Uncountable Infinity (The Set of Reals)

[Agent Smith]

We have the set of naturals $\mathbb{N}$

We have the set of rationals $\mathbb{Q}$

We have the set of irrationals $\mathbb{Q'}$

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We have the set of rational approximations of irrationals $_\text{a} \mathbb{Q'}$

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The set $\mathbb{Q'}$ is in bijection with the set $_\text{a} \mathbb{Q'}$ (every irrational has its own rational approximation)
That is to say $|\mathbb{Q'}| = |_\text{a} \mathbb{Q'}|$ ... (A)

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$_\text{a} \mathbb{Q'} \subseteq \mathbb{Q}$

$|_\text{a} \mathbb{Q'}| \leq |\mathbb{Q}|$ ... (B)

$\therefore |\mathbb{Q'}| \leq |\mathbb{Q}|$ ... from (A) and (B)

The cardinality of the irrationals is at most equal to the cardinality of the rationals.

 

[khansaheb]

We have the set of naturals $\mathbb{N}$

We have the set of rationals $\mathbb{Q}$

We have the set of irrationals $\mathbb{Q'}$

---

We have the set of rational approximations of irrationals $_\text{a} \mathbb{Q'}$

---

The set $\mathbb{Q'}$ is in bijection with the set $_\text{a} \mathbb{Q'}$ (every irrational has its own rational approximation)
That is to say $|\mathbb{Q'}| = |_\text{a} \mathbb{Q'}|$ ... (A)

---

$_\text{a} \mathbb{Q'} \subseteq \mathbb{Q}$

$|_\text{a} \mathbb{Q'}| \leq |\mathbb{Q}|$ ... (B)

$\therefore |\mathbb{Q'}| \leq |\mathbb{Q}|$ ... from (A) and (B)

The cardinality of the irrationals is at most equal to the cardinality of the rationals.

Study - carefully:

Irrational number - Wikipedia

en.wikipedia.org

 

[Agent Smith]

Study - carefully:

Irrational number - Wikipedia

en.wikipedia.org

I know what irrational numbers are.

Am I wrong to say that there's a bijection between irrationals and their rational approximations? Challenge: Give me an irrational number and let's see if I can/can't find its rational approximation.

 

[khansaheb]

rational approximation.

APPROXIMATION ....... you say. What does that prove - if anything?

 

[Agent Smith]

APPROXIMATION ....... you say. What does that prove - if anything?

Nescio. The only way my argument can go wrong is if more than one irrational number maps to the same rational approximation. That can happen if the digits match e.g. 3.14159... and 3.14158... or 3.14160... but rational approximations can be of arbitrary precision i.e. a unique rational approximation can be assigned to any irrational, no matter how many of the digits are identical (there has to be a place value where the digits differ and we can use that to construct distinct rational approximations, no?

 

[Dr.Peterson]

The only way my argument can go wrong is if more than one irrational number maps to the same rational approximation. That can happen if the digits match e.g. 3.14159... and 3.14158... or 3.14160... but rational approximations can be of arbitrary precision i.e. a unique rational approximation can be assigned to any irrational, no matter how many of the digits are identical (there has to be a place value where the digits differ and we can use that to construct distinct rational approximations, no?

No.

Any irrational number has infinitely many rational approximations, and any rational number can be considered an approximation to infinitely many irrational numbers.

If you're serious, start by defining what you mean by "the" (unique) rational approximation of an irrational number, and then what you mean by this:

We have the set of rational approximations of irrationals $_\text{a} \mathbb{Q'}$

As far as I can see, this is the same as $\mathbb{Q}$.

 

[Agent Smith]

@Dr.Peterson, gracias.

It's true that an irrational has infinitely many rational approximations. Doesn't that mean we can construct unique ordered pairs (x, y) where x is an irrational and y is its rational approximation. We have infinite choices (not all of them can be identical with respect to y)? This is an advantage in my view, as it makes bijection possible.

I had to state that the rational approximation is unique because my argument fails if 2/more irrationals map to the same rational, the approximation. That there's an infinite number of rational approximations guarantees unique pairing in my opinion (we don't run out of options for distinct pairs).

 

[Dr.Peterson]

Doesn't that mean we can construct unique ordered pairs (x, y) where x is an irrational and y is its rational approximation.

You show me: try to prove your claim! First you need to say exactly what this means.

I had to state that the rational approximation is unique because my argument fails if 2/more irrationals map to the same rational, the approximation.

You can state it; but can you prove it?

You're implicitly claiming that there is a mapping from each irrational to one approximation. I say there is not.

That there's an infinite number of rational approximations guarantees unique pairing in my opinion (we don't run out of options for distinct pairs).

How does it guarantee this?? You need to prove that.

Above all: Do you know what "unique" means?

I don't think the word means what you think it means.

If you claim that there is a unique rational approximation for any irrational number -- that means, there is only one -- then you have to show how it is determined. You have done no such thing.

 

[Agent Smith]

@Dr.Peterson, good call!!

You show me: try to prove your claim! First you need to say exactly what this means.

Let n be an irrational number such that n =t.abcd... (a,b, c, etc. are its digits)
n's rational approximation with a precision to the 10ths decimal place is ta/10
n's rational approximation with a precision to the 100ths decimal place is tab/100
so on and so forth

Archimedes use the method of exhaustion to approximate $\pi$. I believe it's a general solution we could use for any irrational number.
We also have approximations using sums of series (re Leibniz's formula for $\pi$ and also there's a convergent series sum for $e$ (Euler's number). The summation can be terminated at any arbitrary point, determined by the level of precision one desires, oui?

