mersenneforum.org There are more even numbers than odd numbers
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 2020-08-03, 05:48 #1 retina Undefined     "The unspeakable one" Jun 2006 My evil lair 52·229 Posts There are more even numbers than odd numbers Every odd number is half of an even number. Some even numbers are not double an odd number. Therefore there are more even numbers than odd numbers.
 2020-08-03, 08:26 #2 Nick     Dec 2012 The Netherlands 22×359 Posts Now you understand why number theorists introduce fractional ideals.
2020-08-03, 08:51   #3
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

52·229 Posts

Quote:
 Originally Posted by Nick Now you understand why number theorists introduce fractional ideals.
https://en.wikipedia.org/wiki/Fractional_ideal

I don't understand?

 2020-08-03, 09:47 #4 xilman Bamboozled!     "πΊππ·π·π­" May 2003 Down not across 2·32·569 Posts There is precisely one more even number than there are odd numbers. If a number is odd, so is its negative. All other numbers are even. However the negative of zero is itself zero. Consequently, it does not have a negative counterpart and is the sole exception mentioned above.
 2020-08-03, 09:56 #5 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 35508 Posts I know you are kidding (countable infinities having 1 to 1 relations and all) but I will bite. For every Even number m.2^n for integers m & n where m is odd, there exists one distinct odd integer m^n sooooo they are equal. I am sure there are other equally invalid logics which will result in there being more odd numbers than even ones but can't think of any just yet. ETA scrap that that only is distinct off m is not a power. Last fiddled with by a1call on 2020-08-03 at 10:01
2020-08-03, 10:15   #6
xilman
Bamboozled!

"πΊππ·π·π­"
May 2003
Down not across

2×32×569 Posts

Quote:
 Originally Posted by xilman There is precisely one more even number than there are odd numbers. If a number is odd, so is its negative. All other numbers are even. However the negative of zero is itself zero. Consequently, it does not have a negative counterpart and is the sole exception mentioned above.
I apologise. There is an error in statement of the theorem given above.

It should have read "There is precisely one fewer even numbers than there are odd numbers." The proof itself remains unchanged.

Last fiddled with by xilman on 2020-08-03 at 10:17

2020-08-03, 12:54   #7
a1call

"Rashid Naimi"
Oct 2015
Remote to Here/There

76816 Posts

Quote:
 Originally Posted by a1call I know you are kidding (countable infinities having 1 to 1 relations and all) but I will bite. For every Even number m.2^n for integers m & n where m is odd, there exists one distinct odd integer m^n sooooo they are equal. I am sure there are other equally invalid logics which will result in there being more odd numbers than even ones but can't think of any just yet. ETA scrap that that only is distinct off m is not a power.
Let's make another go at this:

For every Even number m^a*2^n for integers m, a & n where m is odd, there exists two distinct odd integer m^n-/+m^a, sooooo there are twice as many odd numbers as there are even ones.

Counterexamples are likely/appreciated.

2020-08-03, 13:50   #8
kriesel

"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest

23·19·29 Posts

Quote:
 Originally Posted by retina Some even numbers are not double an odd number.
Do tell. Give a list of five of them that are not immediately preceded by an odd number each.

2020-08-03, 13:55   #9
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

52·229 Posts

Quote:
 Originally Posted by kriesel Do tell. Give a list of five of them that are not immediately preceded by an odd number each.
4, 8, 16 ,32, 64

 2020-08-04, 01:06 #10 JeppeSN     "Jeppe" Jan 2016 Denmark 2·7·11 Posts I do see some correct statements above. They should be moved out of this crackpot subforum, by someone. Am I being trolled? /JeppeSN
 2020-08-04, 14:20 #11 storm5510 Random Account     Aug 2009 U.S.A. 2·751 Posts All of this seems to make something very simple into something very complex. It is not. If you take them in pairs, (one of each type), the count will be the same for odds and evens, if the counting stops on an even. (1,2)(3,4)(5,6)(7,8) and so on.

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