If someone insists that the square root of 2 is rational...

[Commie Pinko Dirtbag]

...how many different arguments do you have available to prove them wrong?

[still_one]

1. It all depends how things are defined in your number system /nt

[Commie Pinko Dirtbag]

3. Actually, it doesn't. Irrational numbers are irrational independent of representation.

Maybe you're thinking about infinite periodic decimals. Those can have a finite representation, depending on base. But they are rational numbers.

[Deep13]

5. Weird.

It's the hypotenuse of a right triangle whose other two sides are 1 and 1. Why should that be irrational? It does represent a line segment connecting the 12 and 3 o'clock points on a circle. Seems like anything circular has something irrational about it. Is it because we are trying to measure something inherently round with a linear scale?

[Commie Pinko Dirtbag]

7. A number is ''irrational'' if it can't be expressed as a quotient (aka ''ratio'') of two integers.

http://en.wikipedia.org/wiki/Irrational_numbers#Ancient_Greece

[Deep13]

11. Well, I know that. I'm just wondering why there should be irrational numbers at all.

Obviously all integers are rational because any integer can be expressed as x/1. And tenths of integers and hundredths and thousandths &c. And also halves, quarters, eighths 7c; thirds, sixths, twelfths &c. Yet no matter how fine the measurement, pi is just pi. Weird.

[Jim__]

16. The real numbers are uncountable. The rational numbers are countable.

Given those 2 statements, there has to be irrational numbers.

[Deep13]

21. Well that tells me exactly nothing.

By uncountable, do you mean there is an infinite number of them, because I'm pretty sure there is an infinite number of rational numbers. Do you mean that one cannot have a pi or sq. rt. 2 number of something? The fact that these quantities can be expressed geometrically means that one can have those quantities of something. It is possible to make a hypotenuse for a wooden isosceles right triangle.

[Jim__]

24. Sorry. It's a fairly well-known argument.

Yes, both the reals and the rationals are infinite. In the 19th century, Cantor proved - for instance, his diagonal argument - that there are different levels of infinity. The rational numbers cannot be put into a one-to-one correspondence with the reals - there will always be real numbers that are not included in any mapping process. I'm sure if you search on "Cantor diagonal argument" you can find it.

[struggle4progress]

23. Because a countable dense linear order is not complete, and the attempt to complete

it adds new elements corresponding to proper Dedekind cuts

By a linear order, I mean a collection of mutually comparable elements: given any pair of distinct elements a and b, either a < b or b < a -- that is, for any pair of elements a and b, trichotomy holds: one (and only one) of the assertions a < b, a = b, b < a is valid

By dense linear order, I mean a linear order with the following property: given any pair of distinct elements a < b, there is an element c (not necessarily unique) with a < c < b

By countable dense linear order, I mean a dense linear order whose elements can be put in one-to-one correspondence with the natural numbers 1, 2, 3 ... : that is, the elements can be enumerated a(1), a(2), a(3) ... so each element is associated with one (and only one) natural number (and conversely); of course, this correspondence cannot have much to do with the order relation < of the dense linear order

An element b is called a lower bound for the nonempty set S of elements in a linear order if every s in S satisfies: b < s or b = s. A nonempty set S of elements in a linear order is "bounded from below" if there is an element b such b is a lower bound for S. An element b is called the greatest lower bound for the nonempty set S of elements if b is a lower bound for S and every other lower bound c for S satisfies: c < b or b = c

An element b is called an upper bound for the nonempty set S of elements in a linear order if every s in S satisfies: s < b or b = s. A nonempty set S of elements in a linear order is "bounded from above" if there is an element b such b is an upper bound for S. An element b is called the least upper bound for the nonempty set S of elements if b is an upper bound for S and every other upper bound c for S satisfies: b < c or b = c

A linear order is complete if every nonempty set of elements bounded from below has a greatest lower bound and every nonempty set of elements bounded from above has a least upper bound

A proper Dedekind cut in a linear order is a partition of the elements of the linear order into two disjoint nonempty sets L and U, such that: (1) b < c for every b in L and every c in U, and (2) L has no greatest element and U has no least element

Lemma. If a dense linear order admits a proper Dedekind cut, then the linear order is not complete.
Proof. Suppose L and U define a proper Dedekind cut in a complete linear order. Let b be the least upper bound for L and c be the greatest lower bound for U. Clearly b cannot belong to L (since L has no largest element) and similarly c cannot belong to U. So b belongs to U and c belongs L. Thus c < b. Since the linear order is dense, there is some element d with c < d < b. There must be at least one element f of L between d and b, since otherwise d would be upper bound for L with d < b, contradicting the fact that b is the least upper bound for L; similarly, there must be at least one element g of U between c and d; but then g < f, contradicting the fact that L and U define a proper Dedekind cut

