20200803, 05:48  #1 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}×1,447 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. 
20200803, 08:26  #2 
Dec 2012
The Netherlands
3^{2}·7·23 Posts 
Now you understand why number theorists introduce fractional ideals.

20200803, 08:51  #3  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
13234_{8} Posts 
Quote:
I don't understand? 

20200803, 09:47  #4 
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
10011101111110_{2} 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. 
20200803, 09:56  #5 
"Rashid Naimi"
Oct 2015
Remote to Here/There
17×113 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 20200803 at 10:01 
20200803, 10:15  #6  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
2×3×5×337 Posts 
Quote:
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 20200803 at 10:17 

20200803, 12:54  #7  
"Rashid Naimi"
Oct 2015
Remote to Here/There
17×113 Posts 
Quote:
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. 

20200803, 13:50  #8 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
2^{2}×3^{2}×127 Posts 

20200803, 13:55  #9 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}·1,447 Posts 

20200804, 01:06  #10 
"Jeppe"
Jan 2016
Denmark
2^{5}×5 Posts 
I do see some correct statements above. They should be moved out of this crackpot subforum, by someone. Am I being trolled? /JeppeSN

20200804, 14:20  #11 
Random Account
Aug 2009
U.S.A.
644_{16} 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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