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- his teacher suggested proof by induction
i gave a solution to this problem a few semesters ago in some math class, but my professor didnt like it <.< im going to post it to see what others think. one sec let me see if i can find it.
i saw what was wrong with my answer, never mind. I would just say that it is impossible to solve the equation:\[2n=2m+1\]with natural numbers n and m, so its impossible to have a natural number that is both even and odd. Not too sure how you would do that my induction, since a number being even or odd doesnt really depend on previous numbers being even or odd.
let n one number from set of natural numbers N ,so than odd even 2n+1 2n 1=2*0+1=1 2=2*1=2 3=2*1+1=3 4=2*2=4
Alrighty, lets give induction a go. First, we need to define even and odd. The standard definitions are as follows: An integer n is even if it can be expressed in the form n = 2k, where k is some integer An integer n is odd if it can be expressed in the form n = 2k + 1, where k is some integer We are now equipped to prove the statement via induction. We assume that the natural number n is either even or odd. We seek to show that this implies that the natural number (n+1) is also either even or odd. Case 1: n is an even number If n is an even number, then it can be expressed as n = 2k, with k some integer. Thus, n+1 = 2k+1, and by the definition of an odd number, n+1 is odd. Case 2: n is an odd number If n is an odd number, then it can be expressed as n = 2k + 1, with k some integer. Thus, n+1 = 2k+1 +1 = 2k+2 = 2(k+1). Since k is an integer, k+1 is an integer, and thus by the definition of an even number, n+1 is even. In either of these two cases, n+1 is either even or odd. Finally, we show that n= 1 is either even or odd. Since k=0 is an integer, note that 1 can be expressed as n = 2k+1 -> 1 = 2(0) + 1. Thus n is odd, and more generally, n=1 is either even or odd. By induction, we may say that any natural number n is either even or odd.
thnx guys - number theory is not my thing