Proof that irrational numbers do not exist

[epr2008]

Any number c in the real numbers has the form $$x.{c_1}{c_2}...{c_n}$$, in which x is an integer and $$0 \le {c_n} \le 9$$ is a natural number. From the way that we have enumerated the decimal places, clearly number of decimal places is countable. Then there is a bijection from the indexes of the decimal places onto the set of the first n natural numbers. Consider the irrational numbers, namely $$\mathop {\lim }\limits_{n \to \infty } (x.{c_1}{c_2}...{c_n})$$. The infinity in the enumeration of decimal places must then be the same infinity as that of the natural numbers. Then, $$\mathop {\lim }\limits_{n \to \infty } [\frac{{{{10}^n}}}{{{{10}^n}}}(x.{c_1}{c_2}...{c_n})] = \frac{{\mathop {\lim }\limits_{n \to \infty } ({{10}^n} \times x.{c_1}{c_2}...{c_n})}}{{\mathop {\lim }\limits_{n \to \infty } {{10}^n}}} = {\lim }\limits_{n \to \infty } (x.{c_1}{c_2}...{c_n})$$. Is a rational expression in which the numerator and denominator are both integers.

If we consider Euclid's argument and acknowledge that the factor in the numerator is not unique in yielding an integer multiple then we can understand Euclid's flaw. The numerators and denominators will have infinite prime factorizations and therefore it would be impossible to simplify the fraction.

 

[D H]

Any number c in the real numbers has the form $$x.{c_1}{c_2}...{c_n}$$

No, it doesn't.

Even 1/3 doesn't have such a form. 1/3=0.333..., repeating forever. Irrational numbers also go on forever, but they never repeat.

 

[epr2008]

yes it does x=0 and all the c's are 3

 

[epr2008]

Its just saying that all real numbers have a decimal expansion. Its bad notation, yes I know.

 

[epr2008]

It isn't exactly necessary for the argument though. The argument is that the number of decimal places in countable and therefore there better be a countable natural number big enough to multiply and get a natural number.

 

[D H]

Decimal expansion is not bad notation. Yours is, and your bad notation is why your argument is false.

That certain numbers are irrational such as the square root of 2 was known to the ancients. I also suggest your read up on Cantor's diagonalization argument and Hilbert's hotel.

 

[epr2008]

I have read and yes I know that the square root of 2 is considered to be irrational. The argument is that we can consider any number in it's decimal expansion as just having place holders. Then we can number the place holders, i.e. tenths -->1, hundredths -->2 and so on which is clearly a countable set. Since it is countable then there must be a countable power of 10 to multiply it by and give an integer. It seems you are trying to apply inapplicable ideas to a very simple argument. From what I can the argument isn't false it has nothing to do with any specific irrational but it is satisfied by all reals and therefore all irrationals.

 

[epr2008]

If you see a flaw then by all means point it out and explain.

 

[FroChro]

Your way of argumentation is very unelegant (edit: well, in fact it is quite elegant, but "wrongly worded") and that may a problem why you don't see your error. That is that you assume that limit of rational numbers must be a rational number which of course doesn't hold.

 

[FroChro]

(Unconstructive) Proof that irrational numbers does exist can be following:
Any real number between 0 and 1 in binary notation can be assigned (maped) to exactly one subset of set of natural numbers and vice versa. (for example if on n-th place [in expression like 0.01110...] is 0 then number "n" is not in the assigned subset; if there is "1", it is). Therefore there are as many (concept of "larginess" of infinite sets is nontrivial and I will assume you are familiar with it, because my language abilities are not good enough to explain it more; and I will refer to it simply as "larginess" since I don't know english expression for it) real numbers as there are subsets of natural numbers. Now - there are as many rational number as there are natural numbers, while any set of all subsets of some set is larger than original set itself. This proves the thing. (sry for uncompletness, i could prove both things but it would be needed to define "larginess" of infinite set which I'm not willing to do right now)

Edit: "larginness" is called cardinality, read about it here if interested http://en.wikipedia.org/wiki/Cardinality

 

[Goongyae]

> Then we can number the place holders, i.e. tenths -->1, hundredths -->2 and so on which is clearly a countable set. Since it is countable then there must be a countable power of 10 to multiply it by and give an integer

So you're saying that irrational numbers aren't really irrational. They're just ratios between infinitely large integers. This makes some sense, and if this helps you think of them, go for it; but most people won't find your argument very useful, nor will they understand you without a lot of clarification on your part.

