So are there more even numbers than odd ones?

Hayabusa Rider

Admin Emeritus & Elite Member
Jan 26, 2000
50,879
4,268
126
Given that you select two numbers at random

Even x even = even
Even x odd = even
odd x odd = odd

Evens outnumber odds by 2 to 1?


:p
 

AFB

Lifer
Jan 10, 2004
10,718
3
0
Had a bit of hybrid coding going on :eek:

<?php
$odd = 0;
$even = 0;

for($n = 0; n < 50000; n++)
{

if($n % 2 == 0)
{
$even++;
}
else
{
$odd++;
}

}

echo "odds : " . $odd;
echo "evens : " . $even;
?>
 

littleprince

Golden Member
Jan 4, 2001
1,339
1
81
I think all you've shown is that you can nef, start up new useless threads, and that theres more ways to multiply numbers to get an even number than there is for odd.
 

Chronoshock

Diamond Member
Jul 6, 2004
4,860
1
81
Har har funny. Addressing it seriously: even if there are "twice as many" they'd still be equal. As I stated in the n/o thread, even if one seems to have twice as many numbers (ie natural numbers vs integers) they still are equal infinities (ie they have 1 to 1 correspondence).
 

Goosemaster

Lifer
Apr 10, 2001
48,775
3
81
Originally posted by: Chronoshock
Har har funny. Addressing it seriously: even if there are "twice as many" they'd still be equal. As I stated in the n/o thread, even if one seems to have twice as many numbers (ie natural numbers vs integers) they still are equal infinities (ie they have 1 to 1 correspondence).

what about 0 ? :evil:


I know I know..don;t answer that....
 

RaistlinZ

Diamond Member
Oct 15, 2001
7,470
9
91
Yeah, what about 0? Given that there is no -0 and that 0 is an even number, there will always be one more even number than odd number.
 

Chronoshock

Diamond Member
Jul 6, 2004
4,860
1
81
Originally posted by: RaistlinZ
Yeah, what about 0? Given that there is no -0 and that 0 is an even number, there will always be one more even number than odd number.

While this may seem counterintuitive, you can't have one infinity greater than another infinity by some integer. You can't have one infinity being "one bigger" than another. The way an infinity is bigger than another is if you can prove that after mapping all the numbers of one set to another, you can still find another number in one set that is not mapped. Now you'd think that 0 wouldn't be mapped but since its infinite, you can just pop it in.
To find a number that can't be mapped (in the case of real vs natural) would be to look at the set of natural numbers {n1, n2, ... nk} and assume that you can map each real number to each natural number (hypothesis) then you can say that one can create a real number where the ten's place is the ten's place of n1 + 1, the hundred's place is the hundred's place of n2 +1 .... etc. This would give you a number different than any other number mapped (since it would never exactly match any number). This contradicts the original assumption thus it is false. You cannot map the set of real numbers to the set of natural numbers and it is therefore larger.
 

desteffy

Golden Member
Jul 16, 2004
1,911
0
0
there are more irrationals than rationals. LOTS more. (there are more irrational numbers between 0 and 1 than there are rational numbers all together)
 

SinNisTeR

Diamond Member
Jan 3, 2001
3,570
0
0
Originally posted by: mchammer187
Originally posted by: SinNisTeR
there are the same amount..

what about rationals vs irrationals? ;)

same

sqrt of any number that is not a perfect square

rationals have a countable infinity, while irrationals do not. there are different types of infinity.. :)
 

eLiu

Diamond Member
Jun 4, 2001
6,407
1
0
The even/odd #s both map one-to-one/onto the natural numbers...like for n is an element of the natural #s, even = 2n. odd = 2n+1. Thus we have that even/odd numbers are countable. Countable sets compose the 'smallest infinity' indicating that there are in fact equal #s of even &amp; odd numbers. Note that the rational numbers are also countable...indicating that there are equal numbers of rationals, evens, odds, integers, etc.

