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Old 2020-08-03, 05:48   #1
retina
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Default 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.
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Old 2020-08-03, 08:26   #2
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Now you understand why number theorists introduce fractional ideals.
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Old 2020-08-03, 08:51   #3
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Quote:
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Now you understand why number theorists introduce fractional ideals.
https://en.wikipedia.org/wiki/Fractional_ideal

I don't understand?
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Old 2020-08-03, 09:47   #4
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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.
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Old 2020-08-03, 09:56   #5
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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
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Old 2020-08-03, 10:15   #6
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Quote:
Originally Posted by xilman View Post
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
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Old 2020-08-03, 12:54   #7
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Quote:
Originally Posted by a1call View Post
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.
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Old 2020-08-03, 13:50   #8
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Quote:
Originally Posted by retina View Post
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.
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Old 2020-08-03, 13:55   #9
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Quote:
Originally Posted by kriesel View Post
Do tell. Give a list of five of them that are not immediately preceded by an odd number each.
4, 8, 16 ,32, 64
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Old 2020-08-04, 01:06   #10
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I do see some correct statements above. They should be moved out of this crackpot subforum, by someone. Am I being trolled? /JeppeSN
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Old 2020-08-04, 14:20   #11
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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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