FYI, you do not always even need the quotients of infinitely large integers to produce irrational numbers. Sometimes a single integer will do.

Consider the infinitely large integer
...111111

It is equal to the geometric series 1 + 10 + 100 + ... = 1 / (1-10) = -1/9. In this sense, -1/9 equal to an infinite integer, and it is larger than any finite integer.

This kind of thing shouldn't be shocking for anyone familiar with the two's-complement representation uniquitous in computing.

 

[epr2008]

Thank you and I understand what you are saying but even so that step is irrelevant. The property that I used is commonly referred to as scientific notation and it holds in a base 10 number system. What people are failing to realize is that the nth power of 10 is entirely dependent on the nth decimal. This fact makes them equal (the infinities not the power). This is why the Hilbert's Hotel paradox does not apply. It applies to different magnitudes of infinities. The fact that the infinity is countable is purely based on the fact that you can count it which is why the diagonal problem does not apply.

 

[epr2008]

Googyae this is exactly what I was saying thank you. I couldn't find the right words. And I need this to prove something very interesting about prime numbers.

 

[Goongyae]

Other examples:

pi/4 = 3/4 * 5/4 * 7/8 * 11/12 * 13/12 * ... where each denominator is the prime numerator rounded to a multiple of 4. This clearly shows a quotient of two infinite integers.

And if you admit regularized values for divergent products, then
sqrt(2pi) = 1 x 2 x 3 x 4 x 5 x ... x the rest of the positive integers
4pi^2 = 2 x 3 x 5 x 7 x 11 x ... x all the primes

 

[FroChro]

So you're saying that irrational numbers aren't really irrational. They're just ratios between infinitely large integers. This makes some sense

Except there are not infinitely large integers. In fact you can construct them, but such a thing is way beyond present scope of OP, so it's not of much use.

What you are saying is that any irrational number can be written as a limit of some sequence of rational numbers. That doesn't say so much to OP, i quess.

Consider the infinitely large integer
...111111

It is equal to the geometric series 1 + 10 + 100 + ... = 1 / (1-10) = -1/9. In this sense, -1/9 equal to an infinite integer.
This kind of thing shouldn't be shocking for anyone familiar with the two's-complement representation uniquitous in computing.

Well, i don't really know what you mean, but i would like to point out a fact that property of |q|<1 (q is quocient of geometrical series) is needed to derive formula 1/1-q for it's summation.

 

[Goongyae]

>Except there are not infinitely large integers.

They exist if you want them to. Just like 0, -1, i, infinitesimals and things undreamt of which will also be pooh-poohed.

>This kind of thing shouldn't be shocking for anyone familiar with the two's-complement representation uniquitous in computing.

Clarification: ...11111 is equal to -1/9 because if you multiply it by 9 and then add 1 you get ...99999 + 1 which "overflows" to produce 0, the expected result. ...111 is also equal to -1/9 because if you multiply by 10 and add 1 you get ...110 + 1 = ...111 and the number stays the same.

I believe ...11111 might technically be called a "p-adic" number but I never really learned about it through that route.

>|q|<1 (q is quocient of geometrical series) is needed

Not according to Euler. It's a fact that using it with |q| >= 1 is a tremendous aid to calculation, e.g. for computing values of the Riemann zeta function (pretty badly named, since of course it was Euler who discovered its reflection forumla). Quick example:
1/(1-x)=1+x+x^2+x^3+...
d/dx 1/(1-x) = 1/(1-x)^2 = 1+2x+3x^2+...
Now substitute -1 to obtain
1/4 = 1 - 2 + 3 - 4 + ... = (1 - 4) (1 + 2+ 3 + 4 +...)
-1/12 = 1 + 2 + 3 + 4 + ...

 

[FroChro]

>Except there are not infinitely large integers.

They exist if you want them to. Just like 0, -1, i, infinitesimals and things undreamt of which will also be pooh-poohed.

