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Have you done mathematical induction before, like -Base case -hypothesis for case n -proof for case n+1
i has not mate
so hows do i do that thingy majig your talking about
are you there my fellow good sir
Yes, I was looking for a good reference for you. The following link shows us how to do mathematical inductions, with examples. Example 4 is particularly similar to your problem. http://www.analyzemath.com/math_induction/mathematical_induction.html
ok thnx mate i will chat you up if i have any problems
Good, way to go! :)
Statement P (n) is defined by n 3 + 2 n is divisible by 3 STEP 1: We first show that p (1) is true. Let n = 1 and calculate n 3 + 2n 1 3 + 2(1) = 3 3 is divisible by 3 hence p (1) is true. STEP 2: We now assume that p (k) is true k 3 + 2 k is divisible by 3 is equivalent to k 3 + 2 k = 3 M , where M is a positive integer. We now consider the algebraic expression (k + 1) 3 + 2 (k + 1); expand it and group like terms (k + 1) 3 + 2 (k + 1) = k 3 + 3 k 2 + 5 k + 3 = [ k 3 + 2 k] + [3 k 2 + 3 k + 3] = 3 M + 3 [ k 2 + k + 1 ] = 3 [ M + k 2 + k + 1 ] Hence (k + 1) 3 + 2 (k + 1) is also divisible by 3 and therefore statement P(k + 1) is true.
ok so mate i need more explanation on this it so short worded and down righ confusing
@mathmate could you help me for a sec mate
The first step is to establish that the statement you want to prove is true for some n. So P(1)=n^3+2n=1^3+2(1)=3 is divisible by three. In principle, if we have time to show this for all integers, we're done, but we cannot, because n goes to infinity So the next best thing is to show that, if 3|P(n), then 3|P(n+1). 3|x reads 3 divides x. Why ? Because if the hypothesis 3|P(n) is true, then 3|P(n+1) is true, then 3|P(n+2) is true, and so on. Once we have achieved that, we go back to say that 3|P(1), therefore 3|P(2), 3|P(3)... without having to do ALL of the integers.
ok mate so i get that so how di i do part two of my fiding my answer mate
you there mate
anybody of service here mate
f n is even, then n = 2m for some integer m. then substitute n^2 - n + 2 = (2m)^2 - (2m) + 2 =4m^2 - 2m + 2 = 2 ( 2m^2 - m + 1) . and this is a factor of 2 by definition, since it is 2 times some integer , if n is odd , then n = 2r+1 for some integer r n^2 - n + 2 = (2r + 1) ^2 - (2r + 1) + 2 = 4r^2 + 4r + 1 - 2r -1 + 2 by foiling and distributing = 4r^2 + 2r + 2 = 2 ( 2r^2 + r + 1) . and this is a factor of 2 again. we are done :) ok i tried my best and this is what i got is it true
This is a correct direct proof using "proof by cases". The question asks for proof by mathematical induction, which involves the three steps: 1. Proof a base case (say, n=1). 2. Inductive hypothesis: assume statement is true for case n=k. 3. Inductive proof and conclusion: prove that statement is true for case n=k+1. Done. The steps are extremely similar to example 4 that you have studied. Just have to replace the statement with yours. 1. base case : 2 is a factor of f(n)=n2 - n + 2 when n=1. Well, n^2-n+2 = f(1)= 1^2-1+2 = 2, so 2 is a factor of 2, hence statement is true for n=1 (base case). 2. Inductive hypothesis: Assume that (or IF) statement is true for case n=k: 2 is a factor of n2 - n + 2 for n=k, therefore 2 is a factor of f(k)= k^2-k+2, or f(k)=2m, where m\(\in\)Z 3. Work on case n=k+1 f(k+1)=(k+1)^2-(k+1)+2 = k^2+2k+1-k-1+2 =(k^2-k+2) +2k =(2m)+2k =2(m+1), hence 2 is a factor of f(k+1), given that 2 is a factor of f(k). we have just shown that 2 is a factor of f(k+1), which completes the proof by mathematical induction that 2 is a factor of f(n) for all n\(\in\)Z+.