Irrational numbers

[Sunshine]

Just curious about a thing I've been thinking of:

It's true that that there are numbers that aren't rational... let's say x is such a number. Now we take two integers, a and b where a is the integer if x is rounded up, and b is the integer if x is rounded down.
Forming their arithmetic average, we will get a value y that is closer to x than either a or b. If x is smaller than y, we form the average of b and y, and if it's greater, we use a and y...
Keeping on like this we can come arbitrary close to x, expressed with integrals. So there must be infinitely many rational numbers just between a and b. How can it then come that we still can't write x as division with two integers?

 

[Hurkyl]

Forming their arithmetic average, we will get a value y that is closer to x than either a or b.

Why?

Keeping on like this we can come arbitrary close to x, expressed with integrals.

What do integrals have to do with anything?

So there must be infinitely many rational numbers just between a and b.

True; there are infinitely many rational numbers between any two distinct numbers.

How can it then come that we still can't write x as division with two integers?

Where did this come from?



P.S.: It is true that any irrational number can be written as the limit of a sequence of rational numbers...

 

[robert Ihnot]

Sunshine: Forming their arithmetic average, we will get a value y that is closer to x than either a or b. Hurkyl: Why?

Look at $$\sqrt{.8}$$ = .89... I don't think you mean to round up to 1 and down to 0, but its not clear just how many decimals you are going to use. Suppose you said, "Use x number of digits..." Then if the x+1 decimal was a zero, then what?

 

[Algebracus]

It sems like you (Sunshine) are talking about a method for finding rational numbers $$x_1, x_2, x_3, ...$$ so that the difference between the irrational number $$x$$ and the rational number $$x_i$$ is at most half the difference between $$x$$ and $$x_{i-1}$$ for every $$i$$. Still, there is a difference, and this is the point of it: You can always enbetter an approximation of an irrational number, but that means that the irrational number is not rational, because then you could get an optimal approximation (this is in fact the idea behind the proof of $$e$$ being irrational).

 

[robert Ihnot]

Wouldn't it be easier if we just approximated the number by using its first n decimals, and then using its n+1 decimals?

 

[funkstar]

As to why the infinity of the amount of rationals doesn't imply that every number is rational:

The infinity of the number of reals is bigger than that of the number of rationals, $$|\mathbb{Q}|<|\mathbb{R}|$$.

So even though the rationals are an infinite set and a subset of the reals, they can still be (and indeed are) a true subset. Look up "cardinality" and "Cantor's diagonal method" on MathWorld. Once you deal with infinite sets, the concept of size of these sets becomes a bit more complicated than what one initially thinks. Basically to prove two sets have the same number of elements you have to find injective functions between them - "counting" the number of elements as being infinite is not enough.

The ancient proof of the irrationality of $$\sqrt{2}$$ is quite solid, also :tongue:

Naive set theory is fun!

 

[C1ay]

Close but no cigar

Keeping on like this we can come arbitrary close to x..

Arbitrarily close to x does not equal x. If it would equal x exactly then x would not be irrational since irrational is, by definition, a number that does not terminate or become periodic.

 

[Sunshine]

"Forming their arithmetic average, we will get a value y that is closer to x than either a or b."
Why?

let's say x = pi... a = 4 and b = 3. (3+4)/2 = 3.5, that is closer to pi than a is.

Rpbert Ihnot, you have a point there, but it was just the principal I was trying to explain, so I could have formulated it somehow else, but it didn't seem that relevant. :)

Why?
What do integrals have to do with anything?

Sorry, meant integers, not integrals.

You can always enbetter an approximation of an irrational number...

That's my main thought, and since I asked if it could be THEORETICALLY possible, if we kept doing this infinitely many times...

Once you deal with infinite sets, the concept of size of these sets becomes a bit more complicated than what one initially thinks.

I guess that this kills the question that was just growing in my mind after my previous sentence :D

 

[robert Ihnot]

Sunshine: let's say x = pi... a = 4 and b = 3. (3+4)/2 = 3.5, that is closer to pi than a is.

Rpbert Ihnot, you have a point there, but it was just the principal I was trying to explain, so I could have formulated it somehow else, but it didn't seem that relevant. :)

Sure, now we got 3.5 for Pi, well, how are we to get the next approximation? Well, however we do it, we will need to add another digit sooner or later somehow .

Sure, I know the principal is more imporant, and that's why I suggested just putting down the digits of Pi, one at a time: 3, 3.1, 3.14, 3.141, ...etc.

Getting back to your original entry and question: How can it then come that we still can't write x as division with two integers?

BECAUSE: While you were just "trying to explain the principal," you overlooked that a better approximation comes to need more digits and since the irrational number goes to an infinite number of digits, THAT IS WHY we can not make the irrational rational and find it as the quotent of two integers!

 

[HallsofIvy]

let's say x = pi... a = 4 and b = 3. (3+4)/2 = 3.5, that is closer to pi than a is.

But if f x= .9090090009... where we have one more 0 each time, x is irrational.
Taking the average of 0 and 1 gives 1/2 which is NOT closer to x than 1.

Given any two numbers there exist an infinite set of irrational numbers, as well as an infinite set of rational numbers, between them so that doesn't say anything about the difference between rational and irrational numbers.

It is obviously true, however, that given any irrational x, there exist a sequence of rational numbers that converges to x.