Pi solved

[uncool]

The following is a standard definition of the real numbers.

 

The real numbers are equivalence classes of Cauchy sequences of rational numbers; the sequence (a_n) is equivalent to the sequence (b_n) if for any rational number x, there is an N such that for all n > N, a_n - b_n < x. Addition, subtraction, multiplication, and division are defined by placewise addition, subtraction, multiplication, and division, respectively; these are well-defined (that is, they respect equivalence classes).

 

Let a_n be the sequence of numbers (i_n/(2^n)) where i_n is the smallest natural number such that i_n^2 > 2^(2n + 1). You can check for yourself that a_n is Cauchy, and therefore is a representative of some real number. Then I claim that the square of the equivalence class of a_n is 2. Clearly the square is at least 2. Assume that it is specifically greater than 2. Then for some i, for all sufficiently large n, a_n^2 > 2 + 1/2^i > 2 + 1/2^(i + 1) + 1/2^(2i + 6) = 2*(1 + 1/(2^(i + 3)))^2. Then a_n * (2^(i + 3)/(2^(i + 3) + 1) (which is less than a_n) has a square that is larger than 2, which leads to a contradiction, as eventually a_n must be less than every rational number which has a square larger than 2. Therefore the class of a_n is a square root of 2 in the reals.

=Uncool-

 

Funny, I ask the guy to give me the number which multiplied by itself to equal 2 without rounding, and his response is that 'he can construct it'

 

 

Where is the number?.....cricket...cricket. [/QUOTe]

 

The number is the infimum of all positive rational numbers p/q such that p^2 > 2 q^2. That number is well-defined in the real numbers, and therefore exists in the real numbers.

=Uncool-

[DrRocket]

The following is a standard definition of the real numbers.

 

The real numbers are equivalence classes of Cauchy sequences of rational numbers; the sequence (a_n) is equivalent to the sequence (b_n) if for any rational number x, there is an N such that for all n > N, a_n - b_n < x. Addition, subtraction, multiplication, and division are defined by placewise addition, subtraction, multiplication, and division, respectively; these are well-defined (that is, they respect equivalence classes).

 

Let a_n be the sequence of numbers (i_n/(2^n)) where i_n is the smallest natural number such that i_n^2 > 2^(2n + 1). You can check for yourself that a_n is Cauchy, and therefore is a representative of some real number. Then I claim that the square of the equivalence class of a_n is 2. Clearly the square is at least 2. Assume that it is specifically greater than 2. Then for some i, for all sufficiently large n, a_n^2 > 2 + 1/2^i > 2 + 1/2^(i + 1) + 1/2^(2i + 6) = 2*(1 + 1/(2^(i + 3)))^2. Then a_n * (2^(i + 3)/(2^(i + 3) + 1) (which is less than a_n) has a square that is larger than 2, which leads to a contradiction, as eventually a_n must be less than every rational number which has a square larger than 2. Therefore the class of a_n is a square root of 2 in the reals.

=Uncool-

 

 

 

The number is the infimum of all positive rational numbers p/q such that p^2 > 2 q^2. That number is well-defined in the real numbers, and therefore exists in the real numbers.

=Uncool-

 

That is one way to do it. Another is to use Dedekind cuts.

 

The method you describe, using equivalence classes of Cauchy sequences, generalizes to a method that can be used to complete any metric space.

 

Connecstor could see all of these things done elegantly were he not so averse to reading the books that have been suggested to him.

Edited February 28, 2012 by DrRocket

[uncool]

That is one way to do it. Another is to use Dedekind cuts. [/QUOTe]

Yeah, I was going to start writing that as another method (I was going to include several definitions of the reals originally), but multiplication and division are a bit harder to define properly with Dedekind cuts. You need to make sure to check positive and negative things; Cauchy sequence multiplication is generally nicer. Although I should correct myself from earlier; division isn't quite defined by placewise division. First you have to check that you're far enough away from 0.

The method you describe, using equivalence classes of Cauchy sequences, generalizes to a method that can be used to complete any metric space.

Yes; it's also possible to do it through the completeness axiom (any nonempty set that is bounded below has an infimum) or several others. I just chose the one that seemed easiest to demonstrate the point.

=Uncool-

Edited February 28, 2012 by uncool

[connector]

"The number is the infimum of all positive rational numbers p/q such that p^2 > 2 q^2."

