following:

Let ##a \in## ##\mathbb{Q}## and let ##b \in \mathbb{R}## \ ##\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
 
Last edited:
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