Hi! So Im not allowed to use induction on this one. How do i prove that n^2 + n is even when n is natural number? Not a clue on this one. Would you mind to help a bit?
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Originally Posted by veturi Hi! So Im not allowed to use induction on this one. How do i prove that n^2 + n is even when n is natural number? Not a clue on this one. Would you mind to help a bit? . Therefore one of the factors of is certain to be even. Therefore ....
oh my. so easy. Thanks a lot! I'm a bit ashamed tho...
A number can be represent in two forms: n=2k with or n=2k+1 with First case we have: so, for all even n. Second case: so, for all odd n. thus,
Originally Posted by veturi Hi! So Im not allowed to use induction on this one. How do i prove that n^2 + n is even when n is natural number? Not a clue on this one. Would you mind to help a bit? Hello : There is the solution I'have two cases: 1 case : 2 case:
Originally Posted by veturi Hi! So Im not allowed to use induction on this one. How do i prove that n^2 + n is even when n is natural number? Not a clue on this one. Would you mind to help a bit? By Contradiction: If is even there is nothing to prove, so assume is odd and is also odd. As is odd it can be written for some natural , then: Hence ... a contradiction. CB
Originally Posted by CaptainBlack By Contradiction: If is even there is nothing to prove, so assume is odd and is also odd. As is odd it can be written for some natural , then: Hence ... a contradiction. CB Hello : you have any contradiction , because I’can written : Conclusion : even number
Originally Posted by dhiab Hello : you have any contradiction , because I’can written : Conclusion : even number It is a contradiction, I assume it odd and find it even hence the assumption that odd and is also odd fails, so if is odd is not odd. CB. CB
other proof we know that 1+2+3+...+n= n(n+1)/2 so 2|n(n+1), n(n+1) is even
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