John Henry

Please note that this is not a question about whether numbers like Pi really exist in the real world, it is a question about whether irrational numbers, as they are defined, exist within the formal system of mathematics.

Perhaps this should go in Tilted Knowledge, because I'm pretty sure the idea I'm about to present is wrong, but I don't really understand why.

Irrational numbers are those numbers which cannot be expressed as a ratio of two integers.

Take the numbers between 0 and 1. Let us consider a rational number, such as 0.123. This can be expressed as 123/1000. Likewise:

0.1234=1234/10000
0.12345=12345/100000
0.123457= 123457/1000000
0.1234571= 1234571/10000000
0.1234571= 12345714/100000000
...

Since the set of integers is infinite and passes through every sequence of digits, there can be no sequence of digits which is not included in it, even those which are non-repeating and non terminating. Irrational numbers between 1 and 0 are expressed as a non-terminating, non-repeating sequence of digits after the decimal place. So why can we not extend the pattern above infinitely?

If this pattern cannot be continued infinitely, what is its limit?

If it can, then does that not mean that all the non-repeating, non-terminating decimals that make up irrational numbers can be expressed as a ratio of two integers and as such are simply infinitely long rational numbers and not truly irrational?

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filtherton

I've seen proofs that sqrt(2) and sqrt(3) can't possibly be expressed as a ratio of two integers. I imagine that a proof for e is out there somewhere.
Sqrt(2) exists in about the same way as unicorns, though it is much more useful for doing calculations.

I think the problem with what you're suggesting is that it just isn't possible to nail a irrational number down to one specific sequence of integers. They just don't settle down like that.

zen_tom

An 'infinitely long rational number' sounds fairly irrational to me.

raveneye

Any limit, it doesn't matter. As long as you have to take it out infinitely long, then there doesn't exist a pair of integers that form a fraction that equals the irrational number.

Let's turn it around. Take pi. Go ahead and form your "infinitely long" fraction that is equal to pi. Now tell me what the numerator and denominator are. If they are integers, then you've proven your point.

The problem is they can't be integers, because they will both be infinitely long without any periodicity.

asaris

zen_tom: there are plenty of rational numbers which, when expressed in decimal form, are infinitely long. 1/3, for example.

In any case, it can be proven that the set of real numbers is larger than the set of rational numbers. So there must be irrational numbers.

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phukraut

What you have designed above is a limit---not a ratio of two integers, but a rational expression that approaches a real number. An infinitely long integer then is not an integer proper---it's infinity---and hence not in the set of integers.

Since a rational number is composed of two integers, and what you have above are actually two infinitely long strings of digits composing an irrational number, then we can see that what you have built is not a rational number. In actuality, it's a sequence of rational numbers that get ever and ever larger (but never equals) the irrational number, as per the definition of limit. In other words, irrational numbers are not types of rational numbers or vice versa, because talking about infinite integers doesn't make much sense.

zen_tom

Asaris: That's different to it being expressed by an infinately long ratio of integers. 1/3 is fairly short. But yes, I was being a bit facetious. I'm not interested in our labelling a number rational or irrational - it really makes no difference to the number, only in the way we express and describe the concept of quantity and ratio. To say an irrational number doesn't exist doesn't make any sense, and, if it did, it doesn't change anything, those numbers we are referring to are still the same, we've just decided to use a different label to describe them.

raveneye

Yep. But the unique sequence of digits is never infinitely long. If the unique sequence of digits is infinitely long (i.e. without any periodicity) then the number has to be irrational.

raveneye

I agree, as long as you specify that the numerator never starts repeating. If it starts repeating, then you do indeed have a rational number, because the unique sequence of digits is finite (even though the total number of digits in the numerator is infinite).

In fact you can express the number as the ratio of the unique sequence of digits (numerator) to a long string of 9s (denominator). So, for example, 1/7 can be expressed as .142856 repeated infinitely, or it can be expressed as 142856/999999. As soon as you have a repeat, you have two integers whose ratio is the number.

Or to put it another way, if your sequence doesn't repeat, then the numerator has to be infinite. If the numerator is infinite, then that number is not in the set of integers.

