My Math Forum Dedekind cut of √2

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 August 27th, 2015, 05:53 AM #1 Newbie   Joined: Aug 2015 From: London Posts: 2 Thanks: 0 Dedekind cut of √2 √2 corresponds to the Dedekind cut (A,B) where: A={x∈Q|x<0 or x^2<2} B={x∈Q|x≥0 or x^2≥2} Check that this is a Dedekind cut of Q corresponding to a in R and a≥0,a^2=2 so "a=√2". Is the question asking that in between any two points from A and B, there exists √2? I know that a dedekind cut is two sets of real numbers which contain a rational number in between. Still not sure what the point of a dedekind cut is... And all the explanations I've seen only confuse me more.
 August 27th, 2015, 06:29 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 Math Focus: Mainly analysis and algebra A Dedekind cut is two sets of rational numbers with a real number "in between" (it may be a member of the set $B$ if the real number is also rational. There are three conditions for a partition of the rationals into two sets to be a Dedekind Cut:Both $A$ and $B$ are non-empty; Every member of $A$ is less than every member of $B$; and The set $A$ contains no greatest element. In this case that means that given $a \in A$ there exists $a' \in \mathbb Q$ such that $a \lt a'$ and $a'^2 \lt 2$. If these conditions are satisfied, the partition is a Dedekind Cut. And we can use Dedekind cuts to define the reals. If $B$ contains a least element, the cut corresponds to that rational number. i.e. the rational number is a member of the real numbers. If $B$ does not contain a least element, then the cut defines a unique irrational number that "separates" the two sets. In this case the irrational number is $b' \not \in B$ such that $b'^2 = 2$ or $b' = \sqrt2$. Thus, a proof that $b' = \sqrt2$ is irrational might (in theory) consist of a proof that $b' \not \in B$. In practice we have to prove that $b'$ is irrational in order to show that it isn't in $B$. Your question asks you to prove that the three conditions listed above are satisfied, the most difficult being the third. Thanks from topsquark Last edited by v8archie; August 27th, 2015 at 06:33 AM.
 August 27th, 2015, 08:49 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 Math Focus: Mainly analysis and algebra By the way, you have "or" instead of "and" in your definition of $B$.
August 27th, 2015, 09:22 AM   #4
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Quote:
 Originally Posted by ActSci √2 corresponds to the Dedekind cut (A,B) where: A={x∈Q|x<0 or x^2<2} B={x∈Q|x≥0 or x^2≥2}
This not correct. It should be "x≥0 and x^2≥2".

Quote:
 heck that this is a Dedekind cut of Q corresponding to a in R and a≥0,a^2=2 so "a=√2". Is the question asking that in between any two points from A and B, there exists √2? I know that a dedekind cut is two sets of real numbers which contain a rational number in between.
No, this is not correct. It would be correct to say that a real number lies between the two sets but the whole point of the Dedekind cut is to define the real number system so that would be "circular".

Quote:
 Still not sure what the point of a dedekind cut is... And all the explanations I've seen only confuse me more.
The point is to be able to define the set of all real numbers in terms of sets of rational numbers- in particular to define the irrational numbers.

 August 28th, 2015, 02:29 AM #5 Newbie   Joined: Aug 2015 From: London Posts: 2 Thanks: 0 How can even define the irrational numbers? In terms of decimals they go on forever... Say you have something that's not root 2, like 4.14508732405032903502..... (random decimals) that goes on forever. Using dedekind cuts wouldn't you need an infinite number of conditions to define it? (One for every decimal). And actually constructing it on the number line would take forever.... What's wrong with leaving it as a decimal? Much more useful I think. Dedekind cuts are kind of pointless imo. Why is the example of root 2 only used? Because trying to define more complicated irrationals using dedekind cuts involves infinite amount of work.
August 28th, 2015, 04:23 AM   #6
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Quote:
 Originally Posted by ActSci How can even define the irrational numbers?
There are many ways. $\pi$ can be defined as the circumference of a circle of unit diameter. $\sqrt2$ is the diagonal of the unit square. There are othet systematic ways of doing it too.
Quote:
 Originally Posted by ActSci Using dedekind cuts wouldn't you need an infinite number of conditions to define it?
Why would you need an infinite number of conditions? Do you need an infinite number of conditions to define $1 \over 3$ or $1 \over 7$? Both have only infinite decimal expansions. The decimal expansion of a number is just a representation of that number, not the number itself.
Quote:
 Originally Posted by ActSci What's wrong with leaving it as a decimal? Much more useful I think.
How can it be more useful to define a number in a way that takes an infinite amount of time? In fact, if you can't write the definition down, is the number really defined?
Quote:
 Originally Posted by ActSci Why is the example of root 2 only used? Because trying to define more complicated irrationals using dedekind cuts involves infinite amount of work.
The example of $\sqrt2$ is used because it's a natural one to start with. Just as we rarely show a proof that other rationals than $\sqrt2$ are in fact rational.

It clearly doesn't take an infinite amount of work, I've just demonstrated the definition of $\sqrt2$ in finite time. The power of this system is in the set definition. We get all algebraic numbers using polynomial inequalities as above, but we can use any conditions we like.

In one sense, Dedekind Cuts are not hugely useful. They are rarely, if ever, used outside of this context. On the other hand, if you like your mathematics to be rigorous, you need a method of defining and ordering every real number, just as every rational number is (more simply) defined as the ratio of two integers and ordered by ${a \over b} \lt {c \over d}$ if $ad \lt bc$. Dedekind Cuts can be ordered by the contents of the sets $A$ in the definition. $a \lt a'$ if $A \subset A'$.

There's a question for someone more knowledgeable than myself: do we know that there is a set theoretic definition of every transcendental number? That is, a definition that allows us to form a Dedekind Cut for the number?

August 28th, 2015, 04:36 AM   #7
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Quote:
 Originally Posted by ActSci How can even define the irrational numbers?
It's not guaranteed that you can define every real number. In fact it can be proven that there are infinitely many numbers which cannot be defined with finitely many symbols (drawn from a finite alphabet). But why is this a problem?

August 28th, 2015, 08:36 AM   #8
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 Originally Posted by v8archie There's a question for someone more knowledgeable than myself: do we know that there is a set theoretic definition of every transcendental number? That is, a definition that allows us to form a Dedekind Cut for the number?
If you insist that the definition is finite (as I would), then not every transcendental number has a definition.

 August 28th, 2015, 10:20 AM #9 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,537 Thanks: 108 A)-D) from wicki A) A transcendental number is a real or complex number that is not algebraic. B) An algebraic number is a real or complex number that satisfies a rational (same as integral coefficients) polynomial. (x^2=2) C) A real number satisfies postulates for a complete ordered field. D) Real numbers are cuts of the rational numbers. C) <-> D). It follows from A), B) and D) that all real transcendental numbers are cuts of the rational numbers. Note Hilbert showed pi is transcendental. Is cosx transcendental? - Good luck!
August 28th, 2015, 11:38 AM   #10
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 Originally Posted by zylo Is cosx transcendental?
That depends on the value of $x$.

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# Use dedekind theory to define to define 11 and root 11

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