Return HW
Pay John Stone $1, Ronney Sanchez $1, Matt LeBlanc $1
Discuss the issue about duality: the dual of “p and not-p”
Questions on sections 3.1 through 3.6?
Questions on 3.7?
Group work: 3.7.2
Theorem: If n ≥ 1, 1(1!) + 2(2!) + … + n(n!) = (n+1)! – 1.
Proof: Use mathematical induction.
Base case: 1(1!) = 1 = 2 – 1 = (1+1)! – 1.
Induction step: Suppose n is a natural number for which the
formula is true, so that
1(1!) + 2(2!) + … + n(n!) = (n+1)! – 1.
Then
1(1!) + 2(2!) + … + n(n!) + (n+1)(n+1)!
= ((n+1)! – 1) + (n+1)(n+1)!
= (1)(n+1)! + (n+1)(n+1)! – 1
= (1+n+1) (n+1)! – 1
= (n+2) (n+1)! – 1
= (n+2)! – 1
so that n+1 is a natural number for which the formula is
true.
Since we have verified the base case and the induction step,
the formula holds for all natural numbers.
Group work: 3.7.5
Theorem: If n ≥ 1, 1/(1)(2) + 1/(2)(3) + … + 1/(n)(n+1) =
n/(n+1).
Proof: Use mathematical induction (re-indexed for
convenience, as you’ll see).
Base case: 1/(1)(2) = 1/2.
Induction step: Suppose n–1 is a natural number for which
the formula is true, so that
1/(1)(2) + 1/(2)(3) + … + 1/(n-1)(n) = (n-1)/n.
Then
1/(1)(2) + … + 1/(n-1)(n) + 1/(n)(n+1)
= (n-1)/n + 1/(n)(n+1)
= (n-1)(n+1)/(n)(n+1) + 1/(n)(n+1)
= ((n-1)(n+1)+1)/(n)(n+1)
= n^2 / (n)(n+1)
= n/(n+1)
so that n is a natural number for which the formula is true.
Since we have verified the base case and the induction step,
the formula holds for all natural numbers.
Group work: 3.7.7
Claim: For all positive integers n, the number of strings of
n 0’s and 1’s that contain an even number of 1’s is 2^(n-1).
Proof: Let E(n) be the number of length-n bit-strings with
an even number of 1’s (so that 2^n – E(n) is the number of
length-n bit-strings with an od number of 1’s; that may
seem irrelevant now but it will be useful shortly).
We want to prove that E(n) = 2^(n-1) by induction.
The base case (n=1) is easy: There are just two strings of
length 1, “0” and “1”; only one of them contains an even
number of 1’s, so E(1) = 1, which agrees with the formula.
For the induction step: Suppose E(n) = 2^(n-1).
Let’s count the strings of length n+1 that contain an even
number of 1’s.
There are two kinds of strings of this type: those that have
an even number of 1’s in the first n positions (and then end
with a 0) and those that have an odd number of 1’s in the
first n positions (and then end with a 1).
Of the former, there are E(n); of the latter, there are 2^n –
E(n).
So E(n+1) = E(n) + (2^n – E(n)) = 2^n = 2^((n+1)-1),
which agrees with the formula.
Hence the claim is true for n+1.
Therefore by mathematical induction, the claim is true for
all positive integers n.
(Curiously, we never really use the induction hypothesis in
this proof! But that’s the exception, not the rule.)
Other questions on 3.7?
Collect homework and section summaries.