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Am I assuming too much here?

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- January 30th 2010, 07:07 PMdavismjproduct of rational and irrational is always irrational
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Am I assuming too much here? - January 30th 2010, 10:00 PMBlack
I don't understand why you're setting in your proof if is irrational.

To prove these, you just use the fact that is a*field*, i.e, closed under addition, subtraction, multiplication, and division.

If , then by closure, . Contradiction.

If , then by closure, . Contradiction. - January 31st 2010, 06:40 AMHallsofIvy
He did NOT assume . He showed, first, that

**if**you assume a/t is a rational number,**then**you arrive at t= r/s, a ration number and so a contradiction.

He then showed that**if**you assume at is rational, [b]then[b] you arrive at t= r/s, again a contradiction. His proof is perfectly valid.

Quote:

To prove these, you just use the fact that is a*field*, i.e, closed under addition, subtraction, multiplication, and division.

If , then by closure, . Contradiction.

If , then by closure, . Contradiction.