Download Math 2534 Solutions to Homework 6 Prove the following theorems

Math 2534 Solutions to Homework 6
Prove the following theorems using PMI
Theorem 1: n  N , 2  4  6  ...  (2n)  n2  n
Proof: I will verify that the hypothesis is true for at least one value of n  N
Consider n = 1, we see that 2 = (1)2+(1)
Now consider n = 2 and we see that 2 + 4 = (2)2+2
We will now assume that our hypothesis is true for each natural number between n = 1 up to
some arbitrary natural number k. ie. 2  4  6  ...  (2k )  k 2  k
My intent is to prove true for k+1. I will show that
2  4  6  ...  2(k  1)  (k  1)2  (k  1)  k 2  3k  2
Now for the body of the proof:
Consider the k+1 term below.
2  4  6  ...  2k  2(k  1) 
[2  4  6  ...  2k ]  2( k  1) 
[k 2  k ]  2(k  1) 
By the inductive assumption
k 2  3k  2
which is the form that I intended to produce to verify that the k+1 term works.
Therefore I have assumed true up to k and proved true for k+1, so the hypothesis is true for all
natural numbers.
Theorem 2: n N , 5n  1 is divisible by 4
Proof: I will verify that the hypothesis is true for at least one value of n  N
Consider n = 1, 51 – 1 = 4 and is therefore divisible by 4
Now consider n = 2 and we see that 52 – 1 = 24 which is divisible by 4.
We will now assume that our hypothesis is true for each natural number between n = 1 up to
some arbitrary natural number k. ie. 4 (5k  1)
k 1
that 4 (5
My intent is to prove true for k+1. I will show
 1)
For the body of the proof we will consider the k+1 term,
5k 1  1  5k 5  1  (5k  1)  (4)5k by the inductive assumption we know that 4 divides (5k  1)
k
and 4 also divides (4)5k since 4 is a factor in this term. Therefore 4 (5  1) .
I have assumed true up to k and proved true for k+1. The hypothesis is true for all natural
numbers by PMI.
Theorem 3: n  N  5, n2  2n
Proof: I will first verify that the hypothesis is true for at least one value of n.
Consider n = 5, we have that (5)2 < 25 which is 25 < 32
Now assume that the hypothesis is true up to some arbitrary value k. ie: k 2  2k
I will now prove true for k+1. ie: (k+1)2  2k 1
In the main body of the proof we will consider the k+1 term (k+1)2.
So
(k  1) 2  k 2  2k  1
< 2k  (2k  1) by the inductive assumption
< 2k  2k
Since for all natural numbers greater than 3, 2n + 1 < 2n
= 2(2k )  2k 1
Since I assumed true up to k and proved true for k+1, the hypothesis is true for all
natural numbers greater than 4.
Theorem 4: If a jigsaw puzzle has n pieces, then it is completed in n-1 fits.
(It takes one “fit” to join an additional piece to those that have already been assembled)
See solution in PMI Resource sheet 3 on the Class Information Page.