>This kind of thing shouldn't be shocking for anyone familiar with the two's-complement representation uniquitous in computing.

Clarification: ...11111 is equal to -1/9 because if you multiply it by 9 and then add 1 you get ...99999 + 1 which "overflows" to produce 0, the expected result. I believe ...11111 might technically be called a "p-adic" number but I never really learned about it through that route.

Thanks for reply.

Well, then it is quite a different thing (- and interpretation,which is questionable ), I would say. And it is in no way helpfull to OP who has problems with some basic concepts of mathematics (both formally/"exactly" and intuitionaly) .

>|q|<1 (q is quocient of geometrical series) is needed

Not according to Euler. It's a fact that using it with |q| > 1 is a tremendous aid to calculation, e.g. for computing values of the Riemann zeta function (pretty badly named, since of course it was Euler who discovered its reflection forumla).

I know nothing about it, but it clearly depends on some "non-casual" interpretation. There is no other derivation known to me than this one:

S_n:=1+q+...+q^(n-1)
S_(n+1)=S_(n+1) ; i will write this identity in 2 different ways:
1+q*S_(n)=S_(n) + q^n
S^n=(1-q^n)/(1-q)
and use limit to infinity if possible.

 

[Goongyae]

>There is no other derivation known to me than this one:

There's this one, too:

(1+x+x^2+x^3+x^4+...)(1-x)
= 1 + x + x^2 + x^3 + x^4 + ...
-x -x^2 -x^3 - x^4 - ...
= 1

 

[ramsey2879]

Thank you and I understand what you are saying but even so that step is irrelevant. The property that I used is commonly referred to as scientific notation and it holds in a base 10 number system. What people are failing to realize is that the nth power of 10 is entirely dependent on the nth decimal. This fact makes them equal (the infinities not the power). This is why the Hilbert's Hotel paradox does not apply. It applies to different magnitudes of infinities. The fact that the infinity is countable is purely based on the fact that you can count it which is why the diagonal problem does not apply.

You can't count up to the final "nth decimal" of every irrational number, whether in decimal or scientific notation. Neither can you count up to infinity. Also what do you mean by "the diagonal problem" and why does it not apply?

 

[FroChro]

Sorry, I didn't read your addition to previous post.
I will need think about what you wrote. I'll (maybe :p) post back later.

 

[Goongyae]

The digits of a specific infinitely large integer / quotient of inifnitely large integers / non-terminating irrational number may be countable. But the set of such numbers, each with a countably infinite number of digits, is uncountable. The diagonal argument still applies.

If you try to enumerate your infinite sequences
#1: 12354234523453455...
#2: 45645623523543454...
#3: 56954694056940560...
#4: ...

Then I can always choose a number that won't appear in your list, by having its digits vary from each of the bold digits.

 

[dodo]

One thing I don't understand in this line of arguing is that there is no such thing as an infinitely long integer. There is an infinite number of integers, but each of them has a finite number of digits.

So, there are infinitely long decimal expansions, but there is no infinitely long integer to act as denominator.

 

[FroChro]

>There is no other derivation known to me than this one:

There's this one, too:

(1+x+x^2+x^3+x^4+...)(1-x)
= 1 + x + x^2 + x^3 + x^4 + ...
-x -x^2 -x^3 - x^4 - ...
= 1

I think this example and one in your previous post both are assuming some principles which don't generally hold (like changing ordering of non-absolutely-convergent series in this case and switching derivation with infinity summation/integration in other one).
I'm not telling that it must be wrong (whatever does that word mean), my mathematical knowledge is quite "small" anyway. (<-) But I don't really see what use it can be (i mean your interpretation of some operations), neither what does it really mean intuitively.

 

[Goongyae]

>there is no such thing as an infinitely long integer.

As I mentioned, they are a part of the recognized field of study called p-adic integers. If you don't like them, don't use them.

In any case, I still think it's most useful to think of irrational numbers in the conventional way: as an infinite convergent sequence of rationals, or a non-terminating sequence of digits. And one is still constrained by the diagonal argument however one interprets an irrational number.

