Sample Problems. Lecture Notes Proof by Induction page 1. Prove each of the following statements by induction.

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1 Lecture Notes Proof by Induction page Sample Problems Prove each of the following statements by induction.. For all natural numbers n; n (n + ) a) ::: + n. b) ::: + n n (n + ) (n + ). c) ::: + n 3 n (n + ).. For all natural numbers n; ::: + n (n + ) n n For all natural numbers n; :::: + n n ( + (n ) n ).. For all natural numbers n 5; n n : 5. Prove that for all natural number n; 0 n can be written as a sum of two perfect squares.. Recall (F n ) is the Fibonacci sequence, de ned as F ; F ; and F n+ F n + F n+ Prove each of the following statements for all natural numbers n. a) F + F + F 3 + ::: + F n F n+ b) F + F + F 3 + ::: + F n F n F n+ c) F + F 3 + ::: + F n F n c copyright Hidegkuti, Powell, 0 Last revised: November 3, 03

2 Lecture Notes Proof by Induction page Sample Problems - Solutions. a) For all natural numbers n; ::: + n Part : We check for the rst few values of n. If n, then If n, then If n 3, then So the statement is true for n ; and 3. Part. n (n + ). LHS and RHS LHS + 3 and RHS LHS and RHS Suppose that k is a positive integer for which ::: + k Let us add k + to both sides. and the right hand side becomes k (k + ) RHS k + k + k + k (k + ) + (k + ) ( + ) ( + ) 3 (3 + ) ::: + k + (k + ) k (k + ) k + 3k + So we have proved that for all positive integers k, if 3 This is the inductional hypotheses (k + ) k (k + ) + (k + ) + (k + ) (k + ) (k + ) ((k + ) + ) then ::: + k ::: + k + (k + ) k (k + ) is true (k + ) ((k + ) + ) c copyright Hidegkuti, Powell, 0 Last revised: November 3, 03

3 Lecture Notes Proof by Induction page 3 b) For all natural numbers n; ::: + n Part : We check for the rst few values of n. If n, then If n, then If n 3, then LHS and RHS LHS + 5 and RHS n (n + ) (n + ). ( + ) ( + ) ( + ) ( + ) 30 5 LHS and RHS So the statement is true for n ; and 3. Part. Suppose that k is a positive integer for which 3 (3 + ) ( 3 + ) ::: + k k (k + ) (k + ) This is the inductional hypotheses Let us add (k + ) to both sides ::: + k + (k + ) and the right hand side becomes We factor out k + RHS k (k + ) (k + ) + (k + ) k (k + ) (k + ) + (k + ) RHS k (k + ) (k + ) k + + (k + ) k + k + k + k + k + (k (k + ) + (k + )) k + 7k + and so happens k + 7k + factors as (k + ) (k + 3). So we have RHS k + k + 7k + k + (k + ) (k + 3) (k + ) ((k + ) + ) ( (k + ) + ) So we have proved that for all positive integers k, if (k + ) (k + ) (k + 3) then ::: + k ::: + k + (k + ) k (k + ) (k + ) (k + ) ((k + ) + ) ( (k + ) + ) c copyright Hidegkuti, Powell, 0 Last revised: November 3, 03

4 Lecture Notes Proof by Induction page c) For all natural numbers n; ::: + n 3 n (n + ) Part : We check for the rst few values of n. If n, then If n, then If n 3, then LHS 3 and RHS ( + ) LHS and RHS ( + ) 3 9 LHS and RHS 3 (3 + ) 3 So the statement is true for n ; and 3. Part. Suppose that k is a positive integer for which ::: + k 3 k (k + ) This is the inductional hypotheses Let us add (k + ) 3 to both sides ::: + k 3 + (k + ) 3 and the right hand side becomes RHS k (k + ) + (k + ) 3 k (k + ) + (k + ) k + (k + ) (k + ) So we have proved that for all positive integers k, if then (k + )3 k + k ::: + k 3 k (k + ) factor out (k + ) is true (k + ) ::: + k 3 + (k + ) 3 (k + ) ((k + ) + ) (k + ) (k + ) (k + + ) c copyright Hidegkuti, Powell, 0 Last revised: November 3, 03