John Henry

Why didn't I think of looking that up before? A quick Google turned up multiple proofs of this, including one that generalised to other square roots.

Clearly, then, irrational numbers do exist. Does any more general or generalisable proof exist that all non-terminating, non repeating real numbers are irrational?

I don't entirely understand the assertion that the set of integers contains no infinite numbers, which doesn't mean I don't believe it, but my problem is as follows:

For any finite set of consecutive, positive integers starting with 1, the value of the largest member is equal to the magnitude of the set.

ie The set |A| is the first n consecutive integers greater than zero.

|A|=n
A={1,2,3...n}

The largest element of this set is always equal to n.

eg

n=9
|A|=9
A={1,2,3,4,5,6,7,8,9}

n=10
|A|=10
A={1,2,3,4,5,6,7,8,9,10}

n=1000
|A|=1000
A={1,2,3...1000}

The set of all integers is infinite we can set n as infinite. If we set n infinite, then we have a set of integers which contains n, which is infinite, giving us an infinite integer.

...don't we?

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zen_tom

What's wrong with that?

filtherton

You'd have the nth integer. Infinity isn't a number, it's more of a concept.

raveneye

Yep, the set of integers is infinite. It is countably infinite (as is the set of all rational numbers). The irrationals and reals are not countable, although they are also infinite. So the set of reals is larger than the set of rational numbers, even though both are infinite.

Another way to look at it, in terms of countability, is that every integer is specifiable in arithmetic terms: namely the unique number of times that 1 is added to 0.

So if you were right, then you could tell me two unique numbers, that arithmetically compute two integers, whose ratio equals Pi. Right now, all you're telling me is that those two numbers are "infinity" and "infinity". But infinity is not a number in the sense that it is subject to the laws of arithmetic. So we're back where we started from.

That doesn't mean that you couldn't evaluate the limit (as phukraut points out). It just means that you can't express the limit as the ratio of integers. Rather it's always going to be an infinite sum of some kind, that you can compute as far out as you want.

phukraut

Well, if I understand the idea correctly, once you have established that a real number is non-terminating and non-repeating, then it is by definition irrational. (This can be seen by the fact that the real number set is "larger" than the set of rational numbers. This is used in the proof that the real numbers are uncountable.) The big problem is how to prove something is irrational without necessarily knowing if it is non-repeating or non-terminating. Proving Pi irrational was a big deal for example. We still don't know if Pi^e is irrational, where "e" is the exponential constant.

No, you cannot simply <em>set</em> n to infinity and expect the properties of integers to carry over formally. You may see infinity being used in Complex Analysis as part of a set, like C union {infinity}, but this is just shorthand, and shouldn't imply infinity can be treated as such.

What you are saying above makes intuitive sense, but so does the following process:

1/1=1; 100/100=1; 9999/9999=1; infinity/infinity=1. But this would be wrong. raveneye said it best, two sets can have infinite number of members yet be different sizes. Everything is intuitive up until we reach infinity. I believe this is why the notion of <a href="http://mathworld.wolfram.com/CardinalNumber.html">Cardinality</a> was introduced.

Yakk

Because long yet finite sequences, and infinite sequences, are qualitatively different beasts.

The proof in question relies on contradiction and is non-constructive.

You used proof by example.

3 is prime
5 is prime
7 is prime
thus, all odd numbers are prime.

Proof by example isn't a proof.

There is difficulty in ruling out infinite integers from the set of integers. But the infinite integers you cannot rule out behave more like finite integers than the ones you want to use.

(Godel's incompleteness theorem can be interprited as the impossiblity of restricting the set of integers to only numbers which can be written out finitely)

Once again, this is a non-constructive statement, and requires proof by contradiction to be valid in order for it to be true. There is a logic of mathematics that includes all demonstrateable mathematical truths in which you cannot prove that there are more reals than rationals.

For example, if you insist that all real numbers can be described algorithmically, then you end up with a real number line that behaves a hell of a lot like the real number line you play with normally, yet it contains countably many numbers.

This is a slightly obtuse branch of mathematics, and I'm still learning about it myself. I just thought I'd point out that 'truth' is in the eye of the axiom system.

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