>I think this example and one in your previous post both are assuming some principles which don't generally hold

You're quite right. But I was able to calculate in a few lines that zeta(-1) = -1/12. Divergent series, when used carefully, can be extremely practical. And in quantum field theory they use them all the time. "In a short period of less than a year, two distinguished physicists, A. Slavnov and F. Yndurain, gave seminars in Barcelona, about different subjects. It was remarkable that, in both presentations, at some point the speaker addressed the audience with these words: 'As everybody knows, 1 + 1 + 1 + · · · = −1⁄2'. Implying maybe: If you do not know this, it is no use to continue listening." - http://en.wikipedia.org/wiki/1_+_1_+_1_+_1_+_…

 

[FroChro]

Thank you Goongyae for pointing some things out. I'm reading about p-adic numbers on wikipedia right know. I didn't know about it before, seems quite interesting.

 

[Robert1986]

I wouldn't have argued a single thing against this claim until he can tell me why the proof for the irrationality of sqrt(2) is wrong. So, what say you?

 

[mathwonk]

this belongs in the circle squaring or angle trisecting thread.

 

[willem2]

What was wrong with this proof, is the idea that lim n->inf 10^n is an integer.
Even tough there is an infinite number of integers, all of them are finite.

 

[Robert1986]

What was wrong with this proof, is the idea that lim n->inf 10^n is an integer.
Even tough there is an infinite number of integers, all of them are finite.

Infinity is not an integer. That is the problem. 10^infinity makes no sense in the way that we wants to use it. You can't say infinity + infinity and get anything other than infinity. And furthermore, if lim f(n) = x as n -> a this does not mean that when f(a) = x. Consider, for example, 1/x. As x -> 0 1/x -> infinity but 1/0 isn't defined.

Again, what is wrong with the proof that sqrt(2) is irrational. If his conclusion is correct, this proof has an error. Find it, please.

 

[Robert1986]

Any number c in the real numbers has the form $$x.{c_1}{c_2}...{c_n}$$, in which x is an integer and $$0 \le {c_n} \le 9$$ is a natural number. From the way that we have enumerated the decimal places, clearly number of decimal places is countable. Then there is a bijection from the indexes of the decimal places onto the set of the first n natural numbers. Consider the irrational numbers, namely $$\mathop {\lim }\limits_{n \to \infty } (x.{c_1}{c_2}...{c_n})$$. The infinity in the enumeration of decimal places must then be the same infinity as that of the natural numbers. Then, $$\mathop {\lim }\limits_{n \to \infty } [\frac{{{{10}^n}}}{{{{10}^n}}}(x.{c_1}{c_2}...{c_n})] = \frac{{\mathop {\lim }\limits_{n \to \infty } ({{10}^n} \times x.{c_1}{c_2}...{c_n})}}{{\mathop {\lim }\limits_{n \to \infty } {{10}^n}}} = {\lim }\limits_{n \to \infty } (x.{c_1}{c_2}...{c_n})$$. Is a rational expression in which the numerator and denominator are both integers.

If we consider Euclid's argument and acknowledge that the factor in the numerator is not unique in yielding an integer multiple then we can understand Euclid's flaw. The numerators and denominators will have infinite prime factorizations and therefore it would be impossible to simplify the fraction.

This is even worse than I thought. You can't evaluate this limit:
$$\mathop {\lim }\limits_{n \to \infty } [\frac{{{{10}^n}}}{{{{10}^n}}}(x.{c_1}{c_2}...{c_n})]$$

like you think you can.

 

[Robert1986]

Any number c in the real numbers has the form $$x.{c_1}{c_2}...{c_n}$$, in which x is an integer and $$0 \le {c_n} \le 9$$ is a natural number. From the way that we have enumerated the decimal places, clearly number of decimal places is countable. Then there is a bijection from the indexes of the decimal places onto the set of the first n natural numbers. Consider the irrational numbers, namely $$\mathop {\lim }\limits_{n \to \infty } (x.{c_1}{c_2}...{c_n})$$. The infinity in the enumeration of decimal places must then be the same infinity as that of the natural numbers. Then, $$\mathop {\lim }\limits_{n \to \infty } [\frac{{{{10}^n}}}{{{{10}^n}}}(x.{c_1}{c_2}...{c_n})] = \frac{{\mathop {\lim }\limits_{n \to \infty } ({{10}^n} \times x.{c_1}{c_2}...{c_n})}}{{\mathop {\lim }\limits_{n \to \infty } {{10}^n}}} = {\lim }\limits_{n \to \infty } (x.{c_1}{c_2}...{c_n})$$. Is a rational expression in which the numerator and denominator are both integers.

