How big is infinity?

[PIMPBOT5000]

Ok, I was thinking about this a while back a never tried to get an answer to it so here go's...

On a number line there are an infinite amount of numbers...

There are also an infinite amount of prime numbers, however to a lesser degree of infinity when compared to the entire set of all integers. I have heard some say that since they are both infinitly large they are both equal amounts. However, since every number is not a prime number wouldn't that disprove the idea that they are equal?

 

[RossGr]

There are, indeed, different "sizes" of infinity. The prime numbers should be the same infinity as the integers, this means that a one to one correspondence can be formed between them. This is not true for the real numbers, you cannot "count" the Real numbers with the Integers. This means that you cannot create a numbered list of Real numbers which contains ALL Reals. Thus there are more Real numbers then integers.

 

[dejitaru]

Originally posted by: PIMPBOT5000
Ok, I was thinking about this a while back a never tried to get an answer to it so here go's...

On a number line there are an infinite amount of numbers...

There are also an infinite amount of prime numbers, however to a lesser degree of infinity when compared to the entire set of all integers. I have heard some say that since they are both infinitly large they are both equal amounts. However, since every number is not a prime number wouldn't that disprove the idea that they are equal?

That's only if you're setting bounds.
If you are counting prime numbers to infinity, the series does not terminate.
There are an infinite amount of primes, also an infinite amount of integers. infinity=infinity

You can't view infinity on a number line. It is not a point, nor is it a range.

 

[dejitaru]

1+1-1+1-1...=1+1-1+1-1...
so
1+(1-1)+(1-1)...=1+1+(-1+1)+(-1+1)...
simplified
1+(0)+(0)...=2+(0)+(0)...
therefore
1=2

 

[TerryMathews]

1/0

 

[sao123]

Indeed there are different levels of infinity.

Just to show you an example...opinion from my own mind...
There are infinately many integers. Z in x = {m...-2,-1,0,1,2...n}
There are infinately many distinct noninteger - rational (R) & irrational (S) numbers numbers such that
Z < R < Z+1 & Z < S < Z+1.
So now if there are infinately many Z, Z+1, then there must be infinity2 = (infinity * infinity) total numbers (Z,R,S)


However for a more factual approach...
Perhaps you should read up on Georg Cantor's research (1874) when he discovered that there is more than one level of infinity.
Here is a few examples...
http://www.ii.com/math/ch/
http://www.missioncollege.org/depts/math/clouse2/wendi.htm
http://www.facstaff.bucknell.edu/udaepp/090/w3/bretw.htm
http://www.facstaff.bucknell.edu/udaepp/090/w3/mikeh.htm

 

[isaacmacdonald]

word.

some infinite sets include different infinite subsets. Einstein said some shiznit about that.

eg: infinite 3d set may be comprised of multiple 2d infinite subsets (infinite in different directions). I guess that means that just because a set has no termination doesn't mean it includes all numbers.

 

[PIMPBOT5000]

Originally posted by: sao123
Indeed there are different levels of infinity.

Just to show you an example...opinion from my own mind...
There are infinately many integers. Z in x = {m...-2,-1,0,1,2...n}
There are infinately many distinct noninteger - rational (R) & irrational (S) numbers numbers such that
Z < R < Z+1 & Z < I < Z+1.
So now if there are infinately many Z, Z+1, then there must be infinity2 = (infinity * infinity) total numbers (Z,R,S)


However for a more factual approach...
Perhaps you should read up on Georg Cantor's research (1874) when he discovered that there is more than one level of infinity.
Here is a few examples...
http://www.ii.com/math/ch/
http://www.missioncollege.org/depts/math/clouse2/wendi.htm
http://www.facstaff.bucknell.edu/udaepp/090/w3/bretw.htm
http://www.facstaff.bucknell.edu/udaepp/090/w3/mikeh.htm

I understand now
Cantor's hypothesis makes perfect sense.
Thx for the help guys.

 

[Mday]

you're looking for the word countable. there are countably many integers. this also implies countably many rational number. however, there are uncountably many irrational numbers. also, the word density also comes in. for any interval you choose, bounded by 2 rational numbers, you will find an irrational number. irrational numbers are infinitely dense.

so, there are 2 infinities, a countable one, and an uncountable one. the uncountable one is MUCH larger than the countable one.

think of exponentiation and linear. so, F=x, and G=x^2, and H=2^x all are infinity when x = infinity. however, if you remember your calculus, taking the limit as x approaches infinity, F/G = 0. and G/H = 0 and F/H =0. this is somewhat of a way to show you that infinity is very undefined.

taking infinity/infinity can yield 3 things: 0, some x < infinity, or infinity.