 

Yes, that is what mathematicians call the sqrt of 2, it is not however, a square root of 2. it is a number, it is real, it is quite infinitely long, but it is not the definition of a sqrt of a number, since it doesn't exactly multiply itself to equal 2. I'm not, and have not been, arguing that there is not a number or concept between there, or that this concept isn't useful, or anything of the sort - only that there isn't a square root of 2, even though in laymans terms, many people imagine that there is one because it would be a big hassle to invent a whole new name for only one exception to a rule. Again, for the trillionth time, all of this splitting hairs is solely to point out why the sqrt of 2 is not nice and clean to newbies, for greater understanding of mathematics on the fundamental level - understanding the origins of concepts, rather than blindly accepting whatever your orthodox professor either by lecture or book or online forum is trying to drill in your head without any fundamental explanation.

Edited February 28, 2012 by connector

[uncool]

"The number is the infimum of all positive rational numbers p/q such that p^2 > 2 q^2."

 

Yes, that is what mathematicians call the sqrt of 2, it is not however, a square root of 2. it is a number, it is real, it is quite infinitely long, but it is not the definition of a sqrt of a number, since it doesn't exactly multiply itself to equal 2.

Yes, it does. When you multiply it by itself, you get exactly 2, as I proved earlier. It's clear that you can't get a number that's less than 2, and I proved that you can't get a number that's greater than 2, since if it were, then there would be a rational number p/q with p^2 > 2 q^2 greater than the number. Therefore, as the reals are an ordered field (which you can prove for yourself), you must get 2.

 

ETA: To be slightly more specific, when you multiply the sequence a_n defined earlier, you get another Cauchy sequence. Under the definition of equivalence, it is equivalent to the sequence that is constantly 2 (that is, (2, 2, 2, 2, ...)). Since the real numbers embed into the real numbers with the map a -> (a, a, a, a, ...), that sequence can be considered to be the same 2 as in the rational numbers. Therefore, in the real numbers, the square of the equivalence class of a_n is the equivalence class of 2.

=Uncool-

Edited February 28, 2012 by uncool

[DrRocket]

 

Yes; it's also possible to do it through the completeness axiom (any nonempty set that is bounded below has an infimum) or several others. I just chose the one that seemed easiest to demonstrate the point.

=Uncool-

 

No, completeness of the real numbers is not an axiom. It is a result of the construction of the real numbers themselves -- using Dedekind cuts or equivalence classes of Cauchy sequences. You can show that any two complete Archimedian fields are isomorphic, so the precise method of construction is not important.

 

The least upper bound property is equivalent to completeness of the real numbers with the usual metric that is induced by the order of the reals.

[rktpro]

I remember fairly well that we used to plot under-root of two on a number line.

[uncool]

No, completeness of the real numbers is not an axiom. It is a result of the construction of the real numbers themselves -- using Dedekind cuts or equivalence classes of Cauchy sequences. You can show that any two complete Archimedian fields are isomorphic, so the precise method of construction is not important.

It can be taken as an axiom for real numbers, which you then show that Dedekind cuts and equivalence classes of Cauchy sequences satisfy. That is, you define a real number system to be an ordered field that satisfies the lub property, and then show that the constructions both satisfy that axiom.

The least upper bound property is equivalent to completeness of the real numbers with the usual metric that is induced by the order of the reals.

Yup.

=Uncool-

[Fuzzwood]

http://www.wikihow.com/Recognize-a-Troll-on-the-Internet

 

Trolls do not normally have a social life and try to get attention, any attention, even negative, by annoying people on fora.

[MathIsCool]

Unfortunately mathiscool has absolutely nothing to argue with except trying to attack my character and distract everyone rather than (ironically) overcoming his ignorant bliss and admitting his error in claiming sqrt of 2 exists anywhere but his and other mathematicians imagination.

 

he seems to somehow forget that the Pythagorean theorem only works when you round, or estimate, c. Instead this nutcase believes that because it works out on his calculators or estimates where he rounds it up that somehow sqrt of 2 actually exists and satisfies the equation exactly - it doesn't, it never has.

 

Hahahahahahahahahahahahahahahahahahaha.

[Klaynos]

!

Moderator Note

Please can everyone ensure they refrain from personal attacks.

connector, can you please try and understand what people are trying to tell you. It seems as if you are just disreguarding their comments which is rather frustrating.

MathIsCool, posts like that tend not to help.

Please do not respond to this modnote.

[CaptainPanic]

Even though I completely fail to see the point of this thread ([math]\pi, \sqrt{2}[/math] doesn't exist? wtf? are we seriously discussing that?)...

I am intrigued and entertained. So, I'll join in.

