Math 220 Workshop 4: Induction; graphs
1. Guess a formula for the sum of the first n odd numbers:
n
X
(2k − 1) = 1 + 3 + 5 + · · · + (2n − 1),
k=1
and prove your formula using induction.
Solution: If Sn =
Pn
k=1 (2k
− 1), then
2Sn =
n
X
(2k − 1) + (2(n − k + 1) − 1)
=
k=1
n
X
(2n) = 2n2
k=1
Therefore Sn = n2 . This is a perfectly good proof but lets verify it by induction anyways:
If n = 1 then S1 = 1 = 12 . So the result is true.
Assume Sn = n2 for n ∈ N. Then
Sn+1 = Sn + (2n + 1) = n2 + 2n + 1 = (n + 1)2 .
Therefore the result holds for n + 1. By Mathematical Induction the result holds for all natural
numbers n.
2. All Powers of 2 Are Equal to 1
We are going to prove by induction that,
For all integers n ≥ 0,
2n = 1
The claim is verified for n = 0; for indeed, 20 = 1.
Assume the equation is correct for all n ≤ k, that is
20 = 1, 21 = 1, 22 = 1, ..., 2k = 1.
From these we now derive that also 2k+1 = 1:
2k+1 =
22k
2k × 2k
1×1
=
=
=1
k−1
k−1
2
2
1
Induction is complete.
Where is the error?
Solution: When we carry out the inductive step to prove 21 = 1 ( which corresponds to k = 0 in
the proof), we need to make use of the ”inductive fact” that 2k−1 = 1, which is 2−1 = 1 for k = 0.
Since our inductive hypothesis is only that 20 = 1, we have not actually proved 2−1 = 1, and this
is precisely where the argument fails.
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3. Using induction, prove that the number written as 111...1 (with an even number of 1s) is always
divisible by 11.
Solution: We induct on k where n = 2k is the number of 1’s. Let ak be the number with 2k 1’s.
Consider first k = 1. Then a1 = 11 which is divisible by 11.
Assume now that ak is divisible by 11 and consider ak+1 . By the induction hypothesis ak = 11m
for some natural number m. We have
ak+1 = 100ak + 11 = 100(11m) + 11 = 11(100m + 1).
Therefore ak+1 is divisible by 11. By the Principle of Mathematical Induction, the result holds for
all natural numbers k.
4. Is it possible to draw 9 segments on the plane in such a way that every segment intersects precisely 3
other segments?
Solution: Let us construct a graph representing this situation: put a dot for every segment, and
connect two dots by an edge if the corresponding two segments intersect. Now if every segment
intersects precisely 3 others, then in our graph, every vertex has degree 3. Thus the answer is
”impossible”, and we can prove it by contradiction. Suppose we managed to construct such a
system of segments. Then the corresponding graph would have 9 vertices of degree 3, which is
impossible since the number of odd degree vertices in a graph has to be even.
5. In Wonderland, there are 100 cities, and every city has 50 roads leading into it. Prove that it is possible
to get from any city A to any city B in Wonderland by driving on these roads.
Solution: Since there are 50 cities connected by a road with A, and 50 cities connected with B,
there has to be a common city between these two sets (if all these cities were distinct, there would
have been at least 102 cities altogether!). Thus, it is possible to get from A to be B, and in fact,
with at most one intermediate city!
6. On Asteroid M220, there are a few houses, and every house has 10 paths leading to it. Suppose you
know that it is possible to get from every house to every other house by walking on a path. Now
suppose that one path got blocked by a landslide. Prove that it is still possible to get from every house
to every other house.
Solution: Let us make a graph with the vertices being houses, and the edges – the paths between
them. Then it is given that the degree of every vertex is 10 and our graph is connected. One edge
gets erased (the blocked path), and we need to prove that the graph is still connected. Suppose to
the contrary that it becomes disconnected. Then there are at least 2 connected components. Let
A and B be the vertices that the erased edge used to connect. Note that if it is still possible to get
from A to B somehow, (that is, if A and B are still in the same connected component after that
edge has been erased), then the graph is still connected: just any time the erased edge was used,
we can now use the long alternative way from A to B. So, if we assume that there is more than
one connected component, then A and B would have to belong to different connected components.
This would mean that, say, in the connected component that contains A, there are several vertices
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of degree 10 and one vertex of degree 9 (A itself, since after erasing one edge, A has only 9 left).
This contradicts the fact that the connected component is itself a graph, and thus can only have
an even number of vertices of odd degree.
Solution: Quiz.
Let {Fn } be the Fibonacci sequence. Show that for all n ∈ N,
n−1
X
F2i+1 = F2n .
i=0
Hint: Use induction. Don’t use the explicit formula from homework.
Pn−1
Let Sn = i=0 F2i+1 . Consider first n = 1. Then S1 = F1 = 1 = F2 and so the result is valid for
n = 1.
Assume that Sn = F2n and consider Sn+1 for n ∈ N. We have
Sn+1 = Sn + F2n+1 = F2n + F2n+1 = F2n+2 = F2(n+1) .
Therefore the result holds for n + 1. By the Principle of Mathematical Induction the result holds
for all n ∈ N.
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