If we consider Euclid's argument and acknowledge that the factor in the numerator is not unique in yielding an integer multiple then we can understand Euclid's flaw. The numerators and denominators will have infinite prime factorizations and therefore it would be impossible to simplify the fraction.

As an additional response to this already completely debunked, completely asinine post, I'll point out that it is absolutely no surprise that it is possible to express irrational numbers as an infinite sum, which is exactly what you have done.


As a side note, assuming that what you have written actually makes any sense, I would really like to know what you have allegedly proven about prime numbers

 

[disregardthat]

Not according to Euler. It's a fact that using it with |q| >= 1 is a tremendous aid to calculation, e.g. for computing values of the Riemann zeta function (pretty badly named, since of course it was Euler who discovered its reflection forumla). Quick example:
1/(1-x)=1+x+x^2+x^3+...
d/dx 1/(1-x) = 1/(1-x)^2 = 1+2x+3x^2+...
Now substitute -1 to obtain
1/4 = 1 - 2 + 3 - 4 + ... = (1 - 4) (1 + 2+ 3 + 4 +...)
-1/12 = 1 + 2 + 3 + 4 + ...

-1/12 is not the infinite sum 1 + 2 + 3 + ... The formula will simply not apply if x = -1. What you are doing is extending the domain of 1/(1-x)^2 to x = -1, but it will not correspond to the limit of 1+2x+3x^2+... anymore!

 

[Hurkyl]

Then we can number the place holders, i.e. tenths -->1, hundredths -->2 and so on which is clearly a countable set. Since it is countable then there must be a countable power of 10 to multiply it by and give an integer.

You are confusing two different things.

1. For each decimal place, there exists a power of 10 to multiply it by to move it to the left of the decimal place.
2. There exists a power of 10 to multiply a number by that moves all decimal places to the left of the decimal place

You make an argument that would prove the first point, but you are making the very different (and wrong!) claim that you've proven the second point.

So you're saying that irrational numbers aren't really irrational. They're just ratios between infinitely large integers. This makes some sense, and if this helps you think of them, go for it;

I'm doubt that it is helping him to think of them that way. He may find it appealing to think that way, but I'm pretty sure he isn't qualified to judge on his own that it is improving his understanding rather than taking away from it.

FYI, you do not always even need the quotients of infinitely large integers to produce irrational numbers. Sometimes a single integer will do.

Consider the infinitely large integer
...111111

e.g. this numeral doesn't name an integer, and it is strongly misleading to suggest that it does. This numeral can be used to name a "10-adic integer", but most "10-adic integers" are not integers.

p-adic numbers are quite useful; that's why we've defined them and even have whole fields of study devoted to them -- but it is not useful to pretend that p-adic integers are integers; that is a sure path to get all sorts of wrong ideas stuck in your head.

But I was able to calculate in a few lines that zeta(-1) = -1/12. Divergent series, when used carefully, can be extremely practical.

Yes -- but in doing so you are not computing what is commonly known as "the sum of an infinite series". In this case, you are probably referring to computing a "zeta regularized sum".

 

[epr2008]

So for clarification, my point was that saying there are real numbers that cannot be written in terms of integer ratios is in fact equivalent to the narrow-mindedness of Zeno's Paradox. It is not a simple ratio, but an infinite ratio. It is a limit problem not a finite one.

 

[Robert1986]

I understand exactly what you are saying. But you haven't added anything new the mathematics. It is, in fact, well known that every rational number is the limit of a sequence of irrational numbers. This is pretty much what you have shown. It is not a surprise to anyone who has taken a semester of Real Analysis. The fact that there are reals that cannot be written in terms of integer ratios is, in fact, an important distinction if only because of the fact that they are countable whereas the entire real line isn't. There is nothing trivial about that. Putting this aside, I grant that what you are saying is, in some sense, correct (at least what you wrote in the OP is "correct", even if it is trivial). So, what is the result you have proven with respect to prime numbers?