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July 7th, 2018, 04:08 PM   #11
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Quote:
 Originally Posted by Maschke I did not say that at all. A field is required to have $0 \neq 1$. I merely pointed out that mathman's third line was unjustified.
If $0 \neq 1$, then 1 is NOT the additive identity so $x+ 1\ne x$ is indeed justified.

July 7th, 2018, 04:17 PM   #12
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Quote:
 Originally Posted by Country Boy If $0 \neq 1$, then 1 is NOT the additive identity so $x+ 1\ne x$ is indeed justified.
Right. $x \neq x + 1$ is a CONSEQUENCE of $0 \neq 1$.

But mathman used $x \neq x + 1$ to attempt to PROVE that $0 \neq 1$. He assumed the thing he was trying to prove.

July 7th, 2018, 07:20 PM   #13
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Quote:
 Originally Posted by Maschke I did not say that at all.
I didn't say you did.

July 7th, 2018, 08:53 PM   #14
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Quote:
 Originally Posted by romsek I didn't say you did.
Sorry. Trying to think about 0 = 1 makes my head hurt.

It turns out, by the way, that there's a thing called "the field with one element." Such a thing does not exist; but many abstract areas of math act as if it "should" exist. I don't really understand much about this but here it is.

https://en.wikipedia.org/wiki/Field_with_one_element

This is a pretty good article. It gives the flavor of this amazing out-there stuff whether or not you follow the details. Nobody can construct or even define a field with one element, but if they had one it would do all kinds of great things, even perhaps solve the Riemann hypothesis.

Last edited by Maschke; July 7th, 2018 at 09:00 PM.

July 8th, 2018, 12:50 PM   #15
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Quote:
 Originally Posted by Maschke How do you get line three, $x+1\ne x$? Aren't you assuming the thing you're trying to prove? If $0 = 1$ then $x + 1 = x$ and there is no contradiction.
I guess I overdid it. $x\times 1=x, x\times 0=0$. Both will be true only if $x=0$. I don't think much is done with a field consisting of one element .

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