Mathematics for Computer Science

“mcs” — 2017/3/3 — 11:21 — page 143 — #151
5.3. Strong Induction vs. Induction vs. Well Ordering 143
Proof. We use induction. Let P.n/ be the proposition that 2 C 3 C 4 C C n D
n.n C 1/=2.
Base case: P.0/ is true, since both sides of the equation are equal to zero. (Recall
that a sum with no terms is zero.)
Inductive step: Now we must show that P.n/ implies P.n C 1/ for all n 0. So
suppose that P.n/ is true; that is, 2 C 3 C 4 C C n D n.n C 1/=2. Then we can
reason as follows:
2 C 3 C 4 C C n C .n C 1/ D Œ2 C 3 C 4 C C n C .n C 1/
n.n C 1/
D C .n C 1/
2
.n C 1/.n C 2/
D
2
Above, we group some terms, use the assumption P.n/, and then simplify. This
shows that P.n/ implies P.n C 1/. By the principle of induction, P.n/ is true for
all n 2 N.
Where exactly is the error in this proof?
Homework Problems
Problem 5.8.
The Fibonacci numbers F .n/ are described in Section 5.2.2.
Prove by induction that for all n 1,
F .n 1/ F .n C 1/ F .n/ 2 D . 1/ n : (5.8)
Problem 5.9.
For any binary string ˛ let num .˛/ be the nonnegative integer it represents in binary
notation. For example, num .10/ D 2, and num .0101/ D 5.
An n C 1-bit adder adds two n C 1-bit binary numbers. More precisely, an
n C 1-bit adder takes two length n C 1 binary strings
˛n WWD a n : : : a 1 a 0 ;
ˇn WWD b n : : : b 1 b 0 ;
and a binary digit c 0 as inputs, and produces a length-(n C 1) binary string
n WWD s n : : : s 1 s 0 ;