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How many irrational numbers have names?

Anonymous Thu 21 Nov 12:04:59 2019 No.11165672 View Reply Original Report

Quoted By: >>11165685 >>11165686 >>11166607 >>11168829 >>11169222

There is Pi
e
The Square Roots.
The logarithmic constant. (?)

Anonymous

Anonymous Thu 21 Nov 2019 12:14:59 No.11165685 Report

Quoted By:

>>11165672
Square root of 4 is rational.

Anonymous

Anonymous Thu 21 Nov 2019 12:16:20 No.11165686 Report

Quoted By: >>11165704 >>11165714 >>11165890

>>11165672
Well, there's an uncountably infinite number of irrational numbers, and an uncountably infinite number of them don't have names, so therefore, none of them have names. Final anser: 0.

Anonymous

Anonymous Thu 21 Nov 2019 12:27:33 No.11165704 Report

Quoted By: >>11165729 >>11165732 >>11166604 >>11166617 >>11168984

>>11165686
>there's an uncountably infinite number of irrational numbers
Proof?

Anonymous

Anonymous Thu 21 Nov 2019 12:32:14 No.11165714 Report

Quoted By:

>>11165686
Well, there's an uncountably infinite number of real numbers, and an uncountably infinite number of them are irrational, so therefore, none of them are rational.

Anonymous

Anonymous Thu 21 Nov 2019 12:52:22 No.11165729 Report

Quoted By:

>>11165704
Go ahead and try counting them.
In all seriousness, I'm pretty sure there's a proof that there are an uncountably infinite number of functions which cannot be described. I'd assume the proof would be similar, if not follow from the fact that you can define the reals as the set of equivalence classes of almost homomorphisms over the integers.

Anonymous

Anonymous Thu 21 Nov 2019 12:55:39 No.11165732 Report

Quoted By:

>>11165704
The reals are uncountable. If you take away a countable subset (i.e. the rationals), you're still left with an uncountable set (i.e. the irrationals). Proof: think.

Anonymous

Anonymous Thu 21 Nov 2019 14:37:10 No.11165890 Report

Quoted By:

>>11165686
>so therefore, none of them have names. Final anser: 0.
Actually, it's a set with zero Lesbesgue measure. It is literally zero only if you consider a quotient
I know it was a joke, sry

Anonymous

Anonymous Thu 21 Nov 2019 15:28:40 No.11166014 Report

Quoted By: >>11166440 >>11168620

There are trigonometric quantities (transcendental) like pi and tau and inverse trig functions
There is the exponential and logarithmic function (also transcendental). e, e^2, ln(2), etc
There are algebraic numbers like sqrt(2), sqrt(5), the golden ratio, silver ratio, plastic ratio, etc
There is the Reimann zeta function evaluated at arbitrary values, like Apery's constant=?(3)
There is the Euler-Mascheroni constant, which is the limit of the difference between the harmonic series and the natural log

Q:
Can physical constants be thought of as irrational? Like the coefficient of universal gravitation, boltzmann's constant, the electrical permittivity in a vacuum, etc.

Anonymous

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Anonymous Thu 21 Nov 2019 19:32:31 No.11166440 Report

Quoted By: >>11166615

>>11166014
>There are trigonometric quantities (transcendental) like pi and tau and inverse trig functions
>There is the exponential and logarithmic function (also transcendental). e, e^2, ln(2), etc
>There are algebraic numbers like sqrt(2), sqrt(5),
> The golden ratio, silver ratio, plastic ratio, etc
>There is the Reimann zeta function evaluated at arbitrary values, like Apery's constant=?(3)
>There is the Euler-Mascheroni constant, which is the limit of the difference between the harmonic series and the natural log

So I guess you have 6 classes of irrational numbers. The classes are infinite, but what actually has A NAME is the class. So we have six so far, in terms of what I was originally trying to ask.

> The Boltzmann constant is defined to be exactly 1.380649×10?23

Not irrational. I'll have to look up the others.

