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#1 2006-04-12 14:34:38




I need to prove that the cubed root of 2 is irrational. How would I go about that?

Also give this statement: "If n is any positive integer, then n + n + 41 is always a prime number." I need to either prove it correct or give a counterexample. From looking at it I thought there was no way that was true, however every number I plug in works out so i'm guessing its probably true. However i have no idea how to prove that it is true. Any help?

#2 2006-04-13 00:43:23



Re: Proofs

assume cube root of 2 is rational.

Then √2 = m/n where gcd(m, n) = 1, so 2 = m/n and m=2n.  Thus, m is even so m is even.

Since m is even, let m = 2k.  Then m = 8k.  So 8k = 2n.  So 4k = n and n must be even, so n is even.

But then both m and n are even, so the gcd(m, n) is not 1.  Contradiciton.

For the second one, let's do an informal proof.  If n + n + 41 is always prime for any natural number n, then we have a way to calculuate values of primes that are infinitely large.  We don't have such a way, therefore the statement is not true.

Now if only math teachers would accept that argument wink

41*41 + 41 + 41 = 41(41 + 1 + 1).  So 41 | n + n + 41 when n=41

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

#3 2006-04-13 04:28:54



Re: Proofs


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