- #1

Brad_Ad23

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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?

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- Thread starter Brad_Ad23
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Brad_Ad23

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- #2

mathman

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It's trivially obvious. Why do you ask?

- #3

Brad_Ad23

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- #4

Integral

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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.

- #5

Brad_Ad23

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Yes that is the proofs I came up with...different notation but same message.

- #6

HallsofIvy

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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.

- #7

Brad_Ad23

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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.

- #8

STAii

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Actually, the sum of ANY two odd numbers is an even number (wether they are primes or not).It is indeed obvious the sum of any two odd primes is even

Look at integral's proof.

You can adjust Integral's proof to proove this.Is there a law or theorem somewhere that states the difference between any two odd numbers is even?

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).

- #9

Brad_Ad23

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Merci

- #10

marcus

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Originally posted by Brad_Ad23

Merci

LOL

"It is, after all, the Goldbach Conjecture," is a great line.

I suspect that you set up this thread with that line

in mind from the start

in order to have an opportunity to deliver it.

Perhaps I'm easily amused today but find it difficult

to stop chuckling at this thread.

- #11

Brad_Ad23

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- #12

Messiah

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Originally posted by Brad_Ad23

The difference between the number êÄ and the number Ç{ is DEFINITELY ODD (don't you think?)

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