5 Lecture Notes Proof by Induction page 5. For all natural numbers n; Induction on n: Part. If n ; then If n ; then Part ::: + n (n + ) LHS LHS Suppose that k is a natural number such that n n +. and RHS + and RHS ::: + k (k + ) k k + (Induction hypothesis) We will add (k + ) (k + + ) (k + ) (k + ) to both sides ::: + k (k + ) + (k + ) (k + ) k k + + (k + ) (k + ) Let us simplify the right hand side k k + + (k + ) (k + ) k (k + ) (k + ) (k + ) + k (k + ) + (k + ) (k + ) (k + ) (k + ) k + k + (k + ) (k + ) (k + ) (k + ) (k + ) k + k + Thus we have that from the induction hypotheses ::: + k (k + ) k k + the statement ::: + (k + ) (k + ) k + k + follows. This completes our proof. 3. For all natural numbers n; :::: + n n ( + (n ) n ). Part. If n ; then LHS and RHS + ( ) If n ; then LHS + 0 and RHS + ( ) 5 0 Part. Suppose that k is a natural number such that :::: + k k + (k ) k (Induction hypothesis) Let us add (k + ) k+ to both sides. LHS :::: + k k + (k + ) k+ c copyright Hidegkuti, Powell, 0 Last revised: November 3, 03

6 Lecture Notes Proof by Induction page and the right-hand side becomes RHS + (k ) k + (k + ) k+ + (k ) k + (k + ) k + k k + k k+. For all natural numbers n 5; n n + (k ) k + (k + ) k factor out + k k k + k k + k Note: this is a very interesting example illustrating how induction works, how we need both parts to work together to form a logically sound proof. The statement seems to be true immediately at n but then surprisingly, it will be false for a few values of n. Part. If n ; then If n 3; then If n ; then If n 5; then If n ; then If n ; then LHS and RHS and > is true LHS and RHS and > is false! LHS 3 8 and RHS 3 9 and 8 > 9 is false! LHS and RHS and > is false! LHS 5 3 and RHS 5 5 and 3 > 5 is true LHS and RHS 3 and > 3 is true At this point we have a sense that the statement will stay true because the left-hand side doubles when we go from k to k + while the right-hand side just increases from k to (k + ). The proof in part will formalize this idea, that between the two types of growth, doubling is much faster. All we need to show is that moving from k to (k + ) is a smaller increment than doubling. In short, that (k + ) < k k + k + < k subtract k k + < k This should be easy to prove for most positive integers. We can either solve the quadratic inequality or be a little bit sloppy or generous and say that if k is greater than, then and if k is greater than 3, then < k add k k + < k + k k + < 3k 3 < k multiply by k > 0 3k < k c copyright Hidegkuti, Powell, 0 Last revised: November 3, 03

7 Lecture Notes Proof by Induction page 7 In short, if k > 3, then k + < 3k < k and so This will be the core of the proof in part. Part. k + < k add k to both sides k + k + < k (k + ) < k Suppose that k is a natural number such that k > k (Induction hypothesis) Let us multiply both sides by. LHS k k+ and the right-hand side becomes So we have RHS k If k > 3; then k > k k+ > k k+ > k k + k > k + 3k > k + k + (k + ) and so we have that if k > k is true AND k > 3, then k+ > (k + ). So we could easily prove the inheritance property for all integers greater than 3 but the statement itself is NOT true for n 3: Both components of the proof work together starting at n 5. statement inheritance to the next number n true false n false false n 3 false true (but we didn t prove it) n false true n 5 true true This table illustrates why induction only works here for n 5. For more documents like this, visit our page at and click on Lecture Notes. questions or comments to mhidegkuti@ccc.edu. c copyright Hidegkuti, Powell, 0 Last revised: November 3, 03

Factoring out the GCF and the Difference of Squares Theorem

Lecture Notes Factoring A page 1 Factoring out the GCF and the Difference of Squares Theorem Sample Problems 1. Completely factor each of the following. a) 3x 1 c) 3a 1 b) x y d) 3a 1a e) x 1 f) x + 1