Lemma. If a dense linear order admits no proper Dedekind cuts, then the linear order is complete.
Proof. Suppose there is a nonempty set S with upper bound x but without least upper bound. Then no upper bound for S belongs to S. Let U be the collection of all upper bounds for S; then U is nonempty, since it contains x. Similarly let L be the set of all elements of the linear order than are not upper bounds for S; L is nonempty since it contains all of S. Obviously b < c for any b in L and c in U. Since L and U do not define a proper Dedekind cut, either L has a greatest element f or U has a smallest element g. But if L has a greatest element f, then f is not an upper bound for S, so there some s in S with f < s; therefore s must be in U, and so s must be the least upper bound of S, a contradiction. So L has no greatest element and therefore U has a smallest element g, which is obviously a least upper bound for S, another contradiction. Thus every nonempty set S bounded from above but has a least upper bound. Similarly, one sees that every nonempty set S bounded from below but has a greatest lower bound.

Theorem. A dense linear order is complete if and only if it admits no proper Dedekind cuts.
Proof. The two lemmas above.

Theorem. A countable dense linear order admits a proper Dedekind cut.
Proof. We will construct an increasing sequence of elements with an upper bound but no least upper bound. Begin by listing the elements a(1), a(2), a(3), ... Find the first element a(j) in the list with the property that the linear order contains some element b < a(j); set a(u(1)) = a(j); this will be an upper bound to our sequence. Cross off the list a(u(1)) and all the other a(j) satisfying a(u(1)) < a(j). Find the first remaining a(j) such that a(j) < a(u(1)); set a(j) = a(n(1)); this will be the first element of our increasing sequence. Cross off all the a(j) satisfying a(j) < a(n(1)). Supposing that we have defined a(n(1)) < a(n(2)) < ... < a(n(k)) < a(u(k)) < ... < a(u(2)) < a(u(1))
and have crossed off the list all a(j) such that a(j) < a(n(k)) or a(u(k)) < a(j). We do not want the sequence to have least upper bound a(u(k)). So find the first remaining a(j) such that a(n(k)) < a(j) < a(u(k)), set a(u(k+1)) = a(j) and then cross off the list a(u(k+1)) and all the a(j) satisfying a(u(k+1)) < a(j). We want the sequence to increase. So find the first remaining a(j) such that a(n(k)) < a(j) < a(u(k + 1)), set a(n(k+1)) = a(j) and then cross off the list a(n(k+1)) and all the a(j) satisfying a(j) < a(n(k+1)). Proceeding, we obtain an infinite increasing sequence a(n(1)) < a(n(2)) < a(n(3)) < ... with an infinite decreasing sequence of upper bounds ... < a(u(3)) < a(u(2)) < a(u(1)). Notice that every element gets crossed off the list at some point: at stage k > 1, for example, the element a(j) with smallest index j that still remains on the list will be crossed off, and certainly j is at least k - 1. Now suppose the sequence a(n(i)) has a least upper bound b; then b = a(k) for some k, and b = a(k) has certainly been crossed off the list by the time we are done with stage k + 1. But being an upper bound for the increasing sequence, it cannot have been crossed off the list for being some a(n(j)) or being being smaller than some a(n(j)); and being a least upper bound for the sequence, it cannot have been crossed off the list for being some a(u(j)) or being being greater than some a(u(j)); hence it has not been crossed off the list, a contradiction. Thus the increasing sequence has no least upper bound. (Similarly, we could construct a decreasing sequence of elements with a lower bound but no greatest lower bound.) Thus the countable dense linear order is not complete; hence it admits a Dedekind cut.

To summarize:
Countable dense linear orders (like the rational numbers) have gaps in them. These gaps must be filled if one wants to use analytical tools like the least upper principle

[still_one]

17. No I am not. Math is governed by a set of definitions from which other properties are proven. If I

decide to define a "Real Number" system with different assumptions, my irrational number definition may be different from the one accepted

In fact one could define a whole new system where addition, subtraction, multiplication, and division are entirely different operations

Not unlike the republicans who decide to create their own special mathmetics so to speak. In fact they tend to create their own special brand of science

Forget about radioactive decay, and carbon 14, the earth is 5000 years old by their belief system

[dimbear]

26. Agreed. It doesn't depend on the definition of 'rational', but on

just what you mean by 2.