 

[Sunner]

Originally posted by: PIMPBOT5000
Ok, I was thinking about this a while back a never tried to get an answer to it so here go's...

On a number line there are an infinite amount of numbers...

There are also an infinite amount of prime numbers, however to a lesser degree of infinity when compared to the entire set of all integers. I have heard some say that since they are both infinitly large they are both equal amounts. However, since every number is not a prime number wouldn't that disprove the idea that they are equal?

You do realize that this is the kinda stuff that drives people insane, and/or into suicide, right?

 

[Shalmanese]

Infinity is this: <----------------------> big.

 

[Jeff7]

Originally posted by: dejitaru
1+1-1+1-1...=1+1-1+1-1...
so
1+(1-1)+(1-1)...=1+1+(-1+1)+(-1+1)...
simplified
1+(0)+(0)...=2+(0)+(0)...
therefore
1=2

In the thrid line, 1+(1-1)+(1-1)...=1+1+(-1+1)+(-1+1)...
Where'd that second +1 come from? It appears to have just been stuck in there to confuse us. Not that I'd accuse you of that.

On a number line there are an infinite amount of numbers...

There are also an infinite amount of prime numbers, however to a lesser degree of infinity when compared to the entire set of all integers. I have heard some say that since they are both infinitly large they are both equal amounts. However, since every number is not a prime number wouldn't that disprove the idea that they are equal?

I don't think there can be different degrees of infinity; just as the laws of physics break down in a black hole (area of infinite density - high mass, zero space), standard laws of math break down with infinity.
Infinity (all real numbers) - Infinity (prime numbers) = Infinity

 

[Smilin]

Originally posted by: dejitaru
1+1-1+1-1...=1+1-1+1-1...
so
1+(1-1)+(1-1)...=1+1+(-1+1)+(-1+1)...
simplified
1+(0)+(0)...=2+(0)+(0)...
therefore
1=2

Lol, you tard. you stuck an extra +1 in there.

 

[wviperw]

he did not stick an extra 1 on there. The only reason the second expression has another 1 is because the series is written out further. You could easily write the first expression with the same amount of 1's as the second and still show the point.

 

[brjames]

Originally posted by: Smilin

Originally posted by: dejitaru
1+1-1+1-1...=1+1-1+1-1...
so
1+(1-1)+(1-1)...=1+1+(-1+1)+(-1+1)...
simplified
1+(0)+(0)...=2+(0)+(0)...
therefore
1=2

Lol, you tard. you stuck an extra +1 in there.

See the ellipses? that means the series goes on forever... the proof is still fallacious but not cuz he "stuck an extra +1 in there"

 

[Agent004]

Jeff7 and Smilin, I am afraid dejitaru is indeed correct! He simply used the fact, an infinite sum can be added in many ways* (the Supermarket checkout theorem, tells us the order you put your shopping doesn't change how much it will cost you...... or so it does, with finitely many terms)

Here is a much clearer example, I was taught this in my first calculus I module.... pretty amused by it

Consider the sum

S = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 +... =

infinity
Sigma ((-1)^(n-1))/n = Ln 2 (natural log of 2)
n=1

However, we can also consider S as

S = (1 - 1/2) - 1/4 + (1/3 - 1/6) - 1/8 + (1/5 -1/10) - 1/12 + ... = 1/2 - 1/4 + 1/6 - 1/8 + 1/10 - 1/12 +... = 1/2 ( 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 +... ) = 1/2 Ln 2!!!

So we have: Ln 2 = S = 1/2 Ln 2, <=> 1 = 1/2 ( or 1 = 2, if you prefer )

End of calculation

Conclusion: If you are summing over a finitely many terms, the order you add them doesn't matter. However, if you are summing over infinitely many terms, it does!



Also

Infinity (all real numbers) - Infinity (prime numbers) = Infinity

Is incorrect, as ' Infinity (all real numbers) ' or rather ' infinity ' is a number, whether I can write it out is another thing, but it doesn't change it is a number. Hence if I denote ' Infinity (all real numbers) ' = x.

Infinity (all real numbers) - Infinity (prime numbers) = x - x = 0

 

[Kuroyama]

Here is an easier version of your prime number problem.

There are infinitely many integers. Double all of them and you get infinitely many even numbers. Are there the same number of each?

There are the same number of evens as there are integers, since every integer got mapped to a unique even number (1 -> 2, 2-> 4, etc.). But at first glance there are only half as many even numbers, since 3, 5, etc are not even. As was pointed out, these are both cases of countably infinite.