You can state it; but can you prove it?

You're implicitly claiming that there is a mapping from each irrational to one approximation. I say there is not.

Non liquet, but , you need to disprove me now.

If you claim that there is a unique rational approximation for any irrational number -- that means, there is only one -- then you have to show how it is determined. You have done no such thing.

The (x, y) pair, where x is the irrational and y the rational approximation, is unique.
There are an infinite number of rational approximations for any irrational number (the desired precision decides the satisfactoriness of the approximation). For example $\frac{22}{7}$ as an approximation for $\pi$ is accurate only to the 2nd decimal place, while $\frac{315}{115}$ (Zu's constant) is more accurate.
Any given approximation will be ambiguous so long as the digits of the irrationals it approximates are identical, but they can't all be identical, otherwise they would be the exact same number. So, to disambiguate we use the digit that's different to construct different rational apprximations. So a unique pairing between irrationals and rationals (their approximation) is possible.

P.S. Why don't we use $\frac{314}{100} = \frac{157}{50}$ as an approximation for $\pi$?

 

[khansaheb]

Why don't we use 314100=15750\frac{314}{100} = \frac{157}{50}100314=50157 as an approximation for π\piπ?

We do use that for "engineering" approximations.

Yet

$\displaystyle \pi \ne \frac{314}{100}$

 

[mario99]

We do use that for "engineering" approximations.

Yet

$\displaystyle \pi \ne \frac{314}{100}$

I am as an engineer, I personally prefer to use $\displaystyle \frac{22}{7}$ for all engineering experiments involving $\displaystyle \pi$.

 

[Agent Smith]

I am as an engineer, I personally prefer to use $\displaystyle \frac{22}{7}$ for all engineering experiments involving $\displaystyle \pi$.

There's an arbitrariness there that's disconcerting. Shouldn't we then choose what makes for easier computation? With 314/100 as $\pi$, any calculations involving multiplication by $\pi$ becomes easy; division by $\pi$ becomes harder though, compared to 22/7.

Interestingly this becomes a nonissue in the modern age, either of the two approximations are handled easily by a simple calculator, although 314/100 requires more keypresses.

 

[khansaheb]

There's an arbitrariness there that's disconcerting. Shouldn't we then choose what makes for easier computation? With 314/100 as $\pi$, any calculations involving multiplication by $\pi$ becomes easy; division by $\pi$ becomes harder though, compared to 22/7.

Interestingly this becomes a nonissue in the modern age, either of the two approximations are handled easily by a simple calculator, although 314/100 requires more keypresses.

How about:

π = 3

It all depends on the use - are you using it for laying out a circular playground or are designing the accelerator CERN ?

 

[Agent Smith]

How about:

π = 3

It all depends on the use - are you using it for laying out a circular playground or are designing the accelerator CERN ?

On target. Why would CERN need to be "more circular" than (say) a garden table?

 

[khansaheb]

On target. Why would CERN need to be "more circular" than (say) a garden table?

I give up.....

 

[Dr.Peterson]

There's an arbitrariness there that's disconcerting. Shouldn't we then choose what makes for easier computation? With 314/100 as $\pi$, any calculations involving multiplication by $\pi$ becomes easy; division by $\pi$ becomes harder though, compared to 22/7.

Interestingly this becomes a nonissue in the modern age, either of the two approximations are handled easily by a simple calculator, although 314/100 requires more keypresses.

I gave up long ago, but I might as well say this, for the sake of anyone who reads this and gets confused:

The whole point of an approximation is to be have an inaccurate value that is easy to use, but close enough to be suitable. That's where the arbitrariness lives, and it's entirely expected. Some tasks, such as at CERN (not more circular, but more precise!) need a better approximation, while others could do with just "about 3".

Today, we mostly use the approximation of pi given by our calculators (or computer programs), because that's as good as we can process, and as good as we need, and no harder to use than a closer approximation. And it typically takes only one or two key presses! One press on the Windows calculator, for example, gives me 3.1415926535897932384626433832795. Why do anything else?

Fractional approximations (such as the remarkably accurate and memorable 355/113 = 3.1415929..., which you got entirely wrong, or the more familiar 22/7) are useful only where fractions are suitable (such as some mental arithmetic, or if you lost your calculator and phone).

Now I'll go back to ignoring pointless discussions.

 

[Agent Smith]

The whole point of an approximation is to be have an inaccurate value that is easy to use, but close enough to be suitable.

That's exactly the opposite of arbitrariness. Let me be clear (looks as though you guys have had enough). We have $2$ choices for $\pi$:
1. $\frac{22}{7}$ (traditional value for $\pi$ taught in high school)
2. $\frac{314}{100} = \frac{157}{50}$ (the approximation of $\pi$ I recommend)

When it comes to accuracy (how many digits match between the approximation and the real $\pi$) both are equally accurate. When it computations $\frac{314}{100}$ is easier to work with (because of the 100 in the denominator; all we have to do is multiply with the numerator and then move the decimal point $2$ places to the right). So why use $\frac{22}{7}$ when it's far more convenient to use $\frac{314}{100}$?

Also no one answered my question with regard to how more accurate values of $\pi$ lead to greater circularity[/imath]. In geometry, using a compass to construct an almost perfect circle doesn't require $\pi$ at all

 

[Agent Smith]

Corrigendum:
Move the decimal point 2 places to the left.