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root 2 corresponds to the Dedekind cut $(A,B)$ where:

$A = \{ x \in \Bbb Q|x\lt 0$ or $x^2\lt 2 \}$

$B = \{ x \in \Bbb Q|x\ge 0$ or $x^2\ge2 \}$

Check that this is a Dedekind cut of $\Bbb Q$ corresponding to a in $\Bbb R$ and $a \ge0, a^2 = 2$ so "$a = \sqrt 2$".

I don't quite understand how $A$ and $B$ have been defined, and what exactly I'm supposed to do, and why. I get that Dedekind cut shows there is a real in between two rationals, but I don't get how to prove it.

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  • $\begingroup$ Well, what does "is a Dedekind cut of $\mathbb{Q}$ corresponding to ..." mean in your definitions? The basic point is that (in the language of the real number system that we "don't have yet") $A=(-\infty,\sqrt{2}) \cap \mathbb{Q}$ and $B=(\sqrt{2},\infty) \cap \mathbb{Q}$. Do you see why that is? If not, draw a picture. $\endgroup$ – Ian Aug 24 '15 at 14:01
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    $\begingroup$ You need to change "or" for "and" in the definition of $B$ in order to make it a Dedekind cut. $\endgroup$ – Smurf Aug 24 '15 at 14:13
  • $\begingroup$ "I don't quite understand how A and B have been defined" . Do you mean you do not understand the set notation? A is the set of all rational numbers that are either negative or whose square is less than 2. B is the set of all positive rational numbers whose square is larger than 2. (As Jorge said, you need "and" not "or" in your definition of B.) You are asked to prove this is a "Dedekind Cut". Okay, what is the definition of Dedekind Cut? Show that these sets satisfy all the conditions of that definition. $\endgroup$ – user247327 Aug 24 '15 at 14:29
  • $\begingroup$ Ian - I'm really not sure. My lecture notes don't explain it very well, and I can't find any explanations online that help me understand how to answer the question. $\endgroup$ – Seb Aug 24 '15 at 14:32
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    $\begingroup$ should b e $B = x \in \Bbb Q|x\ge 0 \text{ and } x^2 \ge 2\}$. $\endgroup$ – GEdgar Aug 24 '15 at 14:45
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The way I learned "Dedekind cut" it applied to only one set, your "$A$", not two but that's not a big deal since, given $A, B$ is automatically defined as its complement.

A set of rational numbers, $A$, is a "Dedekind cut" if and only if

1) $A$ is not empty. Here, that's obvious- $A$ contains all negative numbers.

2) $A$ is not all rational numbers. With your $A$ and $B$, that's the same as saying $B$ is not empty. Here, this is true because $2$ is not negative and $2^2= 4$ is larger than $2$. $2$ is not in $A$ (and is in B$$).

3) If $a$ is in $A$ and $b< a$ then $b$ is in $A$. If $a$ is in $A$ either it is a negative number or its square is less than $2$. If $b$ is negative it is in $A$ so we need only look at the case that $0< b< a$. Then $b^2< a^2< 2$ so $b$ is in $A$.

4) $A$ does not have a largest member- that is usually the hardest part to show. I would use a "proof by contradiction. If there exist a largest number, $a$, in $A$, then $a^2< 2$. Let $e= 2- a^2$ which is positive. Look at $a+re$ for some positive rational number, $r$. $(a+ re)^2= a^2+ 2are+ r^2e^2$. That is clearly a rational number. It will be less than $2$ if $2are+ r^2e^2< e$ or $(e^2)r^2+ (2ae)r- 3< 0$. Show that there exist such a number, $r$.

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  • $\begingroup$ So Dedekind cuts in your sense always encode irrationals? (You could perfectly well define Dedekind cuts without your property 4, in which case some Dedekind cuts, e.g. $A=(-\infty,0] \cap \mathbb{Q}$, encode rationals.) $\endgroup$ – Ian Aug 24 '15 at 16:47
  • $\begingroup$ @Ian: You misunderstood. To encode 0, as in your example, use the rational numbers less_than 0. This set satisfies property 4. The point is that rational numbers can be defined your way, but irrational numbers cannot. The method user247327 described works for both rationals and irrationals. $\endgroup$ – user3697176 Aug 24 '15 at 17:10
  • $\begingroup$ @user3697176 Both ways work, but after seeing your point I agree that the alternative (where a rational cut misses one rational number) is cleaner. But irrational numbers can indeed be defined the other way, basically because $(-\infty,r] \cap \mathbb{Q} = (-\infty,r) \cap \mathbb{Q}$ if $r \not \in \mathbb{Q}$. $\endgroup$ – Ian Aug 24 '15 at 17:12
  • $\begingroup$ @Ian. You cannot define $\sqrt{2}$ in this fashion; your definition is circular! (Of course you could define the cut for $\sqrt{2}$ using $A = \{\ldots\le2\}$, but then you need to replace 4) by the requirement that the complement of A has no lower bound, which is distinctly awkward.) $\endgroup$ – user3697176 Aug 24 '15 at 18:25
  • $\begingroup$ @user3697176 As written it's circular, but the point is to encode the interval in some inequality that can be formulated only in the language of rationals (which in this case is easy). But you're right, there is some awkwardness then. $\endgroup$ – Ian Aug 25 '15 at 0:02

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