[damntexdem]

2. It fits the mathematical definition of an irrational number.

No other arguments should be needed. To produce them is to accept the anti-intellectual world view that such things are debatable by nonmathematicians.

[MineralMan]

4. The word "rational" has a number of meanings.

In the world of mathematics, it has just one. The square root of 2 is an irrational number, by definition. There is no argument.

However, the concept of the square root of 2 can be considered by the rational mind, as long as you ignore the mathematical definition of rational.

Many words have a number of different meanings, depending on how they are used.

[MineralMan]

6. Further, the square root of every prime number is irrational.

There's a good proof of that. 2 is a prime number.

[Speck Tater]

8. I never argue with irrational people, especially about irrational numbers. nt

[Commie Pinko Dirtbag]

10. Sigh, yeah, sometimes I think I should do the same.

But then I have a vision of official math books stating that the square root of two is exactly 577/408, and that keeps me going on.

[rug]

9. Have fun applying geometry to theology.

Let me know when you get somehere.

[Deep13]

12. It's easy to apply something real to something imaginary.

You can always change the imaginary conditions to match reality. The problem with theologians is that they try to change reality to match what is imaginary.

[MineralMan]

14. Can God be described using imaginary numbers?

An interesting question...

[Goblinmonger]

29. Geometry?

Swing and a miss!

[laconicsax]

30. +1

[struggle4progress]

31. Don't be rude. The question can indeed be phrased as a geometry question:

Given two perpendicular lines and a mark on each at unit length U from the crossing, let length L be the distance between the two end marks

Now construct two sequences of lengths M(0), M(1), M(2), ... and N(0), N(1), N(2), ... as follows:

M(0) = L, N(0) = U
Assuming M(n) and N(n) have been constructed, we ask: Is N(n) = 0?
If Yes: stop.
If No: define M(n+1) = max(M(n) - L(n), L(n)) and N(n+1) = min(M(n) - L(n), L(n))

(In an ideal world, these calculations could all be done with ruler and compass, and in fact the process just described occurs in Euclid's old geometry text)

The question "Is sqrt(2) irrational?" is the same as the question "Does the above process ever stop?"

[Nihil]

32. It's been done before - many times - and the people involved have certainly felt they got somewhere.

e.g., Freemasonry, Buddhism, ...

:hi:

[RaleighNCDUer]

13. You don't have to prove them wrong. They have to prove themselves right. nt

[laconicsax]

15. Don't be so strident!

:P

[struggle4progress]

18. Offhand I can think of three that are clearly different:

There's the ancient Pythagorean argument by contradiction, which can be recast in various forms

The continued fraction expansion is easily seen to be infinite and periodic. This can probably be recast as an argument, involving the Euclidean algorithm, showing that 1 and sqrt(2) are incommensurable

The slickest argument is that if 2*a*a = b*b were soluble in integers, then computing prime factorizations we'd have an odd number of 2s on the left and an even number on the right

I also see a proof by Eisenstein's criterion: 2 divides all terms (except the first) of x*x - 2 and 2*2 does not divide the last term, so x*x - 2 is irreducible over the rationals, and hence (being quadratic) has no rational root. But perhaps on careful examination this would reduce to the prime factorization argument

I also see a proof by algebraic number theory: since x*x - 2 is monic, its roots are algebraic integers, so if sqrt(2) were rational it would be a rational integer (that is, an ordinary integer), but it is easy to see by simple inequalities that x*x - 2 has no integer roots. But perhaps on careful examination this also would reduce to the prime factorization argument

[dimbear]

19. I do all my thinking and calculating modulus 7.

In that particular arithmetic the square root of 2 is 3.

I can't come up with any proof that 3 is irrational.

[Curmudgeoness]

20. Oh lordy, I stumbled into the math geek forum by accident!

Thought I was in R/T. I am confused. Must be a time/space thing.

[ZombieHorde]

22. I would just show them my math book, or a math themed web site. nt.

[Commie Pinko Dirtbag]

25. Good idea. (Although ''The Ancestor's Tale'' is kinda heavy.)

Then again, I could use the exercise.

[laconicsax]

28. Paperbacks are your friend. n/t

[Odin2005]

27. The Ancient Greek Pythagorans murdered a follower who said that it was irrational.