Mday, there are many more infinities. In fact, as odd as it may sound, it is mathematically proven that we cannot prove that there are (or are not) infinities between countable and uncountable. So if your teacher asks you to prove that the real numbers are the next set bigger than integers then you better give up already. This is the Continuum Hypothesis :

http://mathworld.wolfram.com/ContinuumHypothesis.html

And finally, Digitaru did not do anything wrong. This is something you should have learned when doing infinite series in Calc I or II. If a series is alternating then the order of summation can change what the series adds up to. If I move around the terms then 1+1-1+1-1+1 + ... can be made to add up to ANY integer : 0, 1, 2, or whatever you like! It is NOT a convergent series.

 

[Smilin]

I see what you are saying but I'm not buying it, sorry. You ARE adding an additonal +1 in there. Normally it would appear at the end of the left side term OR the right side term will end up one short at the end. Problem is the 'end' will never be reached. For all practical purposes you're doing infinity + infinity - infinity. I can prove this the easy way by stating that one does not equal two. I can also prove it the hard way but I'll have to get back to you in say..eternity.

 

[syberscott]

I agree with smilin. Two equal infinite series have been ended at two different places and then compared. I don't see what this proves (other than the fact it screws with your head for a minute).

 

[Agent004]

Smilin, okey. Prove my calculations are wrong then. I am not adding 1 (or any 'extra' terms) or anything. If I am correct then so is dejitaru.

Also, it would help you to read up on calculus section of Power series, especially the comparison method.

 

[titanmiller]

If infinity is infinantly long then it must be able to repeat... 14257484547574325043523543.....(6 bizillion digits later)...........14257484547574325043523543 and it starts over. But say 6 bizillion ditits later you get 142574845475743250435235437 then it doesnt repeat until in another ammount of digits later the set does repeat considering the changed digit [7]. And so on. Infinity can neither be proved or disproved. It really boggels your mind though, kind of like thinking about the vasness of the universe
I hope you understand what I am getting at.

 

[PIMPBOT5000]

Originally posted by: titanmiller
If infinity is infinantly long then it must be able to repeat... 14257484547574325043523543.....(6 bizillion digits later)...........14257484547574325043523543 and it starts over. But say 6 bizillion ditits later you get 142574845475743250435235437 then it doesnt repeat until in another ammount of digits later the set does repeat considering the changed digit [7]. And so on. Infinity can neither be proved or disproved. It really boggels your mind though, kind of like thinking about the vasness of the universe
I hope you understand what I am getting at.

In the scope of infinity there is no such thing as a repeating series, because it can't repeat if it never ends I would think?

Maybe im wrong?

 

[Kuroyama]

Since the simple explanation didn't seem to convince, let's try something harder with limits.

Let a_n = 1 + 1 - 1 + 1 - 1 + .... (sum of first 'n' terms), so a_1 = 1, a_2 = 2, a_3=1, etc.

Then 1 + 1 - 1 + 1 - 1 + .... = limit(n -> infinity) a_{2n} = 2

But 1 + 1 - 1 + 1 - 1 + ... = limit(n -> infinity) a_{2n+1} = 1

Both of these are the same infinite sum.

To see this, consider the 100th term (it's +1). It is certainly in a_{2*100} and in a_{2*100+1}, since both add up more than the first 100 terms. Likewise, the 1000001 term (it's -1) is in a_{2*1000001} and in a_{2*1000001+1}.

Therefore, the nth number is always counted in both a_{2n} and in a_{2n+1}. Taking n to infinity shows that the first limit added up all the numbers, but also shows that the second limit also included all the numbers.

Since these are both valid way to add up terms up to infinity then they are equal, and 2 = 1.

 

[Kuroyama]

Infinity need not repeat. Consider the following sequence of numbers.

10

10100

101001000

10100100010000

etc.

Once you see the pattern then you'll see that none of these numbers have repetition. Keep writing the sequence out to 100th term, 1000000th term, .... to the infinite one and you see that infinity need not have repeating segments. But we may as well say that

10100100010000..... = infinity = 12348728973502935....

so it is sort of a meaningless exercise.

 

[dejitaru]

There's no "extra +1". 1+(-1)=1-1

1+1-1+1-1+... has no value, so we can't say that it is equal to something

There are also an infinite amount of prime numbers, however to a lesser degree of infinity when compared to the entire set of all integers. I have heard some say that since they are both infinitly large they are both equal amounts. However, since every number is not a prime number wouldn't that disprove the idea that they are equal?

You could say that there are "more" integers than primes, simply that they are seen more often, but this isn't the case.
There are an infinite amount of primes and an infinite amount of intergers.

In the scope of infinity there is no such thing as a repeating series, because it can't repeat if it never ends I would think?

1/7= 0.142857142857...
repeating, yet true only if infinite
1/3=0.33333...