 

So, if [math]\sqrt{2}[/math] does not exist, then where does it disappear into non-existance? Say we have a triangle with a right angle of 90 degree, and sides of 1 and 1. Maths says that the 3rd side is then length [math]\sqrt{2}[/math]. The other two sides have length 1... they exist. The angle of 90 degrees exists too.

 

Or does it? As an engineer, I know that if you measure carefully, the sides will not be 1. And the angle will probably not be 90 degrees either (it might in reality be pretty close, but no cigar). So, you could argue that when it comes to measurements, all the whole numbers, the entire domain of [math] \mathbb{N} [/math] or even [math]\mathbb{Z}[/math] does not exist. Which is of course b*ll*cks. So, let's dismiss this part.

 

So, maybe somewhere between the one side of the triangle to the other, it disappears into non-existance.

 

So, let's try again... and let's say that 1 and 90 are numbers that exist. So, the 3rd side of the triangle is (1²+1²)^1⁄2. In that, the numbers 1 exist. The formula as a whole, which was found by Pythagoras also exists. So... eeh... since we simply defined [math]\sqrt{2}[/math] as the result of that, why wouldn't that exist?

 

I fail to see where [math]\sqrt{2}[/math] would disappear into non existance. And that's probably because it actually DOES exist.

 

But by all means, please carry on in this thread. It's highly amusing.

[khaled]

Connector,

 

[math]\sqrt{2} = 1.41421356237309504880168872420969807856967187537694807317667973799... \approx \frac{99}{70}[/math]

 

The approximation differs from the correct value by less than [math]\frac{1}{10,000}[/math] --Wikipedia

 

It has a fractional form, you know: [math][1;\bar{2}][/math]

 

and finally, (personally I don't think you're familiar with this):

 

[JohnStu]

Yes, that is the defintion of pi.

 

Pi does have a pattern though and that pattern is that it is completely patternless. This means that you can expect no repeats what so ever in the decimals. My brother used it in his memorization of pi digits contest. He managed to remember 1050 digits all correct in 7 minutes. The trick is to know it is patternless.

[rktpro]

Yes, that is the defintion of pi.

 

Pi does have a pattern though and that pattern is that it is completely patternless. This means that you can expect no repeats what so ever in the decimals. My brother used it in his memorization of pi digits contest. He managed to remember 1050 digits all correct in 7 minutes. The trick is to know it is patternless.

 

Unless you get Feynman point.

[connector]

I will be leaving this forum for good since the moderators and majority of posters here are dead set on continuing with orthodox errors rather than admitting their oversights and helping newbies understand simple math concepts, such as why when you multiply the square root of 2 by itself and DO NOT round, you get 1.9999 etc

 

It doesn't matter in what complicated way you write a formula to get a number that multiplies itself to equal 2, there is always some rounding involved, as sqrt of 2 itself goes on forever. We hopefully can agree that is a fact, and if it's not forever, then the end point has not been found. When you take a number that goes on forever with endless, non pattern digits, you can't multiply it by itself to exactly equal 2 without rounding, no matter how utterly close to 2 it is, it isn't 2. How this has escaped the minds of the people on this thread so far is truly beyond me.

 

 

See ya.

[uncool]

I will be leaving this forum for good since the moderators and majority of posters here are dead set on continuing with orthodox errors rather than admitting their oversights and helping newbies understand simple math concepts, such as why when you multiply the square root of 2 by itself and DO NOT round, you get 1.9999 etc

The simple math concept is that you do get 2, period. You have not proven anything, and as such, your posts are simply laughed at.

It doesn't matter in what complicated way you write a formula to get a number that multiplies itself to equal 2, there is always some rounding involved, as sqrt of 2 itself goes on forever. We hopefully can agree that is a fact, and if it's not forever, then the end point has not been found. When you take a number that goes on forever with endless, non pattern digits, you can't multiply it by itself to exactly equal 2 without rounding, no matter how utterly close to 2 it is, it isn't 2. How this has escaped the minds of the people on this thread so far is truly beyond me.

 

 

See ya.

Are you done throwing that tantrum?

 

Now will you come back and actually explain to me how you think my proof is incorrect? Or will you continue repeating your point again and again after it's been demonstrated to you precisely how it is wrong?

=Uncool-

[ajb]

It doesn't matter in what complicated way you write a formula to get a number that multiplies itself to equal 2, there is always some rounding involved, as sqrt of 2 itself goes on forever.

 

 

(Assuming you have not gone quite yet)

 

You are right in the sense that computationally you will inherit some rounding errors. Your calculator or computer may indeed not give you precisely that [math]\sqrt{2}\sqrt{2}= 2[/math].

