We are trying to prove that a proposition P(n) is true for all n≥0 using mathematical induction. We have proven the Base Case P(0) and also proven that P(k)→P(k+1) for all k≥0 We could have proven the exact same result by proving that: a) P(k−1)→P(k) for all k≥−1 b) P(k)→P(k+1) for all k≥−1 c) P(k+1)→P(k+2) for all k≥−1 d) P(k+1)→P(k+2) for all k≥1

We are trying to prove that a proposition P(n) is true for all n≥0 using mathematical induction. We have proven the Base Case P(0) and also proven that P(k)→P(k+1) for all k≥0 We could have proven the exact same result by proving that: a) P(k−1)→P(k) for all k≥−1 b) P(k)→P(k+1) for all k≥−1 c) P(k+1)→P(k+2) for all k≥−1 d) P(k+1)→P(k+2) for all k≥1

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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For n ≥ 2 , Prove that (2 choose 2 ) + (3 choose 2 ) + (4 choose 2 ) + ....... + ( n choose 2 ) = ( n+1 choose 3 ) [Hint: Use induction and Pascal's rule.]
True or false?
a) For every positive integer n, n! is defined to be n! = 1 · 2 · 3 · … · n. Use mathematical induction to show that 1 ·1! + 2 ·2! + 3 ·3! + … + n · n! = (n + 1)! − 1 for all positive integers. (Hint: note that (n + 1)! = n!(n + 1).) b) Use mathematical induction to show that the product of three consecutive positive integers is always divisible by 6.
Suppose that P(n), Q(n) and R(n) are statements about the integer n. Let S(n) be the statement: “If P(n) is true then Q(n) and R(n) are both true." (a) The contrapositive of S(n) is the statement Olf Q (n) and R(n) are both true then P(n) is false. Olf Q (n) and R(n) are both true then P(n) is true. Olf at least one of Q(n) and R(n) is false then P(n) is false. Olf Q (n) and R(n) are both false then P(n) is false. Olf at least one of Q(n) and R(n) is false then P(n) is true. Olf P(n) is false then at least one of Q(n) and R(n) is false. Olf Q (n) and R(n) are both false then P(n) is true. ONone of the other answers. (b) The statement: "If Q (n) and R(n) are both true then P(n) is true." is OThe converse of S(n). OThe contrapositive of S(n). OThe converse of the contrapositive of S(n). ONone of the other answers.
In this problem, we will prove by induction that all cats are black. Let P(n) be the proposition, "In any group of n cats, if at least one of the cats in the group is black, then all of the cats in the group are black." Claim: (Vn e N)P(n) holds. Proof: (BC) P(1) is clearly true. (IH) Fix n E N and assume P(n) is true. (IS) We prove P(n + 1). Suppose we have a group of n + 1 cats that includes at least one black cat. Call this cat Midnight. Now, fix a different cat in the group. Call this cat Snow. Remove Snow from the group. We are left with a group of n cats that includes a black cat (namely, Midnight). By our inductive hypothesis, all cats in this group are black. Now remove Midnight and return Snow to the group. Again we have a group of n cats. All the cats in this group are black, except perhaps for Snow. Applying our inductive hypothesis to this group, we see that in fact all cats in this group must be black, including Snow. Hence all n+1 of the cats in our original group are…
If the contradiction method is used to prove the assertion "For any positive integer n > 0, if a" is even, then a is even.", then a valid implementation of that method begins by assuming that the negation of the statement is true. Choose all of the following statements that correctly express the negation of the above assertion. Vn E N – {0}, rem(a?, 2) = 0 → rem(a, 2) = 0 Vn EN – {0}, rem(a², 2) = 0 rem(a, 2) = 1 En e N – {0}, rem(a?, 2) = 0 → rem(a, 2) In eN - {0}, rem(a², 2) = 0 ^ rem(a, 2) = 1 For any positive integer n > 0, if a" is even, then a is odd. There exists a positive integer n > 0 where a" is even implies that a is od. There exists a positive integer n > 0 where a" is even and a is odd.