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Proof of $1+2+3+\cdots+n = \frac{n(n+1)}{2}$ by strong induction: Using strong induction here is completely unnecessary, for you do not need it at all, and it is only likely to confuse people as to why you are using it. It will proceed just like a proof by weak induction, but the assumption at the outset will look different; nonetheless, just ...
91. With simple induction you use "if p(k) p (k) is true then p(k + 1) p (k + 1) is true" while in strong induction you use "if p(i) p (i) is true for all i i less than or equal to k k then p(k + 1) p (k + 1) is true", where p(k) p (k) is some statement depending on the positive integer k k. They are NOT "identical" but they are equivalent.
So the steps are. Prove that, if n> 2 n> 2 and every number m m with 2 ≤ m <n 2 ≤ m <n is a product of primes, then also n n is a product of primes. The base case is OK, as 2 2 is prime. Suppose n> 2 n> 2. If n n is prime, we're done. Otherwise n = ab n = a b, with 1 <a <n 1 <a <n and 1 <b <n 1 <b <n.
This also helps draw the distinction between proofs by strong induction and proofs by regular induction, specifically that the latter need a base and an inductive step, while the former only needs an inductive step (but may require special cases).
Bogus Proof by Strong Induction. 1. Strong mathematical induction without basis step. 1.
The proof that strong induction implies regular induction uses the fact that a natural number is either 0 0 or a successor (something that is usually proven from Peano's five axioms using weak induction). So you need that additional property to ensure that weak induction suffices. – Arturo Magidin. Mar 15 at 14:32.
I am having a great deal of trouble grasping the essential difference between Simple induction and Strong Induction proof. I was hoping someone here would do me the honor of making an example. First illustrating simple induction then the same example using strong induction and then point out the essential difference between these two procedures.
Structural Induction proof on binary search trees 2 Demonstrate that every positive integer can be expressed as the sum of distinct non-negative integer powers of 2
It is mainly a matter of presentation. When you speak of a proof by strong induction, it is mainly to bring to the reader's attention that the induction step will not only consider one but potentially several step backwards. Personally I never liked the distinction between strong and simple induction, for me they are the same.
The hockey-stick identity is extremely useful, and you’ll find three different proofs at the link, including a combinatorial proof and a proof by induction. For the proof by induction that’s suggested in your hint, your induction hypothesis should be P(n, k), i.e., that. n ∑ j = 1k − 1 ∏ i = 0(j + i) = 1 k + 1 k ∏ i = 0(n + i).