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O'Factry

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Every natural number greater than 1 is a sum of two lesser natural numbers. There are three possiblilities in addition; an even plus an even equals an even, an even plus an odd equals an odd, and an odd plus an odd equals an even. In two of three cases the result is an even number. Therefore there are twice as many even numbers as odd numbers.
 

StrugglingSceptic

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O'Factry said:
Every natural number greater than 1 is a sum of two lesser natural numbers. There are three possiblilities in addition; an even plus an even equals an even, an even plus an odd equals an odd, and an odd plus an odd equals an even. In two of three cases the result is an even number. Therefore there are twice as many even numbers as odd numbers.
You might be right. Give me a sec; I'm going to count them.
 
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Tangnefedd

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O'Factry said:
Every natural number greater than 1 is a sum of two lesser natural numbers. There are three possiblilities in addition; an even plus an even equals an even, an even plus an odd equals an odd, and an odd plus an odd equals an even. In two of three cases the result is an even number. Therefore there are twice as many even numbers as odd numbers.

I suffer from dyscalculia that is like dyslexia only with numbers, so I hope you don't think I am rude when I say that I haven't a clue what you are talking about, LOL!!!! :clap:
 
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StrugglingSceptic

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ThePhoenix said:
There are four cases
Odd+Odd
Odd+Even
Even+Odd
Even+Even

Four chances, two make odd, two make even. That one was actually a big puzzle to me for a while until I figured it out.
What does it have to do with "how many" odd and even numbers there are? I'm assuming the OP is a joke until told otherwise.
 
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ThePhoenix

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StrugglingSceptic said:
What does it have to do with "how many" odd and even numbers there are? I'm assuming the OP is a joke until told otherwise.
Well lets assume that two numbers were chosen at random. For the proportionality of Odds to Evens to be correct the sum of the two numbers must have a 50/50 chance of being odd or even as the number of random choices approaches infinity (this is called a limit). For each random number selection, those are the four possibilities. If the ratio of odds to evens is 1:1 then each has an equal possibility of occuring (note that this does not demonstrate that the ratio of odds to evens is one, it simply proves that there is no inherint contradiction to the ratio being one). I believe that the proof has to do with divisibility by 2, and the fact that for any non-one divisor if n is divisible by x then n+1 is not evenly divisible by x.
 
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StrugglingSceptic

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ThePhoenix said:
Well lets assume that two numbers were chosen at random. For the proportionality of Odds to Evens to be correct the sum of the two numbers must have a 50/50 chance of being odd or even as the number of random choices approaches infinity (this is called a limit). For each random number selection, those are the four possibilities. If the ratio of odds to evens is 1:1 then each has an equal possibility of occuring (note that this does not demonstrate that the ratio of odds to evens is one, it simply proves that there is no inherint contradiction to the ratio being one). I believe that the proof has to do with divisibility by 2, and the fact that for any non-one divisor if n is divisible by x then n+1 is not evenly divisible by x.
This is an interesting way of looking at it, and intuitive in some respects, but I'm pretty sure it's nonsense (unless my number theory course skipped it). As everyone else is saying in this thread, the set of odd numbers is infinite and the set of even numbers is infinite, and both sets are countable.
 
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StrugglingSceptic

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ThePhoenix said:
The fact that two systems are infinite does not mean that you can't find the ratio between them.
There is no such thing as a ratio between sets. "Countable" is a technical name to describe sets, and doesn't really have anything to do with counting things up.
 
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caravelair

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the easiest way to prove that there are the same number of evens and odds is to find a one-to-one and onto mapping from the evens to the odds. f(x) = x + 1 is such a mapping.

similarly, f(x) = 2x is a one-to-one and onto mapping from the natural numbers to the evens, showing that these 2 sets have the same cardinality!
 
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O'Factry

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This reminds me of the Marilyn VonSavaunt (sp) controversy regarding the "Let's make a deal!" problem. As stated by Ms VS, Monty gives you a choice of three doors, one of which contains a fabulous new car, and the other two contain donkeys and rocks and such. You make a choice, and Monty has one of the other doors opened to reveal a pile of coal. Now he gives you a choice, of whether to stay with your original pick, or switch. Ms. VS (the world's most intelligent woman, according to her) says you stand a better chance of winning the car if you change your decision. I (having barely passed differential equations) have to disagree. I say your chances are even. It is obvious to me that Monty is not going to reveal the car, so you know it is behind one of the two doors. It makes no difference what kind of shell gave has transpired in the past, you are now faced with the choice of two doors, one of which contains the corvette. 50/50 says I. What sez youze?
 
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