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

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SUMMARY

The sum of a rational number and an irrational number is always irrational, as demonstrated through a proof involving Dedekind Cuts. The discussion references Ethan D. Bloch's "The Real Numbers and Real Analysis" and Derek Goldrei's "Classic Set Theory," both of which utilize Dedekind Cuts and Cauchy Sequences in their definitions. The proof shows that if a rational number \( a \) is added to an irrational number \( b \), assuming the result \( r \) is rational leads to a contradiction, confirming that \( r \) must be irrational.

PREREQUISITES
  • Understanding of Dedekind Cuts as defined in real analysis.
  • Familiarity with the properties of rational and irrational numbers.
  • Basic knowledge of proofs by contradiction.
  • Exposure to Cauchy Sequences in the context of real numbers.
NEXT STEPS
  • Study the concept of Dedekind Cuts in detail, particularly in Bloch's "The Real Numbers and Real Analysis."
  • Explore proofs involving irrational numbers and their properties.
  • Learn about Cauchy Sequences and their role in real analysis.
  • Investigate other proofs of the irrationality of sums involving rational and irrational numbers.
USEFUL FOR

Mathematics students, educators, and anyone interested in understanding the properties of real numbers, particularly in the context of rational and irrational number interactions.

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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
    20.7 KB · Views: 790
  • Bloch - Defn 1.7.3 ... ....png
    Bloch - Defn 1.7.3 ... ....png
    14.7 KB · Views: 606
  • Bloch - Lemma 1.6.8 ... ....png
    Bloch - Lemma 1.6.8 ... ....png
    18.7 KB · Views: 709
Last edited:
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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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Math Amateur said:
Is that correct?
Yes.
 
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