 

But pure mathematics tells us that there is such a number [math]\sqrt{2}[/math] and it is perfectly well defined.

 

The same for [math]\pi[/math], you may well get computational errors as you are forced to make some truncation, but pure mathematics can usually avoid such rounding and handle things in a more abstract way.

Edited March 1, 2012 by ajb

[connector]

Ok one last post.

You can't use something not-proven to prove that it exists.

If you say sqrt of 2 x sqrt of 2 = 2, you used the nonproven sqrt of 2 to try and prove it exists.

first find a number to actually mean the sqrt of 2. (you will run into a problem - you cant because the supposed number doesn't end, and no matter how far you go you just get longer 1.9999 results.

 

So, in all of the formulas, abstract methods, and everything else people are using to try and prove that sqrt of 2 exists, they are either rounding, using variables that represent numbers but not using the numbers themselves, therefore subjecting to rounding errors, or errors of using unproven numbers to prove that other unproven numbers or the unproven number in question actually exists. These are all errors, faults, whatever you want to call them, but it's not based on reality. If you can't work something out on the fundamental level, such as (n)(n)=2, then n doesn't exist even if you can 'make it work' on a higher, abstract level.

 

@uncool

the number in your proof is quite real. it is not however the square root of 2, it's just very close to the square root of 2. that is the fundamental problem in it. you said that it's the square root of 2 exactly - it's not, only if you round, or 'force stop' the endless digits it creates. example: if you take the number in your proof and multiply it by itself, you don't get 2, unless you truncate or round

Edited March 1, 2012 by connector

[qsa]

http://en.wikipedia.org/wiki/0.999...

[uncool]

Ok one last post.

@uncool

the number in your proof is quite real. it is not however the square root of 2, it's just very close to the square root of 2.

Then show that the proof is wrong. You can continue saying it's not the square root of 2 until your face is blue, or you can actually address the proof.

 

What is the square of the number used if it is not 2? Is the square of the number greater than 2 or less than 2?

=Uncool-

Edited March 1, 2012 by uncool

[khaled]

Ok connector let's try this with Logic,

 

[math]a \in \mathbb{R}^+, b \in \mathbb{R}^+[/math] :

 

Your proposition (A): [math]\sqrt{2}[/math] does not exist

 

We know this is true (B): [math]2 \;\; exist[/math]

 

[math]IF \; b = (a \times a) \; AND \; b \; exist \; THEN \; (a \times a) \; exist \; \rightarrow \; a \; exist \; \rightarrow \; \sqrt{b} \; exist[/math]

 

Note that in the above statement: [math]a \times a =b \; \rightarrow \; a^2 = b \; \rightarrow \; b^{1/2} = a \; \rightarrow \; \sqrt{b} = a[/math]

 

Now, let's try [math]b = 2[/math],

 

[math]IF \; 2 = (a \times a) \; AND \; 2 \; exist \; THEN \; (a \times a) \; exist \; \rightarrow \; a \; exist \; \rightarrow \; \sqrt{2} \; exist[/math]

 

If you don't want to use the notion [math]\sqrt{2}[/math] .. you can just use this [math]{2}^{1/2}[/math]

 

It's clear "from the above statement" that your proposition (A) contradicts with truth (B)

Edited March 2, 2012 by khaled

[MathIsCool]

You'll love this:

 

Let [math]a=e^{\frac{ln(2)}{2}}=\sum\limits_{n=0}^{\infty}{\frac{(ln(2))^n}{n!2^n}}[/math].

 

Then [math]a^2=(e^{\frac{ln(2)}{2}})^2=e^{ln(2)}=2[/math].

[uncool]

Ok connector let's try this with Logic,

 

[math]a \in \mathbb{R}^+, b \in \mathbb{R}^+[/math] :

 

Your proposition (A): [math]\sqrt{2}[/math] does not exist

 

We know this is true (B): [math]2 \;\; exist[/math]

 

[math]IF \; b = (a \times a)[/MATH]

Your logic is wrong here; you've already assumed that a exists to make this statement.

=Uncool-

 

You'll love this:

 

Let [math]a=e^{\frac{ln(2)}{2}}=\sum\limits_{n=0}^{\infty}{\frac{(ln(2))^n}{n!2^n}}[/math].

 

Then [math]a^2=(e^{\frac{ln(2)}{2}})^2=e^{ln(2)}=2[/math].

I'm assuming that connector doesn't think that ln(2) exists exactly either.

=Uncool-

[John Cuthber]

Does connector actually exist?