Math 283 Spring 2013
Presentation Problems Solutions
1. An integer is even if and only if its square is even.
Solution.
We have two items to prove here. First, if an integer is even, its square is even. Let m
be an even integer. This means that m = 2k for some integer k. Consider its square
m2 = (2k)2
= 4k 2
= 2(2k 2 )
Since 2k 2 is an integer, this shows that m2 is odd.
Now we must prove that if the square of an integer is even, then the integer itself was
even. We prove the contrapositive statement: If an integer is not even (i.e. odd), then its
square is not even (i.e. odd). Let m be an odd integer. This means m = 2k + 1 for some
integer k. Consider its square
m2 = (2k + 1)2
= 4k 2 + 2k + 1
= 2(2k 2 + k) + 1
Since 2k 2 + k is an integer, this shows that m2 is odd.
2. If m + n is an even integer, then either m and n are both odd integers or m and n are
both even integers.
Solution.
[By Contradiction] Suppose that m + n is an even integer and, by way of contradiction,
that m and n are neither both odd, nor both even. This means that one of m or n is even
and the other is odd. Notice that we can assume m is even since if n were the even number,
we could exchange the names of the variables. Thus, we can let m be even and n be odd.
(This is where we could say, instead, “Without loss of generality, let m be even and n be
odd.”). This means that m = 2k for some integer k and n = 2j + 1 for some integer j. Now
consider their sum
m + n = (2k) + (2j + 1)
= 2(k + j) + 1
This means that m + n is odd. However, m + n is supposed to be even. . Therefore, if
m + n is an even integer, then either m and n are both odd integers or m and n are both
even integers.
Notice that the proof of the contrapositive would be essentially the same.
3. If mn is an even integer, then either m or n is an even integer.
Solution.
[By Contrapositive] Consider the contrapositive statement, “if neither m nor n is an even
integer, then mn is not an even integer.” Now, let m and n be odd integers. This means
that m = 2k + 1 for some integer k and n = 2j + 1 for some integer j. Consider the product
mn = (2k + 1)(2j + 1)
= 4kj + 2k + 2j + 1
= 2(2kj + k + j) + 1
Since 2kj + k + j is an integer, mn is odd.
4. If x, y, and z are integers and x + y is odd and y + z is odd, then x + z is odd.
Solution.
This statement is incorrect, since x = 4, y = 5 and z = 2 is a counterexample. The
correct statement would be “then x + z is even”. This is proved below. Let x, y and z be
integers. Also, let x + y and y + z each be odd. This means that x + y = 2k + 1 for some
integer k and y + z = 2j + 1 for some integer j. Summing the two equations, we have
(x + y) + (y + z) = (2k + 1) + (2j + 1)
x + 2y + z = 2k + 2j + 2
x + z = 2k + 2j + 2y + 2
x + z = 2(k + j + y + 1)
Since k + j + y + 1 is an integer, x + z is even.
5. Two opposite corner squares are deleted from an eight by eight checkerboard. Prove that
the remaining squares cannot be covered exactly by dominoes (rectangles consisting of two
adjacent squares).
(Hint: Use the method of contradiction. Further consider the parity [evenness/oddness]
of the white and black squares.)
Solution. Each domino covers precisely one black square and one white square, so a board
that is covered exactly by dominoes has the same number of white squares as black squares.
Removing the opposite corners leaves us 32 black squares but only 30 white squares so it
cannot be covered exactly by dominoes.
6. Two squares from each of two opposite corners are deleted as shown on the right below.
Prove that the remaining squares cannot be covered exactly by copies of the “T-shape” and
its rotations.
(Hint: Use the method of contradiction. Further consider the parity [evenness/oddness]
of the white and black squares.)
Solution.
Method #1: Notice that 60 squares remain (30 white and 30 black) and so 15 T-shapes
must be used in total. Note, each T-shape covers an odd number of squares of each color.
Since the sum of 15 odd numbers is always odd, any board covered exactly by 15 T-shapes
must have an odd number of squares of each color. Our board has 30 squares of each color,
so it cannot be covered by 15 T-shapes.
Method #2: Since the region has the same number of squares of each color, we can
conclude that we must use the same number of each of the two types of tiles. Thus an even
number of tiles must be used, however, this contradicts the total of 60 squares since 60 is
not 4 times an even number.
7. Prove by contradiction that if n is a positive integer, then
n
n
>
.
n+1
n+2
Solution.
[By Contradiction] Suppose that n is a natural number and
n
n
≤
. Since n is
n+1
n+2
a natural number, (n + 1)(n + 2) is positive and so we will preserve the inequality when
multiplying both sides by (n + 1)(n + 2). Doing so shows that n(n + 2) ≤ n(n + 1) which
means n2 + 2n ≤ n2 + n. Moving all of the terms to the left hand side, we have n ≤ 0, but
n ∈ N. .
n
n
Therefore
>
.
n+1
n+2