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**Kazy****Member**- Registered: 2006-01-24
- Posts: 37

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 n² + 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?

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

assume cube root of 2 is rational.

Then ³√2 = m/n where gcd(m, n) = 1, so 2 = m³/n³ and m³=2n³. Thus, m³ is even so m is even.

Since m is even, let m = 2k. Then m³ = 8k³. So 8k³ = 2n³. So 4k³ = n³ and n³ 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 n² + 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

41*41 + 41 + 41 = 41(41 + 1 + 1). So 41 | n² + 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..."

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**Kazy****Member**- Registered: 2006-01-24
- Posts: 37

Thanks!

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