Note that the reals (and thus irrationals) are not countable...the proof for this is kind of annoying so i won't bother.

chrono, have you taken 18.100b?
 

mchammer187

Diamond Member
Nov 26, 2000
9,114
0
76
Originally posted by: SinNisTeR
Originally posted by: mchammer187
Originally posted by: SinNisTeR
there are the same amount..

what about rationals vs irrationals? ;)

same

sqrt of any number that is not a perfect square

rationals have a countable infinity, while irrationals do not. there are different types of infinity.. :)

i dont understand how you can count the rationals

.01 .001
.0001 .00001

isn't there an uncountable infinite number of rationals between .01 and .0100000000000000000001

now if you are talking integers that is one thing but rationals is another
 

Chronoshock

Diamond Member
Jul 6, 2004
4,860
1
81
Originally posted by: mchammer187
Originally posted by: SinNisTeR
Originally posted by: mchammer187
Originally posted by: SinNisTeR
there are the same amount..

what about rationals vs irrationals? ;)

same

sqrt of any number that is not a perfect square

rationals have a countable infinity, while irrationals do not. there are different types of infinity.. :)

i dont understand how you can count the rationals

.01 .001
.0001 .00001

isn't there an uncountable infinite number of rationals between .01 and .0100000000000000000001

now if you are talking integers that is one thing but rationals is another


Its all in the way you organize it. Rationals can be represented as the ratio a/b where a and b are both integers and b != 0. Now simply take all possible combinations of A/B where A and B both represent the set of all integers (but no 0 in B). This is mappable to the natural numbers
 

eLiu

Diamond Member
Jun 4, 2001
6,407
1
0
this is a totally un-rigorous way of looking at it, but rationals are countable b/c we can look at'em this way...

rational = m/n with m,n are integers, n!=0. So for m=1, there are countably many numbers 1/n. For m=2, there are countably many 2/n.

Now you'll just have to take this on faith...but there's a theorem of analysis that says that the infinite union of countable sets (each 1/n, 2/n, 3/n, 4/n....m/n and repeat for negatives is countable) is still countable.

Another way to think about it is that with rationals, we're limited to certain combinations of decimal expansions...irrational numbers fill in all the other combinations (pi, e, sqrt 2, etc). But there is no general expression for irrational numbers, and we cannot write them out in a systematic list the way we can write out 1/n, 2/n, etc.
 

RaynorWolfcastle

Diamond Member
Feb 8, 2001
8,968
16
81
FWIW, the easiest way to (unrigorously) define countable sets is whether you can establish a 1 to 1 mapping from positive integer numbers to the set you are trying to count. In addition, any finite union, multiplication, or intersection of countable sets is also countable.

Integers are countable. Also, rationals are countable since they can be viewed as a 2-D array integers which is a form of multiplying sets (albeit in a particular form); thus rationals are also countable. There are much more formal ways of describing these concepts, but this description is generally sufficient for people who are just looking for a quick and simple intuitive iunderstanding of why rationals are countably infinite.
 

zugzoog

Senior member
Jun 29, 2004
447
0
0
Originally posted by: WinstonSmith
Given that you select two numbers at random

Even x even = even
Even x odd = even
odd x odd = odd

Evens outnumber odds by 2 to 1?


:p


This argument is so flawed.

The first is that one scenario has been forgotten: namely Odd x even = even.

You have also assumed that there are an equal number of odd and even numbers to prove that there are a greater number of even numbers.......

You are stating that the number of even numbers are greater that odd numbers from the results of a calculation? This is plain wrong. to give an example....

24 will result from

Even x even = even four instances (2*12,4*6,6*4,12*2)
Even x odd = even 2 instances (8*3,24*1)
Odd x even = even 2 instances (1*24,3*8)

23 will result from

odd x odd = odd 2 instances (1*23,23*1)

So you will be counting 24 a total of 8 times, but 23 only twice...