# difference between any two odd numbers is even

Tags: difference, numbers
 P: 499 Is there a law or theorem somewhere that states the difference between any two odd numbers is even? Or the difference between 2 even numbers is even?
 P: 499 Well its also trivially obvious that the sum of any two odd primes yields an even integer over 4, yet that has not been proven yet. I just wanted to make sure that these were because someone keeps telling me they aren't. Luckily in the meanwhile I came up with my own proof for them, so it is all good.
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## difference between any two odd numbers is even

Let m= 2n+1
Let r= 2s+1

r and m are arbitrary odd numbers(1 greated then an even number)

so
r+m = (2n +1)+ (2s+1)= 2n+2s+2 = 2(n+s+1)

r+m is even.
QED
You can do something similar for 2 even numbers.
 P: 499 Yes that is the proofs I came up with...different notation but same message.
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PF Gold
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 Well its also trivially obvious that the sum of any two odd primes yields an even integer over 4, yet that has not been proven yet.
What's not trivially obvious is what you MEAN by "an even integer over 4". If you mean, literally, "an even integer divided by 4" then it doesn't make sense because the sum of two odd primes certainly has to be an INTEGER, not a half. If you mean "divisible by 4" then it's not true: 3+ 7= 10.
The fact that the sum of two odd PRIMES is even is "trivially obvious" because the sum of two odd numbers is always even. That certainly has been proven, in fact, I've seen it given as an exercise in high-school algebra texts.
 P: 499 I forgot that mathematicians need to be exact here. Every even integer greater than 4 is the sum of two odd primes. It is indeed obvious the sum of any two odd primes is even (after all, they are just special odd numbers), but that seems to be obvious as well at first glance, but not proven. It is, after all, the Goldbach Conjecture.
P: 353
 It is indeed obvious the sum of any two odd primes is even
Actually, the sum of ANY two odd numbers is an even number (wether they are primes or not).
Look at integral's proof.
 Is there a law or theorem somewhere that states the difference between any two odd numbers is even?
You can adjust Integral's proof to proove this.
Let A be an odd number, B another odd number
A = 2k + 1
B = 2n + 1
(where both k and n are integers)
A - B = (2k + 1) - (2n + 1) = 2k-2n + (1-1) = 2(k+n)
Since k and n are integers, k+n is an integer too, and A-B is even (since it can be expressed as 2*integer).
 P: 499 Merci
Astronomy
PF Gold
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