Anonymous

Anonymous Thu 21 Nov 2019 21:33:49 No.11166604 Report

Quoted By:

>>11165704
>tfw you chat shit about university but can't even do the basic shit they teach you

Anonymous

Anonymous Thu 21 Nov 2019 21:35:25 No.11166607 Report

Quoted By:

>>11165672
Fuck off mathbrains

Anonymous

Anonymous Thu 21 Nov 2019 21:41:44 No.11166615 Report

Quoted By:

>>11166440
Most constants are defined to a finite number of decimal places and as such are rational. All others are defined as some function of those other constants taken to a finite number of decimal places. As such, they are rational. One could come up with a constant that is theoretically irrational, like the Stefan-Boltzmann constant, but there's not really much of a point in taking its value in applications to have an infinite decimal expansion, what with there always existing some kind of uncertainty in system state or the value of some observable.

Anonymous

Anonymous Thu 21 Nov 2019 21:43:36 No.11166617 Report

Quoted By:

>>11165704
Diagonalization to show the reals are uncountable
Explicit construction to show the rationals are countable
Proving the union of finitely many countable sets is countable is trivial.
Put them together, and the irrationals must be uncountable.

Anonymous

Anonymous Fri 22 Nov 2019 17:47:04 No.11168620 Report

Quoted By:

bump. I want to know if there are more named irrational numbers than in >>11166014

Anonymous

Anonymous Fri 22 Nov 2019 19:48:20 No.11168829 Report

Quoted By: >>11168851 >>11168862

>>11165672
None of them, since we don't know what any of them are because the set is uncountable. All named irrationals are approximations of a number that can not be named by definition.

Anonymous

Anonymous Fri 22 Nov 2019 19:57:46 No.11168851 Report

Quoted By: >>11168855

>>11168829
sqrt(2) has been known to be irrational for more than 2000 years

Anonymous

Anonymous Fri 22 Nov 2019 19:58:40 No.11168855 Report

Quoted By: >>11168967

>>11168851
Which is just a label for an algorithm, not the number itself.

Anonymous

Anonymous Fri 22 Nov 2019 20:04:24 No.11168862 Report

Quoted By: >>11168866

>>11168829
The set of reals is also uncountable, so everything you said also applies to named rational constants.

Anonymous

Anonymous Fri 22 Nov 2019 20:07:23 No.11168866 Report

Quoted By: >>11168869 >>11168962

>>11168862
Not so, because we can take the intersection of rationals over the reals. Which is just another identity for the rationals. Then, this operation clearly changes the carnality the new set to be countable. If we take the intersection of the irrationals over the reals it can be proved the carnality of this identity is still uncountable.

Anonymous

Anonymous Fri 22 Nov 2019 20:08:31 No.11168869 Report

Quoted By:

>>11168866
cardinality* I should really stop phoneposting.

Anonymous

Anonymous Fri 22 Nov 2019 20:55:50 No.11168962 Report

Quoted By:

>>11168866
Ok, so take the intersection of the computables and the reals. Almost all our named irrationals fall in this countable set, so your argument is still worthless.

Anonymous

Anonymous Fri 22 Nov 2019 20:58:20 No.11168967 Report

Quoted By:

>>11168855
I dont know what this means. sqrt(2) absolutely cannot be represented with a ratio of integers a/b. This is the definition of irrational. Therefore we know at least one irrational number.

Anonymous

Anonymous Fri 22 Nov 2019 21:07:18 No.11168984 Report

Quoted By: >>11168997 >>11169081

>>11165704

we know that $\sqrt{2}$ is irrational. Considering x any real number, $x\sqrt{2}$ is also irrational. Since there are infinite real numbers, we conclude there are infinite irrational numbers of the form $x\sqrt{2}$ .

Anonymous

Anonymous Fri 22 Nov 2019 21:11:41 No.11168997 Report

Quoted By: >>11169081 >>11169199

>>11168984
>Considering x any real number, $x\sqrt{2}$ is also irrational
you meant to say "x rational", "since there are infinitely many rational nmbers"

Anonymous

Anonymous Fri 22 Nov 2019 21:44:45 No.11169081 Report

Quoted By: >>11169199

>>11168984
>>11168997
Let x=0.

Anonymous

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Anonymous Fri 22 Nov 2019 22:35:14 No.11169199 Report

Quoted By:

>>11168997
>>11169081

It was supposed to be a shitpost, stop pointing holes on my rigor bros

Anonymous

Anonymous Fri 22 Nov 2019 22:43:30 No.11169222 Report

Quoted By:

>>11165672
usually the ones that are transcendental/non-alegrbraic

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