Proof Question (1 Viewer)
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bobthebuilder94
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can i just ask where are you getting these questions from? like a textbook? or are these just past papers or what?
apollo1
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past papers. this one is a common question because its a proof. but im just getting all these questions out of past CSSA and independent papers.
largarithmic
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The second part goes something like this:
a = n!e - (some integer). Just expand n! (1+1/1! + 1/2! + ... + 1/n!) and its obviously an integer, right? lets call this integer F[n].
Now suppose e were rational; e = p/q in lowest form (i.e. the gcd of p and q is 1). Then you have:
}{q})
So clearly the denominator of a is AT MOST q (because there could be a common factor with the numerator which cancels).
Note this is true regardless of the choice of n: so as we increase n to infinity, the denominator of a is bounded. That clearly however contradicts 0 < a < 1/n right? Becuase you can choose n large enough (in fact just choose n > q) to produce a contradiction.
a = n!e - (some integer). Just expand n! (1+1/1! + 1/2! + ... + 1/n!) and its obviously an integer, right? lets call this integer F[n].
Now suppose e were rational; e = p/q in lowest form (i.e. the gcd of p and q is 1). Then you have:
So clearly the denominator of a is AT MOST q (because there could be a common factor with the numerator which cancels).
Note this is true regardless of the choice of n: so as we increase n to infinity, the denominator of a is bounded. That clearly however contradicts 0 < a < 1/n right? Becuase you can choose n large enough (in fact just choose n > q) to produce a contradiction.