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I was recently studying the construction of real numbers. I read that the sum of 2 reals using the left Dedekind sets was the set of sum of all the rational numbers contained within those two sets.

What I am not able to understand is how the sum of the rational numbers of those two sets contains all the rational numbers less than the real number associated in the sum. In other words, how can I be sure that there is no rational number greater than the greatest sum of the rationals in the two sets but at the same time lesser than the resulting real number ? [ I am aware that I should not be thinking about the greatest rational in the set, but I am inclined to think about the greatest number when taking the sum as it should represent the largest of the rationals in the resulting set ]

Any help would be greatly appreciated. Thanks in advance.

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  • $\begingroup$ Please capitalize the people's names. $\endgroup$ – Yves Daoust Apr 23 at 6:58
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    $\begingroup$ Done. Thanks for the advice!! $\endgroup$ – Guru Aravind Apr 23 at 7:09
  • $\begingroup$ 'how can I be sure that there is no rational number greater than the greatest sum of the rationals in the two sets but at the same time lesser than the resulting real number?' You can be sure, because it follows from the definition: C=A+B is a set of all sums of elements from A and B. For any $c$ either $\exists(a\in A, b\in B)\ c=a+b$ and then $c\in C$, or there is no such $(a,b)\in A\times B$ and then $c\notin C$. $\endgroup$ – CiaPan Apr 23 at 8:54
  • $\begingroup$ Possibly you had a feeling, that the definition operating on rational numbers only can miss some irrational numbers in A and B, whose sum might be greater than the sum of rationals, but less than the value of considered 'sum of cuts'. But that impression would be false – remember, A & B do not have irrational elements. Irrationals are not defined yet! They will appear soon, as $\mathbb R\setminus\mathbb Q$. But you will need $\mathbb R$ first. As far as you are now, A & B contain rational numbers only. And you consider only sums of those rationals in the definition of 'the sum'. $\endgroup$ – CiaPan Apr 23 at 8:59
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When you say "lesser than the resulting real number", you think as if that real number (i.e. the sum of the two given reals) was already known/obtained by other means.

This is not the case, it is only known from the definition, which you don't have to prove.

To illustrate, $\sqrt2+\sqrt3$ is the set of rational numbers $p+q$ such that $p^2\le2$ and $q^2\le3$, and nothing else. And obviously, no larger rational belongs to that set by definition.


As long as you have not defined the Dedekind cuts, there are no reals, and as long as you have not defined the addition of Dedekind cuts, the sum of two reals doesn't exist.


A philosophical remark:

You might be tempted to say, yeah, but we know that the reals are there independently of us, and $\sqrt2+\sqrt3$ exists and could be such that Dedekind addition fails.

That's right, there could exist some real number $\sqrt2+\sqrt3$ in some real number theory, such that there is a rational that invalidates the inequality. But we need to use definitions to specify which number theory we want to deal with, namely one where Dedekind addition is foolproof (if that exists).

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  • $\begingroup$ But we do know that the sum of 'real 2 and real 3' is to be the same as 'rational 2 and rational 3' , right? In which case we cannot assure that the sum of real 2 and real 3 to be real 5... it could be a number slightly less than real 5, am i correct? $\endgroup$ – Guru Aravind Apr 23 at 7:04
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    $\begingroup$ The sum of reals $2+5$ is defined to be the set of rationals $p+q$ such that $p\le2$ and $q\le5$. Full stop. There is no "other" sum of reals $2+5$. [From the definition, you can write a theorem that states that the sum of two reals that coincide with two rationals does coincide with the sum of these rationals, but this is a theorem, not a property you can take for granted. Stop believing that real numbers exist and have properties before you define them.] $\endgroup$ – Yves Daoust Apr 23 at 7:08
  • $\begingroup$ Very nice answer. Since people are so much used to using real numbers without defining them these things look a bit strange / difficult. $\endgroup$ – Paramanand Singh May 16 at 6:49
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The left set of a Dedekind cut is a subset $A$ of $\Bbb Q$ such that

  • $A\ne\emptyset$
  • $A\ne\Bbb Q$
  • $A$ has no maximal element
  • If $a\in A$ and $q\in\Bbb Q$ with $q<a$, then $q\in A$.

If $A_1$ and $A_2$ have these properties, then certainly so does $A_1+A_2$. Moreover, if the $A_i$ are defined in terms of rational numbers (i.e., $A_i=\{\,x\in\Bbb Q\mid x<a_i\,\}$ for some $a_i\in\Bbb Q$), then it turns out that $A_1+A_2=\{\,x\in\Bbb Q\mid x<a_1+a_2\,\}$. What else could one wish for?

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