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May 8th, 2018, 12:37 AM   #1
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Each even number is equal to zero

Dear Friends
I have come to know an interesting fact that every even number is equal to zero and every odd number is equal to 1; what's your view?

Last edited by skipjack; May 8th, 2018 at 05:06 AM.
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May 8th, 2018, 01:34 AM   #2
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Originally Posted by asitsikary View Post
Dear Friends
I have come to know an interesting fact that every even number is equal to zero and every odd number is equal to 1; what's your view?
Okay, I'll bite. This is true if we are working in $\displaystyle \mathbb{Z} _2$ (or anything isomorphic to it.) But something tells me you are talking about something different.

Please tell us why you are saying that.

-Dan
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Last edited by skipjack; May 8th, 2018 at 05:06 AM.
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May 8th, 2018, 03:22 AM   #3
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I think that it's completely uninteresting until you explain yourself properly.
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May 8th, 2018, 04:27 AM   #4
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Hmmm.....I wonder if any "integration" is involved....
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May 8th, 2018, 05:18 AM   #5
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I wonder how old he is? Zero or one?
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May 8th, 2018, 05:19 AM   #6
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Goodness knows.

But if $2a=2a+2$, then $a=a+1$ if dividing by $2$ makes any sense. So $0=1$.
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May 8th, 2018, 07:29 PM   #7
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Originally Posted by v8archie View Post
Goodness knows.

But if $2a=2a+2$, then $a=a+1$ if dividing by $2$ makes any sense. So $0=1$.
To elaborate on this, what it essentially proves is that any field satisfying the OP's claims must have characteristic 2. This is another way of saying that division by 2 must not make any sense since it leads to a contradiction as you point out.

While I know you are aware of the distinction, I want to point out that the usual computation doesn't assume anything about invertibility.
The argument goes like:

If $2a = 2a + 2$, then $2a = 2(a+1)$. Since it holds for any $a$, it must hold in the case that $a$ is a unit and it follows immediately from uniqueness of inverses that $2a = 0 = 2(a+1)$. In a field, every element is invertible, hence the field has characteristic 2. The argument does not require any mention of inverting 2.

In fact, everything in this argument goes through equally well for arbitrary local rings. In this case, the characteristic is replaced by the Jacobson radical, but otherwise, there is no reason to mention inverting elements or requiring more than just the ring structure.
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May 8th, 2018, 07:48 PM   #8
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Good comments, but I just wonder when we'll hear from asitsikary. Unless, of course, he's a troll.

-Dan
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May 8th, 2018, 10:25 PM   #9
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Originally Posted by topsquark View Post
Unless, of course, he's a troll.
Or a spammer (There's a writing that something was edited by skipjack, probably a link).
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