Tilted Forum Project Discussion Community  

Go Back   Tilted Forum Project Discussion Community > The Academy > Tilted Philosophy


 
 
LinkBack Thread Tools
Old 03-28-2005, 02:21 AM   #1 (permalink)
Addict
 
Location: Grey Britain
Do irrational numbers exist?

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?
__________________
"No one was behaving from very Buddhist motives. Then, thought Pigsy, he was hardly a Buddha, nor was he a monkey. Presently, he was a pig spirit changed into a little girl pretending to be a little boy to be offered to a water monster. It was all very simple to a pig spirit."

Last edited by John Henry; 03-28-2005 at 02:32 AM..
John Henry is offline  
Old 03-28-2005, 04:17 AM   #2 (permalink)
Junkie
 
filtherton's Avatar
 
Location: In the land of ice and snow.
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.
filtherton is offline  
Old 03-28-2005, 04:53 AM   #3 (permalink)
zen_tom
Guest
 
An 'infinitely long rational number' sounds fairly irrational to me.
 
Old 03-28-2005, 05:29 AM   #4 (permalink)
Born Against
 
raveneye's Avatar
 
Quote:
If this pattern cannot be continued infinitely, what is its limit?
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.
raveneye is offline  
Old 03-28-2005, 09:42 AM   #5 (permalink)
Mad Philosopher
 
asaris's Avatar
 
Location: Washington, DC
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.
__________________
"Die Deutschen meinen, daß die Kraft sich in Härte und Grausamkeit offenbaren müsse, sie unterwerfen sich dann gerne und mit Bewunderung:[...]. Daß es Kraft giebt in der Milde und Stille, das glauben sie nicht leicht."

"The Germans believe that power must reveal itself in hardness and cruelty and then submit themselves gladly and with admiration[...]. They do not believe readily that there is power in meekness and calm."

-- Friedrich Nietzsche
asaris is offline  
Old 03-28-2005, 10:05 AM   #6 (permalink)
Addict
 
Quote:
Originally Posted by John Henry
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
...

<snip>

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?
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.
phukraut is offline  
Old 03-28-2005, 10:07 AM   #7 (permalink)
zen_tom
Guest
 
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.
 
Old 03-28-2005, 10:23 AM   #8 (permalink)
Born Against
 
raveneye's Avatar
 
Quote:
zen_tom: there are plenty of rational numbers which, when expressed in decimal form, are infinitely long. 1/3, for example.
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 is offline  
Old 03-28-2005, 10:44 AM   #9 (permalink)
Born Against
 
raveneye's Avatar
 
Quote:
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.

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.
raveneye is offline  
Old 03-28-2005, 11:58 AM   #10 (permalink)
Addict
 
Location: Grey Britain
Quote:
Originally Posted by 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.
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?
__________________
"No one was behaving from very Buddhist motives. Then, thought Pigsy, he was hardly a Buddha, nor was he a monkey. Presently, he was a pig spirit changed into a little girl pretending to be a little boy to be offered to a water monster. It was all very simple to a pig spirit."
John Henry is offline  
Old 03-28-2005, 12:01 PM   #11 (permalink)
zen_tom
Guest
 
What's wrong with that?
 
Old 03-28-2005, 12:08 PM   #12 (permalink)
Junkie
 
filtherton's Avatar
 
Location: In the land of ice and snow.
Quote:
Originally Posted by 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?

You'd have the nth integer. Infinity isn't a number, it's more of a concept.
filtherton is offline  
Old 03-28-2005, 01:07 PM   #13 (permalink)
Born Against
 
raveneye's Avatar
 
Quote:
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.
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.
raveneye is offline  
Old 03-28-2005, 02:24 PM   #14 (permalink)
Addict
 
Quote:
Originally Posted by John Henry
Clearly, then, irrational numbers do exist. Does any more general or generalisable proof exist that all non-terminating, non repeating real numbers are irrational?
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.

Quote:
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.
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.

Last edited by phukraut; 03-28-2005 at 02:27 PM..
phukraut is offline  
Old 03-29-2005, 01:44 PM   #15 (permalink)
Wehret Den Anfängen!
 
Location: Ontario, Canada
Quote:
Originally Posted by John Henry
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?
Because long yet finite sequences, and infinite sequences, are qualitatively different beasts.

Quote:
Originally Posted by asaris
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.
The proof in question relies on contradiction and is non-constructive.

Quote:
Originally Posted by John Henry
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?
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)

Quote:
Originally Posted by 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.
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.
__________________
Last edited by JHVH : 10-29-4004 BC at 09:00 PM. Reason: Time for a rest.
Yakk is offline  
 

Tags
exist, irrational, numbers

Thread Tools

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are On
Pingbacks are On
Refbacks are On



All times are GMT -8. The time now is 12:54 AM.

Tilted Forum Project

Powered by vBulletin® Version 3.8.7
Copyright ©2000 - 2024, vBulletin Solutions, Inc.
Search Engine Optimization by vBSEO 3.6.0 PL2
© 2002-2012 Tilted Forum Project

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360