Solution. The proof will follow by induction on the n. We
prove the base case,
n = 1, follows because n^5 - n = 1^5 - 1 = 0 is divisible by 5.
Now we assume that k^5-k is divisible by 5 for some
positive integer k. Consider,
The first term is divisible by 5 because of the induction
hypothesis and the second
term is divisible by 5 because is contains a factor of 5. Thus the sum is
by 5. This proves the induction hypothesis and completes the proof.
The proof is by induction on n. Base. (note that there are two base cases since the recurrence uses two
previous values f (n - 1) and f(n - 2)).
and we have f(0) = 0 by
definition of f.
and we have f(1) = 1 by
definition of f.
Inductive Step. Assume n > 1. Assume
for all j such
that 0 ≤ j < n. Let IH(j) be the statement so
we are assuming
IH(j) for 0 ≤ j < n. We need to show that
Since n > 1, we can use the recursive definition of f to
f(n) = 5f(n - 1) + 6f(n - 2).
We can use IH(n-1) and IH(n-2) to rewrite f(n-1) and
f(n-2). So from the
inductive hypothesis we get.
which is what we needed to show.
Problem 6. Define the following recurrence
Solution. Proof by induction on n. Note that there
will need to be two base cases
for this induction proof. One (n = 0) is not sufficient because then our
inductive step would have to cover 1, . . . , n. This presents a problem because our
is not defined for n = 1. In general when there are two recursive references to
function like F , namely F(n - 1 and F(n - 2), two base cases are required for a
proof by induction.
Base Step. n = 0. By definition
again we have
Inductive Step. Let n≥2. Assume
the inductive hypothesis. Our goal is to show that this assumption implies that
Consider the recursive definition of F.
by the inductive hypothesis. Note the inequality , and that
here we have applied the
inductive hypothesis twice-once for F(n - 1) and once for F(n - 2). Continuing
Recall that we are deaing with n≥2. In this case we have
that 3n - 4≥n. Of
course, this can also be proved by induction.
as desired. By the principle of mathematical induction we
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