Math Is Fun Forum

EMPhillips1989

Member

Registered: 2008-01-21

Posts: 40

proof by contradiction

let x be a real numer such that

and that for any n there is a natural number N such that

explain carefully what x must be

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: proof by contradiction

Assume x>0.

Then your second inequality can be rewritten as n < 1/x, for all natural n.
However, the Archimedes Postulate says that for any real number, a natural number exists that is bigger than it.
1/x is a real number, so there exists an n such that n > 1/x. This contradicts what we had before, so x>0 is false.

Therefore, x=0.

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Why did the vector cross the road?
It wanted to be normal.